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Math Help - Directional derivative

  1. #1
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    Directional derivative

    Find a,b,c constants so that the directional derivative of f(x,y,z)=axy^2+byz+cz^2x^3 on (1,2,-1) has a maximum value of 64 on a parallel direction to the z axis.

    I think we can calculate the directional derivative by using \langle\nabla f(x_0),x_0\rangle where x_0=(1,2,-1), but a maximum value is asked which I don't get, and I don't either get when it says "on a parallel direction to the z."
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Use:

    \nabla f(1,2,-1)=k(0,0,1)

    so,

     \left\|{\nabla f(1,2,-1)}\right\|=|k|=64



    Fernando Revilla
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  3. #3
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    Quote Originally Posted by FernandoRevilla View Post
    Use:

    \nabla f(1,2,-1)=k(0,0,1)

    so,

     \left\|{\nabla f(1,2,-1)}\right\|=|k|=64
    Okay but, that would imply that k=64 ?

    Then how to get a,b,c?

    Thanks!
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  4. #4
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    The directional derivative to the direction to the axis is

    <br />
\nabla f \cdot (0,0,1)=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})\cdot (0,0,1)=\frac{\partial f}{\partial z} \; | \; _{at \; point \; (1,2,-1)}=64.<br />
    Last edited by zzzoak; February 5th 2011 at 01:14 PM.
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  5. #5
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    The deriviative in the direction of the z axis is simply the partial derivative with respect to z, by+ 2czx^4, evaluated at (1, 2, -1). That will give you as single equation for b and c. a can be anything.
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