4
$\begingroup$

Consider the function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ given by $$f(x,y)=\begin{cases} \frac{xy}{x^2+y^2} &\text{if} (x,y)\neq (0,0)\\ 0 &\text{if} (x,y)=(0,0) \end{cases}$$ For which vectors $v=(v_1,v_2)\neq(0,0)\in\mathbb{R}^2$ does the directional derivative $D_vf(0,0)$ exist? Evaluate the directional derivative wherever it exists.

I've managed to get this down to $$D_vf(0,0)=\lim_{h\to 0}\frac{v_1v_2}{h(v_1^2+v_2^2)}$$ and as far as I'm aware this can only exist if one of $v_1$ or $v_2$ is $0$, but not both by the original condition. If this is the case does that mean the directional derivative only exits for $v=(v_1,0) \; \text{or} \; v=(0,v_2)$? If that is the case, how would I find the value of the directional derivative as it comes down to $D_vf(0,0)=\frac{0}{0}$?

$\endgroup$
  • $\begingroup$ simply observe that for $v=(v_1,0) \; \text{or} \; v=(0,v_2)$ we have $\frac{v_1v_2}{h(v_1^2+v_2^2)}=0$ then $D_vf(0,0)=\lim_{h\to 0}\frac{v_1v_2}{h(v_1^2+v_2^2)}=0$ $\endgroup$ – user Apr 1 '18 at 6:55
5
$\begingroup$

If this is the case does that mean the directional derivative only exits for $v=(v_1,0)$ or $v=(0,v_2)$?

Yes, you are right. Because the limit doesn't exists for $v_1\neq 0\neq v_2$, the directional derivate of $f$ doesn't exists to for $v_1\neq 0\neq v_2$.

If that is the case, how would I find the value of the directional derivative as it comes down to $D_vf(0,0)=\frac00$?

Where does $\frac00$ appear? Consider $v\neq 0$. Therefore, if $v_1=0$, then $v_2\neq 0$ and $$ \frac{v_1v_2}{h(v_1^2+v_2^2)}=\frac{0\cdot v_2}{h(0^2+v_2^2)}=\frac{0}{hv_2^2}=0. $$


The result is less surprising if we consider for $(x,y)\neq (0,0)$ and $r>0$. $$ f(rx,ry)=\frac{rxry}{(rx)^2+(ry)^2}=\frac{xy}{x^2+y^2}. $$ So $f$ is constant along the rays starting from the orgin. If $x=0$ xor $y=0$, then $f$ goes continuously to $0$ for $r\to 0$. Since $f$ is constant on the ray, the directional derivative is obviously $0$. But if $x\neq 0\neq y$, the function $f$ is constant $\frac{xy}{x^2+y^2}\neq 0$ for $r>0$ and $0$ for $r=0$. Hence $f$ has a jump here. Therefore, the directional derivative becomes $\pm\infty$ depending if the jump goes up or down.

$\endgroup$
3
$\begingroup$

Note that for $v=(v_1,0) \; \text{or} \; v=(0,v_2)$

$$\frac{v_1v_2}{h(v_1^2+v_2^2)}=0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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