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Here's the question:

Given the function
\begin{equation} f(x,y)=\begin{cases} x^2\sin(\frac{1}{x^2})+3x+4y\space, & \text{$x\ne0$}.\\ 4y, & \text{$x=0$}. \end{cases} \end{equation} For what values of $m$ the inequality: $D_{\hat n}f(0,0)<m$ holds for every unit vector $\hat n$.
Find a vector $\hat n$ such that $D_{\hat n}f(0,0)=0$.

I have tried two different approaches so far and finished the solution with the second approach, but I'm a little bit struggling with the logic behind first solution approach, I would appreciate approval of my work for the second approach and tips on how to keep going with the first approach.
Thanks in advance.

My work after Vajra's hint (not complete):
$D_nf(0,0) =lim_{t \to 0} \frac{f(n_1t, n_2t)-f(0,0)}{t}=lim_{t \to 0} \frac{(n_1t)^2 \sin (\frac{1}{(n_1t)^2}) + 3n_1t + 4n_2t - 0}{t}=lim_{t \to 0} (n_1t \sin (\frac{1}{(n_1t)^2}) +3n_1 + 4n_2 )= 3n_1 + 4n_2$
Now I know that $\sqrt {n_1^2 + n_2^2}=1 \Rightarrow n_1 = \pm \sqrt {1-n_2^2}$
I'm not sure how to continue, according to what I did in my second approach, I found the maximum of the general directional derivative, and moved from there, but here I can choose $n_1$ to be $\pm$ the square root, which one should I take and why?



(Complete) approach of: $\hat n = \sin(t)i+ \cos(t)j$
$D_nf(0,0) =lim_{h \to 0} \frac{f(sin(t)h, cos(t)h)-f(0,0)}{h}=lim_{h \to 0} \frac{(sin(t)h)^2 \sin (\frac{1}{(sin(t)h)^2}) + 3sin(t)h + 4cos(t)h - 0}{h}=lim_{h \to 0} (sin^2(t)h \sin (\frac{1}{(sin(t)h)^2}) +3sin(t) + 4cos(t) )= 3sin(t) + 4cos(t)=u(t)$.
In order to find the max value I decided to take derivative :
$u'(t) = 3cos(t)-4sin(t)=0 \Rightarrow 3cos(t)=4sin(t) \Rightarrow $ For every $t\ne \frac{\pi}{2} \Rightarrow tan(t)=\frac{3}{4} \Rightarrow t=arctan(\frac{3}{4})$, and I get that the maximum value for $u(t)$ is $5$, so for every $m>5$ this inequality holds for every unit vector $\hat n$.
For second part I tried to find a $t$ such that $3cos(t)+4sin(t)=0$.
and moving from there found that $\hat n = -\frac{4}{5}i + \frac{3}{5}j$.

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    $\begingroup$ Sorry but I find that $\dfrac{\partial f}{\partial \hat n}$ is a unhappy notation. You're not differentiating respect to the variable $\hat n$... I think it's better the notation $D_{\hat n}f$, which represents the directional derivative of $f$ respect to the versor $\hat n$ $\endgroup$ – Vajra May 3 at 13:27
  • $\begingroup$ In general if $\hat n=(n_1,n_2)$ is a versor, you can calculate the directional derivative of a function $f$ differentiable in $(\bar x,\bar y)$ as $$D_{\hat n}f(\bar x,\bar y)=\langle\nabla f,\hat n\rangle=f_x(\bar x,\bar y)\cdot n_1+f_y(\bar x,\bar y)\cdot n_2.$$ $\endgroup$ – Vajra May 3 at 13:36
  • $\begingroup$ @Vajra Sorry about the notation, I've seen it like this just today in the question I didn't know it's problematic, seems like my course goes with it, about the second comment, it's what I meant - the method with the gradient vector that needs $f$ to be differentiable near the point $\endgroup$ – Pwaol May 3 at 14:34
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Hint: try to use the definition: $$D_\textbf v f(\textbf{a}):=\lim_{t\to0}\dfrac{f(\textbf a+t\textbf v)-f(\textbf a)}{t}$$ In your case you'd have, given $\textbf n=(n_1,n_2)$ and $\textbf a=(0,0)$ $$D_{\textbf n}f((0,0))=\lim_{t\to0}\dfrac{f((0,0)+t(n_1,n_2))-f(0,0)}{t}=\lim_{t\to0}\dfrac{f(tn_1,tn_2)-f(0,0)}{t}.$$

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  • $\begingroup$ Thanks for the hint, I added my progress in an edit in my question by following your hint, I'm still trying to figure out how to keep going $\endgroup$ – Pwaol May 3 at 18:07
  • $\begingroup$ Ok, if your calculations are correct, imposing the result you obtained equal to zero you find for which versors $\textbf n$ the directional derivative is zero. You should also look at the case $\textbf n=(n_1,n_2)$, with $n_1=0$. $\endgroup$ – Vajra May 3 at 18:25
  • $\begingroup$ I'm not getting the idea of $n_1=0$ I know that in that case it's $f'_y$ if I'm not mistaken, but how does it help me reach the maximum value of $3n_1 + 4n_2$? $\endgroup$ – Pwaol May 3 at 19:00

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