Directiontal derivative

[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...

[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.

[faust9]

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

[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.

[Chen]

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