Directional derivatives

**ottmar0** · May 10th 2009, 08:19 PM

Calculate the directional derivatives of $f(x,y,z)=x^2y+y^2z+z^2x$ at (1,2,-1) in the direction $u=<2/3,-2/3,1/3>.$

**mr fantastic** · May 10th 2009, 08:29 PM

Originally Posted by ottmar0

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: $\frac{d f}{d \vec{u}} = \nabla \cdot \hat{\vec{u}}$ .

**Chris L T521** · May 10th 2009, 08:30 PM

Originally Posted by ottmar0

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?

**ottmar0** · May 10th 2009, 08:31 PM

you can solve the problem and show the steps because my test is in about 8 hrs.

**mr fantastic** · May 10th 2009, 08:36 PM

Originally Posted by ottmar0

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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