# Homework Help: Vector Functions

1. Oct 5, 2008

### kehler

1. The problem statement, all variables and given/known data
A particle moves along the curve y=ex, z=xex in such a way that z'(t)=2 for all times t, and is at (0,1,0) at time t=0.
Find the velocity and acceleration at time t=0. (Hint: Differentiate implicitly)

3. The attempt at a solution
Here's what I have so far:
r(t) = [x(t), ex(t), x(t)ex(t)]
Since r(0) = [0,1,0], x(0) = 0

v(t) = r'(t) = [x'(t), x'(t)ex(t), x'(t)x(t)ex(t) + x'(t)ex(t)]
Taking out the common factor x'(t), I get v(t) = x'(t) [1, ex(t), x(t)ex(t) + ex(t)]
But z'(t) = 2 for all t. So z'(0) = 2.
Thus, x'(0) (x(0)ex(0) + ex(0)) = 2,
Substituting x(0) = 0 into the equation, I find that
x'(0) = 2/1 = 2
Using this, I find that v(0) = 2 [1, 1, 1] = [2, 2, 2]

a(t) = r''(t) = [x''(t), x''(t)ex(t) + (x'(t))2ex(t), x''(t)(ex(t)) + x(t)x''(t)ex(t) + (x'(t))2x(t)ex(t) +2x'(t)x'(t)ex(t)]
a(0) = [x''(0), x''(0)+ 4, x''(0) +8]

My problem is that I don't know how to find x''(0) from here :(

2. Oct 5, 2008

### Mute

You used the fact that z'(t) = 2 at all times to find x'(0), and then you differentiated to find a(t) = r"(t), but you forgot to use the fact that since z'(t) = 2, z"(t) = 0

Hence, z"(0) = x"(0) + 8 = 0.

3. Oct 5, 2008

### kehler

Oh right. It was that simple.. Silly me. Thanks! :)