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Math Help - Directional derivatives

  1. #1
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    Directional derivatives

    Calculate the directional derivatives of f(x,y,z)=x^2y+y^2z+z^2x at (1,2,-1) in the direction u=<2/3,-2/3,1/3>.
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    Quote Originally Posted by ottmar0 View Post
    Calculate the directional derivatives of f(x,y,z)=x^2y+y^2z+z^2x at (1,2,-1) in the direction u=<2/3,-2/3,1/3>.
    Start by applying the basic definition: \frac{d f}{d \vec{u}} = \nabla \cdot \hat{\vec{u}}.
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  3. #3
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    Quote Originally Posted by ottmar0 View Post
    Calculate the directional derivatives of f(x,y,z)=x^2y+y^2z+z^2x at (1,2,-1) in the direction u=<2/3,-2/3,1/3>.
    First find the Gradient:

    \vec{\nabla} f\!\left(x,y,z\right)=\frac{\partial}{\partial x}\left(x^2y+y^2z+z^2x\right)\mathbf{i}+\frac{\par  tial}{\partial y}\left(x^2y+y^2z+z^2x\right)\mathbf{j} +\frac{\partial}{\partial z}\left(x^2y+y^2z+z^2x\right)\mathbf{k}=\left(2xy+  z^2\right)\mathbf{i}+\left(x^2+2yz\right)\mathbf{j  }+\left(y^2+2zx\right)\mathbf{k}

    Thus, the gradient at \left(1,2,-1\right) would be \vec{\nabla} f \!\left(1,2,-1\right)=5\mathbf{i}-3\mathbf{j}+2\mathbf{k}

    Now, the directional derivative is equal to \vec{\nabla} f\!\left(1,2,-1\right)\cdot\mathbf{u}=\left<5,-3,2\right>\cdot\left<\tfrac{2}{3},-\tfrac{2}{3},\tfrac{1}{3}\right>=\tfrac{10}{3}+2+\  tfrac{2}{3}=\boxed{6}

    Does this make sense?
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    If

    you can solve the problem and show the steps because my test is in about 8 hrs.
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    Quote Originally Posted by ottmar0 View Post
    you can solve the problem and show the steps because my test is in about 8 hrs.
    You have already been shown exactly how to do it.

    If you refering to my post, I suggest in future that to avoid confusion you quote the post you're replying to.
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