Question

The directional derivative of $\varphi =3x^{2}y-4y{z}^2+6z^{2}x$ at point (1,1,1) in the direction of line
$\frac{x-1}{2}=\frac{y-4}{2}=\frac{z}{3}$

• A. $2\sqrt{17}$
• B. $\frac{2}{\sqrt{17}}$
• C. $\frac{29}{\sqrt{17}}$
• D. $\frac{40}{\sqrt{17}}$

Answer 1

The correct option is A $2\sqrt{17}$

$\overline{▽}\varphi =(\frac{\partial \varphi }{\partial x}\hat{i}+\frac{\partial \varphi }{\partial y}\hat{j}+\frac{\partial \varphi }{\partial z}\hat{k})[3x^{2}y-4y{z}^2+6z^{2}x]$


$\Rightarrow {\overline{▽}}_{\varphi }=(6xy+6z^2)\hat{i}+(3x^2-4z^2)\hat{j}+(-8yz+12zx)\hat{k}$

Now at (1,1,1)

$\overline{▽}\varphi =12\hat{i}-\hat{j}+4\hat{k}$ .... (1)

Also direction of line ,$\hat{A}=\frac{2\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{17}}$ ....(2)

$\Rightarrow$
Directional derivative using (1) & (2)


$\overline{▽}\varphi .\hat{A}=(12\hat{i}-\hat{j}+4\hat{k})\left(\frac{2\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{17}}\right)$

$=\frac{24-2+12}{\sqrt{17}}=\frac{34}{\sqrt{17}}=2\sqrt{17}$