Parametric Equations, Tangent Lines, & Arc Length SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should reference Chapter 10.1 of the rec- ommended textbook (or the equivalent chapter in your alternative textbook/online resource) and your lecture notes. EXPECTED SKILLS: • Be able to sketch a parametric curve by eliminating the parameter, and indicate the orientation of the curve. • Given a curve and an orientation, know how to find parametric equations that generate the curve. dy d2y • Without eliminating the parameter, be able to find and at a given point on a dx dx2 parametric curve. • Be able to find the arc length of a smooth curve in the plane described parametrically. PRACTICE PROBLEMS: For problems 1-5, sketch the curve by eliminating the parameter. Indicate the direction of increasing t. 8 < x = 2t + 3 1. y = 3t − 4 : 0 ≤ t ≤ 3 3 17 y = x − from (3; −4) to (9; 5) 2 2 1 8 < x = 2 cos t 2. y = 3 sin t : π ≤ t ≤ 2π x2 y2 + = 1 from (−2; 0) to (2; 0) 4 9 8 x = t − 5 < p 3. y = t : 0 ≤ t ≤ 9 p y = x + 5 from (−5; 0) to (4; 3) 8 > x = sec t < y = tan2 t 4. π > 0 ≤ t < : 2 y = x2 − 1 for x ≥ 1 2 8 > x = sin t < y = cos (2t) 5. π π > − ≤ t ≤ : 2 2 y = 1 − 2x2 from (−1; −1) to (1; −1) For problems 6-10, find parametric equations for the given curve. (For each, there are many correct answers; only one is provided.) 6. A horizontal line which intersects the y-axis at y = 2 and is oriented rightward from (−1; 2) to (1; 2). 8 < x = t y = 2 : −1 ≤ t ≤ 1 7. A circle or radius 4 centered at the origin, oriented clockwise. 8 < x = 4 sin t y = 4 cos t : 0 ≤ t ≤ 2π 8. A circle of radius 5 centered at (1; −2), oriented counter-clockwise. 8 < x = 5 cos t + 1 y = 5 sin t − 2 ; Detailed Solution: Here : 0 ≤ t ≤ 2π 9. The portion of y = x3 from (−1; −1) to (2; 8), oriented upward. 8 < x = t y = t3 : −1 ≤ t ≤ 2 3 x2 y2 10. The ellipse + = 1, oriented counter-clockwise. 4 16 8 < x = 2 cos t y = 4 sin t : 0 ≤ t ≤ 2π dy d2y For problems 11-13, find and at the given point without eliminating the dx dx2 parameter. 8 < x = 3 sin (3t) 11. The curve y = cos (3t) at t = π : 0 < t < 2π 2 dy d y 1 = 0; 2 = dx t=π dx t=π 9 8 2 < x = t 12. The curve y = 3t − 2 at t = 1 : t ≥ 0 2 dy 3 d y 3 = ; 2 = − dx t=1 2 dx t=1 4 8 > x = 2 tan t < y = sec t π 13. The curve π at t = > 0 ≤ t ≤ 4 : 3 p p 2 dy 2 d y 2 = , 2 = ; Detailed Solution: Here dx t=π=4 4 dx t=π=4 16 8 p < x = pt 14. Consider the curve described parametrically by y = 3 t + 1 : t ≥ 0 dy (a) Compute without eliminating the parameter. dx t=64 dy 1 = dx t=64 3 4 (b) Eliminate the parameter and verify your answer for part (a) using techniques from differential calculus. The curve is equivalent to y = x2=3 + 1, x ≥ 0. And, t = 64 corresponds to x = 8. dy dy 1 Thus, = = dx t=64 dx x=8 3 (c) Compute an equation of the line which is tangent to the curve at the point cor- responding to t = 64. 1 y − 5 = (x − 8) 3 8 < x = 2 cos t 15. Consider the curve described parametrically by y = 4 sin t : 0 ≤ t ≤ 2π dy (a) Compute without eliminating the parameter. dx t=π=4 dy = −2 dx t=π=4 (b) Eliminate the parameter and verify your answer for part (a) using techniques from differential calculus. x2 y2 π The curve is equivalent to the ellipse + = 1. And, t = corresponds p p 4 16 4 to the point (x; y) = ( 2; 2 2). Thus, you can use implicit differentiation and dy dy = = −2 p p dx t=π=4 dx (x;y)=( 2;2 2) (c) Compute an equation of the line which is tangent to the curve at the point cor- π responding to t = . 4 p p y − 2 2 = −2 x − 2 (d) At which value(s) of t will the tangent line to the curve be horizontal? π 3π t = and t = 2 2 For problems 16-18, compute the length of the given parametric curve. 8 x = t > > <> 2 16. The curve described by y = t3=2 > 3 > > : 0 ≤ t ≤ 4 5 p 2 10 5 − + 3 3 8 x = et > > <> 2 17. The curve described by y = e3t=2 > 3 > > : ln 2 ≤ t ≤ ln 3 p 16 −2 3 + ; Detailed Solution: Here 3 8 1 > x = t2 > > 2 > < 1 18. The curve described by y = t3 > 3 > > p :> 0 ≤ t ≤ 3 7 3 19. Compute the lengths of the following two curves: 8 8 < x = cos t < x = cos (3t) C1(t) = y = sin t C2(t) = y = sin (3t) : 0 ≤ t ≤ 2π : 0 ≤ t ≤ 2π Explain why the lengths are not equal even though both curves coincide with the unit circle. The length of C1(t) is 2π and the length of C2(t) = 6π. Notice that C2(t) is the just curve C1(t) traversed three times. 20. This problem describes how you can find the area between a parametrically defined curve and the x-axis. The Main Idea: Recall that if y = f(x) ≥ 0, then the area between the curve and Z b Z b the x-axis on the interval [a; b] is f(x)dx = y dx. Now, suppose that the same a a curve is described parametrically by x = x(t), y = y(t) for t0 ≤ t ≤ t1 and that the Z b Z t1 curve is traversed exactly once on this interval. Then, A = y dx = y(t)x0(t) dt. a t0 6 8 > x = sin t < y = cos (2t) Consider the curve π π > − ≤ t ≤ : 4 4 (a) Compute the area between the graph of the given curve and the x-axis by evalu- Z t1 ating A = y(t)x0(t) dt. t0 p Z π=4 2 2 A = cos (2t) cos t dt = −π=4 3 (b) After eliminating the parameter to express the curve as an explicitly defined Z b function (y = f(x)), calculate the area by evaluating A = f(x) dx. a p p Z 2=2 2 2 2 A = p 1 − 2x dx = − 2=2 3 7.