Directional Derivative of z^3-yx^2 at (-4,-5,5) in dir. of v

[think.ms.tink]

Find the directional derivative of f(x,y,z)=(z^3)-yx^2 at the point (-4, -5, 5) in the direction of the vector v=(-5,-4.-1).

I do not have any work to show because I do not even know how to start. Thank you .

 

[stapel]

I do not even know how to start.

We don't know how many classes you've missed, so you'll need to specify how much of the material is unknown to you, so that helpers can try to find useful lesson links. For instance, are you familiar with vectors? Have you done derivatives at all? How about directional derivatives?

Thank you!

 

[think.ms.tink]

i am familiar with it all. I do not know how to began the equation. I know I am having trouble with what comes after you take the derivative, which is 3x^2-2x. after that i am lost.

 

[galactus]

First, find the partials w.r.t x, y, z:

\(\displaystyle f_{x}(x,y,z)=-2xy\)

\(\displaystyle f_{y}(x,y,z)=-x^{2}\)

\(\displaystyle f_{z}(x,y,z)=3z^{2}\)

\(\displaystyle {\nabla}f(x,y,z)=(-2xy)i-x^{2}j+3z^{2}k\)

\(\displaystyle {\nabla}f(-4,-5,5)=-40i-16j+75k\)

\(\displaystyle u=\frac{-5i-4j-k}{\sqrt{(-5)^{2}+(-4)^{2}+(-1)^{2}}}=\frac{-5i-4j-k}{\sqrt{42}}\)

Now, you can do the last step. Find \(\displaystyle d_{u}f\)

That is defined as \(\displaystyle D_{u}f={\nabla}f(x,y,z)\cdot u\)

You have all you need to finish.

 

[think.ms.tink]

Thank you...Thank you so much. i appreciate it.