# Math Help - Directional Derivative

1. ## Directional Derivative

Hey guy I'm having a problem with directional derivative, can any one give me a clue about how to solve this

$f(x,y) = x^3*y^2$ at (-1,2) in the direction parallel to a vector $\frac{pi}{3}$ clockwise from the positive x-axis

2. ## Re: Directional Derivative

Can you at least get a vector going in the right direction?

3. ## Re: Directional Derivative

Hey I can get the derivative of y and x and put them to f (d/dx, d/dy) and sub (-1, 2) in. Im not sure what to do in regard to finding direction

4. ## Re: Directional Derivative

Start by drawing a diagram. Draw a set of axes, and then draw a right angle triangle that makes an angle of 60 degrees to the positive x axis. Fill in the lengths. You should be able to get the direction vector from that...

5. ## Re: Directional Derivative

The directional derivative of function f in direction $\theta$ is the dot product of $\nabla f$ with the unit vector in direction $theta$. In two dimensional Cartesian coordinates, $\nabla f= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}$ and the unit vector in the $\theta$ direction is $cos(\theta)\vec{i}+ sin(\theta)\vec{j}$ so that dot product is just
$cos(\theta)\frac{\partial f}{\partial x}+ sin(\theta)\frac{\partial f}{\partial y}$.

Surely you have seen that before?

6. ## Re: Directional Derivative

Oh so that is what it mean, thank prove it.i have seen it before but as a vector instead of theta.

7. ## Re: Directional Derivative

Originally Posted by junkwisch
Oh so that is what it mean, thank prove it.i have seen it before but as a vector instead of theta.
Well you still have to get a vector from that though...