Question

Find the greatest value of the directional derivative of the function f=x^2yz^2 at (2,1,-1)​

Answer 1

Directional derivative

Solution.

Here the given function is

$\quad\quad f=x^{2}yz^{2}$

Now $grad\:f$

$=\hat{i}\frac{\partial}{\partial x}(x^{2}yz^{2})+\hat{j}\frac{\partial}{\partial y}(x^{2}yz^{2})\\ \quad+\hat{k}\frac{\partial}{\partial z}(x^{2}yz^{2})$

$=2xyz^{2}\hat{i}+x^{2}z^{2}\hat{j}+2x^{2}yz\hat{k}$ at the point $(x,\:y,\:z)$

$=4\hat{i}+4\hat{j}-8\hat{k}$ at the point $(2,\:1,\:-1)$

Therefore the greatest value of the directional derivative of the given function $f$ is

$\quad\quad =|grad\:f|$

$\quad\quad =|4\hat{i}+4\hat{j}-8\hat{k}|$

$\quad\quad =4|\hat{i}+\hat{j}-2\hat{k}|$

$\quad\quad =4\sqrt{1^{2}+1^{2}+(-2)^{2}}$

$\quad\quad =4\sqrt{1+1+4}$

$\quad\quad =4\sqrt{6}$

Answer: The greatest value of the directional derivative of the function $f=x^{2}yz^{2}$ at $(2,\:1,\:-1)$ is $\bold{4\sqrt{6}}$.