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

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... )