Part (a) Acceleration is the 2nd derivative with respect to time. We already have dx/dt and dy/dt, so we’ll simply take another derivative of each and evaluate them at t=2. 1 1 – (1-2e-t)2 2e-t x”(t) = y”(t) = x”(2) = .395 or .396 4(1+t3) – (4t)(3t2) ( 1 + t3 )2 y”(2) = -.740 or -.741 a(2) = 0.395 or 0.396, -0.740 or -0.741

The formula for the speed at t=2 is… Part (a) The formula for the speed at t=2 is… speed = x’(2)2 + y’(2)2 x’(2) = .8173413107 y’(2) = 8/9 speed = (.8173413107)2 + (8/9)2 speed = 1.207 or 1.028

So the curve has a vertical tangent at t = ln 2 or t = 0.693 Part (b) The vertical tangent will occur when dx/dt = 0 and dy/dt ≠ 0. sin-1( 1 - 2e-t ) = 0 1 - 2e-t = 0 2e-t = 1 e-t = ½ dy/dt ≠ 0 when t = ln 2. ln e-t = ln ½ So the curve has a vertical tangent at t = ln 2 or t = 0.693 -t = ln ½ t = ln 2

Part (c) = 0 4t 1 + t3 1 sin-1( 1 – 2e-t ) dy/dt dx/dt = 4t 1 + t3 1 m(t) = Using L’Hopital’s Rule, this comes out to zero. 1 sin-1( 1 – 2e-∞ ) 1 sin-1( 1) lim m(t) = t∞ = 0

Integrating dy/dt would give us the change in the height of the graph. Part (d) y=c Integrating dy/dt would give us the change in the height of the graph. (6,-3) @ t=2 At t = 2, the graph has an initial height of -3 units before it rises up to y = c. c = -3 + dt 2 ∞ 4t 1+t3