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For $(x,y)^2\in \mathbb{R}^2$, let

$f(x,y)=\begin{cases} [(2x^2-y)(y-x^2)]^{1/4}&x^2\leq y \leq 2x^2\\ 0& \text{otherwise}\\ \end{cases} $

show that all directional derivative of $f$ exist at $(0,0)$, but $f$ is not differentiable at $(0,0)$.

My attempt: Firstly, I observed that the curve become linear when it approaches to zero.

Let $u=(u_1,u_2)\in \mathbb{R}^2$ be a unit vector.

$$D_uf(0,0)=\lim_{t \rightarrow 0}\frac{f(tu_1,tu_2)-f(0,0)}{t}=\lim_{t \rightarrow 0}\frac{0-0}{t}=0.$$

This implies all the directional derivatives of $f$ exist at $(0,0)$.

I want to improve the more justification, why $f(tu_1,tu_2)=0$. I understand with graph of the curve. Can anyone suggest me how I improve my justification in this question.

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  • $\begingroup$ Is it possible you miss something in conditions or task itself? $\endgroup$ – zkutch Jul 7 '20 at 6:27
  • $\begingroup$ Which condition? $\endgroup$ – User124356 Jul 7 '20 at 6:33
  • $\begingroup$ Hello likely IU student. If you are preparing for Tier 1, feel free to email me for the resources I used for it. $\endgroup$ – Alfred Yerger Jul 7 '20 at 7:16
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Let $g(t)=(t,\frac32t^2)$. Then $g(t)$ is differentiable at $t=0$ and $g(0)=(0,0)$. If $f(x,y)$ were differentiable at $(0,0)$ then its derivative there would be $0$ and by the chain rule: $$\frac{{\rm d}f(g(t))}{{\rm d}t\qquad}\left|_{t=0}\right.=\nabla f(0,0)\cdot g'(0)=0$$

However if you work out $\frac{{\rm d}f(g(t))}{{\rm d}t\quad}\left|_{t=0}\right.$, it will not be $0$.

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  • $\begingroup$ This is same as my answer except that you write $1.5$ as $\frac 3 2$. What is the point in repeating an answer. $\endgroup$ – Kavi Rama Murthy Jul 7 '20 at 11:59
  • $\begingroup$ Its a slightly different perspective - exploiting the differential as a linear map, rather than for its implications for the speed of growth at the origin. I found it appealing to write it this way. $\endgroup$ – tkf Jul 7 '20 at 12:22
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Hint: The inequalities $t^{2}u^{2}\leq tv \leq 2t^{2} u^{2}$ will automatically fail for $|t|$ sufficiently small and hence $f(tu,tv)=0$ for such $t$. [I will add more details if you cannot justify this].

Since the partial derivatives vanish at the origin the only candidate for the derivative at that point is $0$. So, is $f$is differentiable at $(0,0)$ then we must have $\frac {f(x,y)} {\sqrt {x^{2}+y^{2}}} \to 0$ as $(x,y) \to 0$. Get a contradiction by taking $y=(1.5)x^{2}$ and letting $x \to 0$.

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  • $\begingroup$ Yes. I got it. Can you give me some hint for 2nd part. $\endgroup$ – User124356 Jul 7 '20 at 6:34
  • $\begingroup$ @User124356 I have edited my answer. $\endgroup$ – Kavi Rama Murthy Jul 7 '20 at 7:07

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