Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.