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Vector Functions

  1. Oct 5, 2008 #1
    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. jcsd
  3. Oct 5, 2008 #2

    Mute

    User Avatar
    Homework Helper

    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.
     
  4. Oct 5, 2008 #3
    Oh right. It was that simple.. Silly me. Thanks! :)
     
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