1. 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) = or .396
4(1+t3) – (4t)(3t2)
( 1 + t3 )2
y”(2) = or -.741
a(2) = or 0.396, or

2. 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) =
y’(2) = 8/9
speed = ( )2 + (8/9)2
speed = or 1.028

3. 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

4. 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

5. 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.
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