GATE 2022 EE Q.50: solution and answer

Official answer: BSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

1. A scalar field increases fastest in the direction of its gradient, $\nabla f=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}$, evaluated at the point. The directional derivative along a unit vector $\hat{u}$ is $\nabla f\cdot \hat{u}=|\nabla f|\cos \theta$, and this is largest when $\theta =0$, that is, when $\hat{u}$ points along $\nabla f$.

2. Partial derivative with respect to $x$: $\frac{\partial f}{\partial x}=8x+7y+3z^2$. At $P=(1,0,2)$: $8(1)+7(0)+3(2{)}^2=8+0+12=20$.

3. Partial derivative with respect to $y$: $\frac{\partial f}{\partial y}=7x$. At $P$: $7(1)=7$.

4. Partial derivative with respect to $z$: $\frac{\partial f}{\partial z}=6xz$. At $P$: $6(1)(2)=12$.

5. So $$\nabla f{|}_P=20\hat{i}+7\hat{j}+12\hat{k},$$ which is option (B). The unit vector in this direction is $\frac{20\hat{i}+7\hat{j}+12\hat{k}}{\sqrt{593}}$, and the maximum rate of increase is $|\nabla f|=\sqrt{400+49+144}=\sqrt{593}$. Options (A), (C) and (D) each leave out a component that is not zero at $P$, so none of them is parallel to $\nabla f$.