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Let $f: \mathbb R^2 \rightarrow \mathbb R $ be defined as $$ f(x,y) = \begin{cases} \dfrac {xy^2}{x^2+y^4} & x\ne 0 \\\\ 0 & x=0 ~ \end{cases} $$

Let $D_u f(0,0)$ denote the directional derivative of $f$ at $(0,0)$ in the direction $u = (u_1,u_2) \ne (0,0).$ then $f$ is :

$(i) $ continuous at $(0,0)$ and $D_uf(0,0)$ exists for all $u$.

$(ii) $ continuous at $(0,0)$ but $D_uf(0,0)$ does not exist for some $u \ne (0,0)$

$(iii) $ not continuous at $(0,0)$ and $D_uf(0,0)$ exists for all $u$.

$(iv) $ not continuous at $(0,0)$ and $D_uf(0,0)$ does not exist for some $u \ne (0,0)$

Attempt:

This function is not continuous at $(0,0)$ as along the curve $x = my^2 : f(x,y) = \dfrac {m}{1+m^2}$ which depends on the value of $m$.

In some direction $\theta : x =a \cos \theta, y = a \sin \theta$ and hence the directional derivative at $(0,0)$ in the direction of $\theta$ along a curve of length $a \rightarrow 0 $ will be given as :

$D_u(0,0) =\lim_{a \rightarrow 0} \dfrac {a \cos \theta \cdot a^2 \sin^2 \theta} {a \big (a^2 \cos^2 \theta + a^4 \sin^4 \theta \big )} = \lim_{a \rightarrow 0} \dfrac {\cos \theta \sin^2 \theta}{\cos^2 \theta + a^2 \sin^4 \theta} = \dfrac{\sin^2 \theta}{\cos \theta}$

which does not exist for $\theta = \dfrac {\pi} {2}$

Hence, the function is not continuous at $(0,0)$ and for $u=(0,1), D_u(0,0)$ does not exist either.

Hence, the correct option should be $(iv)$.

Could someone tell me if I am correct?

Thank you!

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  • $\begingroup$ $f: \mathbb R^2 \rightarrow \mathbb R$ $\endgroup$ – Bumblebee Jun 16 '15 at 9:44
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Your argument that $f$ is not continuous at $(0,0)$ is fine.

Your computation of $D_\theta f(0,0)$ is not applicable when $\theta=\pm{\pi\over2}$. You have to treat these cases separately and will obtain $D_{\pm\pi/2}f(0,0)=0$.

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  • $\begingroup$ When $\theta= \dfrac {\pi}{2}$ from origin, then $x=0.$ substituting in the given function, we get $f(x,y)= 0.$ hence directional derivative along y axis is $0$. That means directional derivative exists everywhere? $\endgroup$ – MathMan Jun 16 '15 at 10:01
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Hint: Note that the definition of the directional derivative in the direction $\mathbf{u} = (u_1,u_2)$ is given by $$D_\mathbf{u}f(x,y) \equiv \lim_{h \to 0} \frac{f(x+hu_1,y+hu_2)-f(x,y)}{h}. $$ Try to work out this expression using the function at hand.

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