# 2 simliar questions: Am I right?

1. Oct 7, 2010

### Buri

1. First problem
Let f: R^2 -> R be defined by setting f(0) = 0 and f(x,y) = xy/(x^2 + y^2) if (x,y) is not equal to zero.

For which vectors u not equal to zero, does the directional derivative f'(0;u) exist?

I have that it exists for all u = (h,k) and is equal to hk/(h^2 + k^2)

Do the partial derivatives exist at (0,0)? Yes and they're both equal to 0.

Is f differentiable at 0? No because if it were, its differential at u would equal the f'(0;u), but f'(0;u) is non linear. So its impossible.

Is it continuous at 0? No because arbitrarily close to the origin the function has value 1/2 (this is along the line y = x) but at the origin the function is equal to 0. So it can't be continuous at 0.

2. Second Problem

Its the same thing, just with a differnent function:

f: R^2 -> R f(x,y) = |x| + |y|

I have its direction derivatives don't exist.

I'll write what I did for this one since I'm not exactly sure if its right.

f'(0;u) = lim [t->0] 1/t[f((0,0) + t(h,k)) - f(0,0)] = lim [t->0] 1/t(|th| + |tk|)

See here is where I'm not exactly sure if I'm doing things right. Here I guess I must check that the "right and left" limits exists, but that doesn't really make sense anymore since its a multivariable function. So I basically consider t > 0 and t< 0 and I get |h| + |k| and -|h| - |k|. So they aren't equal and therefore, it doesn't exist.

So directional derivatives at (0,0) don't exist.

So neither do partial derivatives then.

Can't be differentiable. Since all the directional derivatves at 0 don't exist it implies that it is not differerentiable at 0.

Now is it continuous at 0?

I'm trying to show that:

lim [ (x,y) -> (0,0) ] f(x,y) = 0.

I tried proving this directly from the epsilon-delta definition but its not working for me.

I have:

(x² + y²)^(1/2) < delta then (x^2)^(1/2) (y^2)^(1/2) < epsilon

But I can't seem to connect the two. So what I did is I broke up the limit into two parts:

lim [ (x,y) -> (0,0) ] f(x,y) = lim [ (x,y) -> (0,0) ] |x| + lim [ (x,y) -> (0,0) ] |y|

I know the above isn't valid yet since I don't know whether the limit exist, but I'm showing my reasoning.

Now proving that lim [ (x,y) -> (0,0) ] |x| = 0 is basically setting delta equal to epsilon, so its easy. Therefore, the individual limits exists which implies the summation of the limits exist and hence the limit of the summation exists. So it is continuous at 0.

If someone could tell me if I did these right would be awesome!

Thanks for the help!

Anyone? I'm sure you guys all know how to do these lol And besides, I've shown my work! :) So I think I deserve some help :)

Last edited: Oct 8, 2010