# Directiontal derivative

1. Mar 19, 2004

### faust9

OK, I having a small problem understanding how my text book came about an answer to an example problem.

$$f(x,y)=4-x^2-\frac{1}{4}y^2$$
at P(1,2)

This next step is the one that's bugging me:

$$u^\rightarrow=\cos(\frac{\pi}{3})\imath+\sin(\frac{\pi}{3})\jmath$$

This is one of those instances where something magic happens because right now I have little to no clue where the $$\frac{\pi}{3}$$ came from.

Thanks...

2. Mar 19, 2004

### jamesrc

I may be wrong, but isn't that just part of the given information? I mean, you're being asked to find the directional derivative, meaning the rate of change of the function at a point in a given direction. So you know the function z=f(x,y), you're given a point P(1,2), and you're given the unit vector of the direction you're interested in.

3. Mar 19, 2004

### faust9

Yeah your correct... The question is poorly written (or at least poorly formated). Thanks for showing me my stupid mistake.

4. Mar 20, 2004

### HallsofIvy

Staff Emeritus
Specifically, $$u^\rightarrow=\cos(\frac{\pi}{3})\imath+\sin(\frac{\pi}{3})\jmath$$ is the unit vector point at an angle $$\pi/3$$ radians from the positive x-axis.

5. Mar 20, 2004

### Chen

(Psst. Use "\vec v" in LaTeX to display a vector... )