1. x’(t) = (-e-t)(sin t) + (e-t)(cos t)
Part (a) u v
x(t) = e-t sin t, 0  t  2p
“Farthest to the left” means we need to minimize the position function, x(t).
x’(t) = (-e-t)(sin t) + (e-t)(cos t)
-e-t (sin t - cos t) = 0
(-1/et) (sin t - cos t) = 0
sin t = cos t
This can never be zero, so we’ll just ignore it!
t = p/4 or 5p/4

2. The particle is farthest to the left at t = 5p/4.
Part (a)
x(t) = e-t sin t, 0  t  2p
x(p/4) =
ep/4 1 * 2
0.322
x(p/4) = e-p/4 sin (p/4)
x(5p/4) =
e5p/4 1 * 2 -
-0.014
The particle is farthest to the left at t = 5p/4.
x(5p/4) = e-5p/4 sin (5p/4)
We also need to consider the endpoints:
x(0) = e0 sin (0)
x(0) = 0
x(2p) = e2p sin (2p)
x(2p) = 0
t = p/4 or 5p/4

3. x’(t) = (-e-t) (sin t - cos t)
Part (b) u v
x’(t) = (-e-t) (sin t - cos t)
x’’(t) = (e-t)(sin t - cos t) + (-e-t)(cos t + sin t)
x’’(t) = e-tsin t - e-tcos t - e-tcos t - e-t sin t
x’’(t) = -2e-tcos t
A (-2e-tcos t) + (-e-t)(sin t - cos t) + e-tsin t = 0
-e-t (2A cos t + sin t - cos t - sin t) = 0
This can’t be zero.
cos t (2A – 1) = 0
A = 1/2