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a(2) = 0.395 or 0.396, -0.740 or -0.741. Part (a). Acceleration is the 2 nd 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. x”(t) =. x”(2) = .395 or .396. 2e -t. 1 – (1-2e -t ) 2.
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a(2) = 0.395 or 0.396, -0.740 or -0.741 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 x”(t) = x”(2) = .395 or .396 2e-t 1 – (1-2e-t)2 4(1+t3) – (4t)(3t2) y”(t) = y”(2) = -.740 or -.741 (1+t3)2
speed = x’(2)2 + y’(2)2 speed = (.8173413107)2 + (8/9)2 speed = 1.207 or 1.028 Part (a) The formula for the speed at t=2 is… y’(2) = 8/9 x’(2) = .8173413107
Part (b) The vertical tangent will occur when dx/dt = 0 and dy/dt ≠ 0. 1-2e-t= 0 sin-1(1-2e-t) = 0 2e-t= 1 0 e-t= ½ dy/dt ≠ 0 when t = ln 2. ln e-t=ln ½ So the curve has avertical tangent att = ln 2 or t = 0.693 -t = ln ½ t = ln 2
4t 4t 1 1 sin-1(1–2e-t) sin-1(1–2e-t) 1+t3 1+t3 1 1 sin-1(1–2e-∞) sin-1(1) lim m(t) = t∞ Part (c) dy/dt m(t) = = dx/dt Using L’Hopital’s Rule,this comes out to zero. 0 = 0 0
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. ∞ 4t dt c =-3 + 1+t3 2
