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Directional Derivative Problem

  1. Oct 28, 2005 #1
    Hello! I have a question about a problem I am trying to work out.

    The question asks to find the directional derivative of f(x,y,z)=x^2+yz at the point (1,-3,2) in the direction of the path r(t)=t^2i + 3tj+(1-t^2)k.

    Ok, first do I find the derivative of r(t) to get 3j +t(2i-2k) and then use 2i-2k as the direction vector to find the unit vector? Then the unit vector would be
    1/sqrt[2] (2i-2k) and then find the scalar product of the unit vector with the gradient of the f function?

    my final answer was 5*sqrt[2]. Did I work this out right? I am studying for an exam and that is one of the problems I think will be on it.
    Thanks!
     
  2. jcsd
  3. Oct 29, 2005 #2

    HallsofIvy

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    Yes and no. Yes, the "derivative in the direction of the path" is in the direction of the tangent to the path at that point. However, r(t) does go through (1, -3, 2). In order that t2 = 1 and 3t= -1, we must have t= -1 but then z= 1-t2= 0, not 2. Are you sure you copied the proglem correctly? Is it possible that r(t)= t2i+ 3tj+(1+t2)k?
    Also, while 3j+ t(2i-2k) is the tangent vector at t, you cannot then get 2i-2k at any point from that. Taking t= 1, you would get 2i+ 3j- 2k (although, of course, r(1)= i+ 3j or (1, 3, 0), not (1, -3, 0).
    Assuming that r(t)= t2i+ 3tj+(1+t2)k so that the path goes through (1, -3, 2) when t= -1, then r'= 2ti+ 3j+ 2tk. At t= -1, that is -2i+ 3j- 2k which has length [itex]\sqrt{17}[/itex] so a tangent vector is [itex]\frac{-2}{\sqrt{17}}i+ \frac{3}{\sqrt{17}}j- \frac{2}{\sqrt{17}}k[/itex]. Take the dot product of that with the gradient of f.

     
  4. Oct 29, 2005 #3
    Thanks!

    Thank you so much!! First of all, I did copy the problem wrong...the last part of the equation was (1-t^3)k. So, it works out! I don't know why I thought that just taking the derivative of r(t) and then using that would work!

    Thanks again!
     
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