Question

The magnitude of the directional derivative of the function $f(x,y)=x^2+3y^2$ in a direction normal to the circle $x^2+y^2=2,$ at the point $(1,1),$ is

• A. $4\sqrt{2}$
• B. $5\sqrt{2}$
• C. $7\sqrt{2}$
• D. $9\sqrt{2}$

Answer 1

The correct option is A $4\sqrt{2}$

$f(x,y)=x^2+3y^2$ than $gradf=2x\hat{i}+6y\hat{j}$
$i.e.,▽f=2x\hat{i}+6y\hat{j}$
$\Rightarrow (▽f{)}_{(1,1)}=2\hat{i}+6\hat{j}$
Let $\varphi =x^2+y^2-2$ than $grad\varphi =2x\hat{i}+2y\hat{j}$
So, $\hat{n}=\frac{grad\varphi }{\left|grad\varphi \right|}=\frac{2\hat{i}+2\hat{j}}{2\sqrt{2}}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$
The required directional derivative $=(▽f{)}_{(1,1)}.\hat{n}$
$=\frac{2}{\sqrt{2}}+\frac{6}{\sqrt{2}}=\frac{8}{\sqrt{2}}=4\sqrt{2}$