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I am given the following function:

$$ f(x,y)=\sqrt[3]{x^2 y } $$ at (0,0), and need to find the directions $\vec{v}$ for which the directional derivative $D_\vec{v} f (0,0)$ is maximal.

I know the answer should be $ (\pm \sqrt{\frac{2}{3}}, \frac{1}{\sqrt{3}}) $.

Since this function is not even differentiable (I think), I should differentiate according to the definition: $$ D_\vec{v} f (0,0) = \lim_{h\to 0} \frac{\sqrt[3]{cos^2 (\theta) sin(\theta )} h}{h} = \sqrt[3]{cos^2 (\theta) sin(\theta )} $$ and it seems like the only thing I should do now is to find the maximum value of $\sqrt[3]{x^2 y }$. Is there any simple way to do it?

Will you please help me?

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  • $\begingroup$ Your approach is sound. You have found the directional derivative at the origin along the direction $\hat x\cos \theta+\hat y\sin \theta$. Now, simply find the value(s) of $\theta$ for which the directional derivative is a maximum. $\endgroup$
    – Mark Viola
    Sep 3, 2015 at 20:41

1 Answer 1

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The directional derivative at a point $(x,y)$ and along the direction $\hat r=\hat x\cos \theta+\hat y \sin \theta$, where $\theta$ is the angle the unit vector $\hat r$ makes with the $x$-axis, is given by

$$\lim_{h\to 0}\frac{f(x+\cos \theta h,y+\sin \theta h)-f(x,y)}{h}$$

Let $f$ be the function defined by $f(x,y)=x^{2/3}y^{1/2}$. Then, the directional derivative at $(0,0)$ along $\hat r=\hat x\cos \theta+\hat y \sin \theta$ is

$$\lim_{h\to 0}\frac{(h\cos \theta)^{2/3}(h\sin \theta)^{1/3}}{h}=\cos^{2/3}\theta \sin^{1/3}\theta$$

Local extrema of the directional derivative occur at values of $\theta$ where its derivative is zero. Taking a derivative as setting to zero reveals

$$\begin{align} \frac{d(\cos^{2/3}\theta \sin^{1/3}\theta)}{d\theta}&=-\frac23 \cos^{-1/3}\theta \sin^{4/3}\theta +\frac13\sin^{-2/3}\theta \cos^{5/3}\theta\\\\ &=0\implies \tan \theta =\pm \frac{\sqrt{2}}{2} \end{align}$$

Since we seek the maximum directional derivative, we exclude those directions for which $\sin \theta<0$. Therefore, the directional derivative is maximum along

$$\theta =\arctan\left(\frac{\sqrt{2}}{2}\right)$$

$$\theta =\pi -\arctan\left(\frac{\sqrt{2}}{2}\right)$$

which gives for $\hat r$

$$\bbox[5px,border:2px solid #C0A000]{\hat r=\pm \hat x\frac{\sqrt{6}}{3}+\hat y \frac{\sqrt{3}}{3}}$$

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  • $\begingroup$ You're welcome. My pleasure! $\endgroup$
    – Mark Viola
    Sep 4, 2015 at 14:27

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