Math 214 — Solutions to Assignment #5 11.1 10. Let x = t2 , y = t3 . (a) Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as t increases. (b) Eliminate the parameter to find a Cartesian equation of the curve. Solution. (a) t x y −3 −2 −1 0 1 2 3 9 4 1 0 1 4 9 −27 −8 −1 0 1 8 27 y (4,8) 0 t=2 t=1 t=−1 (4,−8) x t=−2 √ (b) t = ± x =⇒ y = ±x3/2 , x ≥ 0, y ∈ R. Or x = y 2/3 , x ≥ 0, y ∈ R. For #12 and 16 (a) Eliminate the parameter to find a Cartesian equation of the curve. (b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases. 12. x = 4 cos θ, y = 5 sin θ, −π/2 ≤ θ ≤ π/2. 1 Solution. y2 x2 + =1 (a) Since cos2 θ + sin2 θ = 1, we get (x/4)2 + (y/5)2 = 1, i.e., 16 25 which is an ellipse with the x-intercepts x = ±4, the y-intercepts y = ±5. But since −π/2 ≤ θ ≤ π/2, we have 0 ≤ cos θ ≤ 1 so the graph consists of only the portion on the right side of the y-axis. (b) y 5 θ= π 2 4 0 x θ =0 −5 θ =− π 2 16. x = ln t, y = √ t, t ≥ 1. Solution. (a) x = ln t =⇒ t = ex =⇒ y = (b) √ ex = ex/2 , x ≥ 0. y 1 0 x 22. Describe the motion of a particle with position (x, y) as t varies in the given interval: x = cos2 t, t = cos t, 0 ≤ t ≤ 4π. Solution. x = y 2 is a parabola opening to the right with the vertex (0, 0). The particle starts at the point (1, 1) (t = 0). It then moves along the parabola to 2 the point (0, 0) (t = π/2) down to (1, −1) (t = π), then back to (0, 0) (t = 3π/2) and to (1, 1) (t = 2π). The same motion is repeated from 2π to 4π. y (1,1) x (1,−1) 11.2 8. Find an equation of the tangent to the curve x = tan θ, y = sec θ at the point √ (1, 2) by 2 methods: (a) without eliminating the parameter and (b) by first eliminating the parameter. Solution. (a) dy dy/dθ sec θ tan θ tan θ = = = = sin θ, dx dx/dθ sec2 θ sec θ √ √ π at (1, 2), 1 = tan θ, 2 = sec θ =⇒ θ = (or π/4 + 2nπ), 4 (since tan θ > 0 in quadrant I and III while sec θ > 0 in quad. I and IV). Hence the √ slope of the tangent at (1, 2) is √ π 1 or 2 0 π y = sin = √ = 4 4 2 2 and the equation is y− √ √ 2 2= (x − 1). 2 (b) Since tan2 θ + 1 = sec2 θ =⇒ x2 + 1 = y 2 , differentiating implicitly we obtain √ x 1 dy = =⇒ y 0 (1, 2) = √ 2x = 2yy 0 =⇒ dx y 2 which gives the same equation as in part (a). 16. Find dy/dx and d2 y/dx2 if x = cos 2t, y = cos t, 0 < t < π. For which values of t is the curve concave upward? 3 Solution. dy dy/dt − sin t sin t 1 1 = = = = = sec t, dx dx/dt −2 sin 2t 4 sin t cos t 4 cos t 4 d dy 1 ( ) sec t tan t d2 y 1 sin t dt dx 4 = = =− · 2 2 dx dx/dt −2 sin 2t 16 cos t sin t cos t 1 1 or =− = − sec3 t. 3 16 cos t 16 The curve is concave upward if y 00 (x) > 0 on 0 < t < π, i.e., 1 − sec3 t > 0 ⇐⇒ sec t < 0 ⇐⇒ cos t < 0 ⇐⇒ t ∈ (π/2, π). 16 18. Find the points on the curve x = 2t3 + 3t2 − 12t, y = 2t3 + 3t2 + 1 where the tangent is horizontal or vertical. Solution. We find the derivative dy 6t2 + 6t 6t(t + 1) t(t + 1) = 2 = = . dx 6t + 6t − 12 6(t + 2)(t − 1) (t + 2)(t − 1) Horizontal tangents occur when y 0 = 0, i.e., t = 0, −1. So the points are t = 0 =⇒ x = 0, y = 1, i.e., (0, 1), t = −1 =⇒ x = −2 + 3 + 12 = 13, y = −2 + 3 + 1 = 2, i.e., (13, 2). Vertical tangents occur at t = −2, 1 (y 0 does not exist there). Then the points are: t = −2 =⇒ x = −16 + 12 + 24 = 20, y = −16 + 12 + 1 = −3, i.e., (20, −3), t = 1 =⇒ x = 2 + 3 − 12 = −7, y = 2 + 3 + 1 = 6, i.e., (−7, 6). 34. Find the area of the region enclosed by the astroid x = a cos3 θ, y = a sin3 θ. Solution. The graph of the astroid is y a −a 0 −a 4 a x Using symmetry A=4 Z a y dx = 4 0 = −12a 2 Z 0 π 2 Z 0 a sin3 θ(−3a cos2 θ sin θ) dθ π 2 4 2 sin θ cos θ dθ = −12a 2 Z Z 0 π 2 1 1 (1 − cos 2θ)2 · (1 + cos 2θ) dθ 4 2 0 3 (1 − 2 cos 2θ + cos2 2θ)(1 + cos 2θ) dθ = − a2 π 2 2 Z π2 3 2 1 1 3 = a − cos 2θ − cos 4θ + cos 2θ dθ 2 2 2 0 π2 Z π 2 1 1 3 2 1 2 θ − sin 2θ − sin 4θ + cos 2θ} cos 2θ dθ = a {z | 2 2 2 8 0 0 (1−sin2 2θ) 3 2 a 2 3 = a2 · 2 = 1 π 1 · −0−0 + 2 2 2 π 3 = a2 π. 4 8 Z 0 0 (1 − u2 ) du (u = sin 2θ) 44. Find the length of the curve x = et + e−t , y = 5 − 2t, 0 ≤ t ≤ 3. Solution. dx/dt = et − e−t and dy/dt = −2. So dx dt 2 + dy dt 2 ∴ = (et − e−t )2 + 4 = e2t − 2 + e−2t + 4 = e2t + 2 + e−2t = (et + e−t )2 s Z 3 2 2 Z 3 dy dx (et + e−t ) dt L= + dt = dt dt 0 0 3 = et − e−t = e3 − e−3 − 1 + 1 = e3 − e−3 . 0 60. Find the area of the surface obtained by rotating the curve x = 3t−t3 , y = 3t2 , 0 ≤ t ≤ 1 about the x-axis. Solution. S = 2π Z b y ds = 2π a Z 1 3t 0 5 2 s dx dt 2 + dy dt 2 dt where dx dt 2 + dx dt dy dt dy dt 2 2 2 = (3 − 3t2 )2 = 9 − 18t2 + 9t4 = (6t)2 = 36t2 = 9 − 18t2 + 9t4 + 36t2 = 9 + 18t2 + 9t4 = (3 + 3t2 )2 ∴ S = 2π Z = 18π 1 2 2 3t (3 + 3t ) dt = 2π 0 Z 1 1 9t2 (1 + t2 ) dt 0 3 1 1 1 48π t t5 = (t + t ) dt = 18π + = 18π + . 3 5 0 3 5 5 2 0 Z 4 66. Find the surface area generated by rotating the given curve about the y-axis. x = et − t, y = 4tt/2 , 0 ≤ t ≤ 1 Solution. S = 2π ∴ dx dt 2 + dx dt dy dt ∴ 2 2 R1 x ds where ds = 0 t 2 2t p (dx/dt)2 + (dy/dt)2 dt. t = (e − 1) = e − 2e + 1, dy dt 2 = (2et/2 )2 = 4et = e2t − 2et + 1 + 4et = e2t + 2et + 1 = (et + 1)2 S = 2π = 2π Z Z 1 (et − t)(et + 1) dt 0 1 0 (e2t + et − tet − t) dt 1 t2 1 2t t t t e + e − (te − e ) − = 2π 2 2 0 1 1 1 2 e + e − e + e − − − 1 − 1 = π(e2 + 2e − 6). = 2π 2 2 2 (Here to do R tet dt, integration by parts was used.) 6