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Given this function:

$$f(x,y) = \left\{\begin{matrix} \frac{x^3 + 2y^3}{x^2 + y^2} & (x,y) \neq 0 \\ 0 & (x,y) = (0,0) \end{matrix}\right.$$

  • Find the directional derivative $\frac{\partial f}{\partial n} (0,0)$ for each unit vector $n$.

  • In which direction the directional derivative is the biggest?


I know that $f_{\vec{n}}(0,0) = \nabla f(0,0) \cdot \frac{\vec{n}}{||\vec{n}||} $

And because $\vec{n}$ is a unit vector: $||\vec{n}|| = 1$ and thus we have that the directional derivative is: $f_{\vec{n}}(0,0) = \nabla f \cdot \vec{n}$

But I don't know how to continue from here... how does the vector $\vec{n}$ comes into play in this question? If the function takes $0$ at the point $(0,0)$ ... I would appreciate your kind help, thanks!

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If $n$ is a unit vector, then $n=(\cos\theta,\sin\theta)$, for some $\theta\in\Bbb R$. And the directional derivative of $f$ at $(0,0)$ in the direction given by $n$ is\begin{align}\lim_{h\to0}\frac{f(hn+(0,0))-f(0,0)}h&=\lim_{h\to0}\frac{h^3\cos^3\theta+2h^3\sin^3\theta}{h^3}\\&=\cos^3\theta+2\sin^3\theta.\end{align}It is not hard to prove that the maximum value of this expression is $2$, attained when $\theta=\frac\pi2$.

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  • $\begingroup$ Hey! Thanks for your awesome answer! - I have a question - is $cos^3 {\theta} + 2 sin^3 {\theta}$ as a final answer fine? or we need to compute it back to algebraic form (not polar) Thanks! $\endgroup$
    – CSch of x
    Aug 18, 2020 at 15:48
  • $\begingroup$ All I can say is that I would accept that as an answer, if it is stated along with it that the unit vector is $(\cos\theta,\sin\theta)$. $\endgroup$ Aug 18, 2020 at 15:49
  • $\begingroup$ Thanks you sir, one last question, I can take the derivative to get the critical points of the trigonometric function $cos^3 ( \theta ) + 2 sin^3( \theta) $ - and get it is indeed at 90 deg. (= $\frac{ \pi}{2}$ rads) - but what stops me from taking $ 2 \pi + \frac{ \pi}{2}$ for example? which is still the critical point. Thanks! $\endgroup$
    – CSch of x
    Aug 18, 2020 at 15:58
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    $\begingroup$ Sure, but it's the same direction! After all$$\left(\cos\left(\frac\pi2\right),\sin\left(\frac\pi2\right)\right)=\left(\cos\left(2\pi+\frac\pi2\right),\sin\left(2\pi+\frac\pi2\right)\right).$$ $\endgroup$ Aug 18, 2020 at 16:01
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    $\begingroup$ I'm glad I could help. $\endgroup$ Aug 18, 2020 at 16:03

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