Question

\[f(x) =\begin{cases}\frac{xy}{\sqrt{x^2+y^2}} & (x,y)\ne 0 \\0 & (x,y) = 0\end{cases}\]
a) Determine if f is differentiable at (0,0)
b) determine for which unit vectors v = vi + wj the directional derivative \(D_vf(0, 0)\) exists.

Answer 1

\[h(0,0) = f(0,0) + f_x(0,0)(x-0)+f_y(0,0)(y-0) = 0\]
\[\lim_{(x,y) \rightarrow(0,0)}\frac{f(x,y)-h(x,y)}{||(x,y)-(0,0)||} \]
What I've got is limit does not exist, so it is not differentiable. Is it correct?

Answer 2

Differentiable with respect to all directions?

Answer 3

I think so.

Answer 4

I'm not sure what definition you're using for directional derivative...

Answer 5

\[D_vf(a) = \lim_{h\rightarrow 0}\frac{f(a+hv) - f(a)}{h}\]

Answer 6

\[D_{\langle a,b\rangle }f(x, y) = \lim_{h\rightarrow 0}\frac{f(x+ah, y+bh) - f(x,y)}{h}\]

Answer 7

Oh wait, so it is a single variable function then?

Answer 8

a and v are vectors.

Answer 9

\[
\lim_{h\to 0}\frac{f(ah,bh)-f(0,0)}{h}
\]

Answer 10

Are you talking about (a) or (b)?

Answer 11

I'm saying \(v=\langle a,b \rangle\)

Answer 12

I'm starting with a)

Answer 13

\[\Huge
\lim_{h\to 0} \frac{\frac{abh^2}{\sqrt{a^2h^2+b^2h^2}}}h
\]

Answer 14

That gives\[\frac{ab}{\sqrt{a^2+b^2}}\]

Answer 15

\[
a^2+b^2>0
\]

Answer 16

The \(\mathbf 0\) vector is only direction that it wouldn't be defined I guess? But technically that sort of vector isn't to be used for a directional derivative.

Answer 17

I think so, or else, we'll have a new function for this question.

But isn't the definition of differentiability of f at (a,b) like this:
\[h(x,y) = f(a,b) + f_x(a,b)(x-a)+f_y(a,b)(y-b)\]
\[\lim_{(x,y)\rightarrow(a,b)} \frac{f(x,y)-h(x,y)}{||(x,y)-(a,b)||}=0\]
if f is differentiable at (a,b)?

Answer 18

I got mine from http://tutorial.math.lamar.edu/Classes/CalcIII/DirectionalDeriv.aspx

Answer 19

I've always learned that differentiablility just means that the limit of the difference quotient exists at the point in question.

Answer 20

The first one isn't asking for directional derivative?

Answer 21

Hmm, I see, I'm not sure how differentiability would be defined here.

Answer 22

So it would be that there is a unique tangent plane then..?

Answer 23

I think I understand your definition now. So what is the problem?

Answer 24

Just to confirm if the limit doesn't limit.

Answer 25

for part a

Answer 26

I'm not so sure. You're getting a \(0/0\) which is an indeterminant form?

Answer 27

\[\huge\lim_{(x,y) \rightarrow(0,0)}\frac{\frac{xy}{\sqrt{x^2+y^2}}}{\sqrt{(x-0)^2+(y-0)^2}} \]

Answer 28

Is this what you went through?

Answer 29

\[
\lim_{(x,y)\to (0,0)}\frac{xy}{x^2+y^2}
\]

Answer 30

Then you realized this limit doesn't exist?

Answer 31

Maybe I'm retracing your steps here. Since the problem isn't obvious I have to work it out as well.

Answer 32

From that, change it to \[\lim_{r \rightarrow 0} sin\theta cos\theta \ne 0\]So, it is not differentiable at (0,0)

Answer 33

Yes, that looks right.

Answer 34

Wow! *relieved*

Answer 35

Different thetas (directions paths of the limit) are giving different values. Not a good sign for a limit.

Answer 36

They don't necessarily have to be 0, they have to be constant.

Answer 37

No, it gives different values, then limit doesn't exist. When limit exists, it should approach to a certain value...

Answer 38

And from my textbook, it should be 0... o_o

Answer 39

I mean it should be 0 for it to be differentiable.

Answer 40

Nope. Unless they are using some other premise to suggest it couldn't be constant as some non 0 value and then asserting it'd have to be 0

Answer 41

Here's the definition from the book

Answer 42

Oh, the book said it had to be \(=0\). I didn't know that yet.

Answer 43

I was thinking the limit just had to exist

Answer 44

Has it been shown that \[D_{\mathbb{u}} f(x,y)=\nabla{f}\cdot \mathbb{u}?\]

Answer 45

It has been mentioned in the book that
\[D_{v}f(a) = \nabla f(a)\cdot v\]

Answer 46

That is useful when assuming a function is differentiable.

Answer 47

Hmm.. But it is not differentiable in this case?

Answer 48

I mean differentiable in terms of the directional derivative, not in terms of that weird definition we used in a)

Answer 49

Honestly I'm not completely sure what the use for that definition is exactly.

Answer 50

We should use this

Answer 51

I already did it.

Answer 52

Hmm, I guess we are done with this question! Thanks a lot!!

Answer 53

Yeah.