Question

A scalar field is given by $f=x^{2/3}+y^{2/3},$ where $x$ and $y$ are the Cartesian coordinates. The derivative of ${}^{\prime }{f}^{\prime }$ along the line $y=x$ directed away from the origin at the point (8,8) is

• A. $\frac{\sqrt{2}}{3}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{2}{\sqrt{3}}$
• D. $\frac{3}{\sqrt{2}}$

Answer 1

The correct option is A $\frac{\sqrt{2}}{3}$

$\overrightarrow{OP}=8\hat{i}+8\hat{j}$
Unit vector along $y=x$ is
$\widehat{OP}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

$f=x^{2/3}+y^{2/3}\Rightarrow ▽f=\frac{2}{3}{x}^{\frac{-1}{3}}\overrightarrow{i}+\frac{2}{3}{y}^{\frac{-1}{3}}\overrightarrow{j}$
At $(8,8)▽f=\frac{1}{3}i+\frac{1}{3}j$
$\therefore$ Directional derivative
$=▽f.\hat{a}=(\frac{\hat{i}}{3}+\frac{\hat{j}}{3}).\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)$=$\frac{\sqrt{2}}{3}$