# Solution verification:Finding Directional derivative with parameters that satisfy an inequality.

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: $$D_{\hat n}f(0,0) holds for every unit vector $$\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.

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 $$\hat n = -\frac{4}{5}i + \frac{3}{5}j$$.

• 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 $\hat n$ – Vajra May 3 at 13:27
• 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 $$D_{\hat n}f(\bar x,\bar y)=\langle\nabla f,\hat n\rangle=f_x(\bar x,\bar y)\cdot n_1+f_y(\bar x,\bar y)\cdot n_2.$$ – Vajra May 3 at 13:36
• @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 – Pwaol May 3 at 14:34

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 $$\textbf a=(0,0)$$ $$D_{\textbf n}f((0,0))=\lim_{t\to0}\dfrac{f((0,0)+t(n_1,n_2))-f(0,0)}{t}=\lim_{t\to0}\dfrac{f(tn_1,tn_2)-f(0,0)}{t}.$$
• 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 $n_1=0$. – Vajra May 3 at 18:25
• 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 $3n_1 + 4n_2$? – Pwaol May 3 at 19:00