(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A particle moves along the curve y=e^{x}, z=xe^{x}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), e^{x(t)}, x(t)e^{x(t)}]

Since r(0) = [0,1,0], x(0) = 0

v(t) = r'(t) = [x'(t), x'(t)e^{x(t)}, x'(t)x(t)e^{x(t)}+ x'(t)e^{x(t)}]

Taking out the common factor x'(t), I get v(t) = x'(t) [1, e^{x(t)}, x(t)e^{x(t)}+ e^{x(t)}]

But z'(t) = 2 for all t. So z'(0) = 2.

Thus, x'(0) (x(0)e^{x(0)}+ e^{x(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)e^{x(t)}+ (x'(t))^{2}e^{x(t)}, x''(t)(e^{x(t)}) + x(t)x''(t)e^{x(t)}+ (x'(t))^{2}x(t)e^{x(t)}+2x'(t)x'(t)e^{x(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 :(

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# Homework Help: Vector Functions

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