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My first question is:

1) Is $f(x,y)=\frac {xy(x^{2}-y^{2})}{{(x^{2}+y^{2})}^{3/2}}$ differntiable at (0,0)?

Considering polar co-ordinates: $x=r\cos \theta$ and $y=r\sin \theta$.

$\Rightarrow$ $f(x,y)= \frac {r\cos \theta r\sin \theta (r^{2}\cos ^{2}\theta-r^{2}\sin ^{2}\theta)}{{(r^{2}\cos ^{2}\theta+r^{2}\sin ^{2}\theta)}^{3/2}}$ = $\frac {r^{4}\cos\theta\sin\theta(\cos ^{2}\theta-\sin ^{2}\theta)}{r^{3}}$ = $r\cos\theta\sin\theta(\cos ^{2}\theta-\sin ^{2}\theta)$

Hence linear in r, therfore there doesn't exist and unique tangent plane at (0,0), therefore not differnetiable there.

2) Considering $f(x,y)= \frac {x^{3}y}{x^{6}+y^{2}}$ (x,y) $\neq$ 0 and 0 if (x,y)=(0,0). If you were to plot the function $\theta \rightarrow f(r\cos\theta, r\sin\theta)$ for $\theta \in [0,2\pi]$ what might the plot look like. Justify your answer.

Bit ensure on what this plot would be and the reason. Im guessing at the crinkle function?

Many thanks in advance.

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up vote 1 down vote accepted

$1)$ It is not differentiable at (0,0). The function does not exist at this point. When you plug in (0,0) you get$\frac{0}{0}$ which is undefined. If you are still not convinced you can take the partial derivative of $f(x,y)$ with respect to $x$ and with respect to $y$ and plug in (0,0): $$ \frac{\partial f}{\partial x}=\frac{5 x^2 y^3 - y^5}{(x^2 + y^2)^{5/2}} \\\frac{\partial f}{\partial y}=\frac{x^5 - 5 x^3 y^2}{(x^2 + y^2)^{5/2}} $$ Now, plug in (0,0) and again in both cases you will get $\frac{0}{0}$ which means the slope of the tangent line is undefined and $f(x,y)$ is not differentiable at (0,0).

$2)$ I could not come up with a convenient way of plotting $f(x,y)=\frac {x^{3}y}{x^{6}+y^{2}}$ without using any software. The only thing I can say about it is that its domain is $x^6 +y^2>0$. This is how Function $f(rcos \theta,rsin\theta)$ will look like: $$f(rcos \theta,rsin\theta)=\frac{r^4cos^3\theta sin\theta}{(r^6cos^6\theta + r^2 sin^2\theta)}$$

and its graph:

enter image description here

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