Question

The derivative of $f(x,y)$ at point (1,2) in the direction of vector $i+j$ is $2\sqrt{2}$ and in the direction of the vector $-2j$ is $-3$. Then the derivative of $f(x,y)$ in direction $-i-2j$ is

• A. $2\sqrt{2}+3/2$
• B. $-7/\sqrt{5}$
• C. $-2\sqrt{2}-3/2$
• D. $1/\sqrt{5}$

Answer 1

The correct option is B $-7/\sqrt{5}$

Let $\overrightarrow{a}=\hat{i}+\hat{j}$
$\overrightarrow{b}=-2\hat{j}$
$\overrightarrow{c}=-\hat{i}-2\hat{j}$
$▽f.\hat{a}=2\sqrt{2}...\left(1\right)$
$▽f.\hat{b}=-3...\left(2\right)$
$\left(1\right)\Rightarrow ▽f.\frac{(\hat{i}+\hat{j})}{\sqrt{2}}=2\sqrt{2}$
$\Rightarrow ▽f.(\hat{i}+\hat{j})=4...\left(3\right)$
$\left(2\right)\Rightarrow ▽f.\frac{(-2\hat{j})}{2}=-3$
$▽f.\hat{j}=3...\left(4\right)$
If $▽f=x\hat{i}+y\hat{j}$
$\left(3\right)\Rightarrow x+y=4$
$\left(4\right)\Rightarrow y=3$
$\Rightarrow x=1$
$▽f=\hat{i}+3\hat{j}$
Directional derivative in the direction of $\overrightarrow{c}=▽f.\hat{c}$
$=(\hat{i}+3\hat{j}).\frac{(-\hat{i}-2\hat{j})}{\sqrt{5}}=\frac{-1-6}{\sqrt{5}}=\frac{-7}{\sqrt{5}}$