Check differentiability in function

[Sux]

Homework Statement

Hello, I can't find any way to prove if this funtion is or isn't differentiable in if $$(x,y)=(0,0)$$ :
$${f(x,y)=\displaystyle\frac{x^{3}}{x^{2}+y^{2}}}$$ if $$(x,y) \neq(0,0)$$

$$f(x,y)=0$$ if $$(x,y)=(0,0)$$

The Attempt at a Solution

Partial derivatives don't exist in (0,0), so i have calculate definition of differentiability, and i get:

$$\displaystyle\lim_{x,y \to{0,0}}{} \displaystyle\frac{x^{3}}{(x^{2}+y^{2})^{3/2}}$$

Which limit doesn't exit, so it is not differentiable in (0,0).

Is this right?

Thanks!

 

[snipez90]

If we approach the origin along the x-axis, we find that the limit is 0. But if we approach the origin along the curve y = x^(3/2), the limit is 1. Hence the function is not continuous at the origin, which of course means it's not differentiable there.

If you were trying to apply the definition of the partial derivative to show non-differentiability at (0,0), then the approach would fail since this is an example of a function which has a directional derivative in every direction at the origin (and of course partial differentiability is the specific case where the unit vector in the definition of the directional derivative is taken to be a standard basis vector). You should demonstrate this.

 

[LCKurtz]

If we approach the origin along the x-axis, we find that the limit is 0. But if we approach the origin along the curve y = x^(3/2), the limit is 1. Hence the function is not continuous at the origin, which of course means it's not differentiable there.

That isn't correct. You may be thinking of the limit as x goes to infinity, not zero. The function is continuous at (0,0).

To the OP: Try looking at the partial f_y along y = 0 and y = x.

 

[snipez90]

Sorry you're right, I was thinking about infinity; the epsilon and deltas check out. Incidentally, I found an error in my untested lecture notes.

As for your hint, presumably you want to show that the partial derivative of y is not continuous at (0,0), but this doesn't guarantee non-differentiability right? The theorem that all the partial derivatives of a function exist and are continuous implies differentiability does not have a converse, as can be demonstrated via the following function:

$$\[ f(x,y) =\begin{cases}x^2\sin(\frac{1}{x}) + y^2 &\text{if }x \neq 0 \\<br /> y^2 &\text{if }x = 0 \end{cases} \]$$

 

[LCKurtz]

Yes, you're correct about the converse. On looking at it a bit more I think it is differentiable at (0,0). One can calculate f_x(0,0) and f_y(0,0) then check the condition for differentiability:

$$f(0+h,0+k) - f(0,0) = f_x(0,0) h + f_y(0,0)k +\varepsilon_1h + \varepsilon_2k$$

where ε₁ and ε₂ → 0 as h and k → 0 directly by writing it out.

 

[Sux]

Sorry I said partial derivatives don't exist, I wanted to say they are not continuous in (0,0).

To the OP: Try looking at the partial f_y along y = 0 and y = x.

I have calculated $$f_{y}$$ and it is not continuous in (0,0).

How can you calculated the condition for differentiability that have written?


Thanks!

 

[Altabeh]

Homework Statement

Hello, I can't find any way to prove if this funtion is or isn't differentiable in if $$(x,y)=(0,0)$$ :
$${f(x,y)=\displaystyle\frac{x^{3}}{x^{2}+y^{2}}}$$ if $$(x,y) \neq(0,0)$$

$$f(x,y)=0$$ if $$(x,y)=(0,0)$$

The Attempt at a Solution

Partial derivatives don't exist in (0,0), so i have calculate definition of differentiability, and i get:

$$\displaystyle\lim_{x,y \to{0,0}}{} \displaystyle\frac{x^{3}}{(x^{2}+y^{2})^{3/2}}$$

Which limit doesn't exit, so it is not differentiable in (0,0).

Is this right?

Thanks!

This function is both continuous and differentiable at (0,0). o But let's check these two conditions:

1- Continuity: since the numerator x³ in $${f(x,y)=\displaystyle\frac{x^{3}}{x^{2}+y^{2}}}$$ tends to zero faster than the denominator, so $$\displaystyle\lim_{x,y \to{(0,0)}}{} \displaystyle\frac{x^{3}}{x^{2}+y^{2}} = 0$$ and by f(0,0)=0, this condition is met by our function.

2- Differentiablity: first start with the derivative of $${f(x,y)=\displaystyle\frac{x^{3}}{x^{2}+y^{2}}}$$ wrt x and after calculating it, find the value for $$(x=\epsilon,y=0)$$. Repeat this calculation for y and then substitute $$(x=0,y=\epsilon)$$ in the result. This value must match the preceding one and of course f'(x=0,y=0) which is clearly zero from the second part of the definition of f(x,y), BUT IT DOESN'T.

[Edit]: f_x($$\epsilon$$,0) = 1 so it does not match f_y(0,$$\epsilon$$)=0 and therefore the function is NOT differentiable at (0,0). For a better understanding of why this condition does not hold for f, see its graph:

http://img149.imageshack.us/img149/627/plotg.jpg

Sounds like we have a cusp there.


AB

 

[LCKurtz]

Yes, you're correct about the converse. On looking at it a bit more I think it is differentiable at (0,0). One can calculate f_x(0,0) and f_y(0,0) then check the condition for differentiability:

$$f(0+h,0+k) - f(0,0) = f_x(0,0) h + f_y(0,0)k +\varepsilon_1h + \varepsilon_2k$$

where ε₁ and ε₂ → 0 as h and k → 0 directly by writing it out.

Sorry I said partial derivatives don't exist, I wanted to say they are not continuous in (0,0).



I have calculated $$f_{y}$$ and it is not continuous in (0,0).

How can you calculated the condition for differentiability that have written?


Thanks!

The first step is to calculate f_x and f_y at (0,0). I get that f_x(0,0)=1 and f_y(0,0)=0; you can check this. Then substituting everything into the equation:

$$f(0+h,0+k) - f(0,0) = f_x(0,0) h + f_y(0,0)k +\varepsilon_1h + \varepsilon_2k$$

gives:

$$\frac {h^3}{h^2+k^2} = 1h + 0k + \varepsilon_1h + \varepsilon_2k$$

Solve this for the epsilon terms:

$$\varepsilon_1h + \varepsilon_2k = \frac {h^3}{h^2+k^2} -h= \frac{hk^2}{h^2+k^2}= 0h+\left(\frac{hk}{h^2+k^2}\right)k$$

That will do it if you can show

$$lim_{(h,k)\rightarrow(0,0)}\frac{hk}{h^2+k^2} = 0$$

Can you take it from there?

[Edit] And then again, maybe that doesn't go to zero