GATE 2018 EE Q.12: solution and answer

Official answer: ASource: Official answer key, GATE 2018 EE (IIT Guwahati)

1. The directional derivative of $\Phi$ along $\mathbf{p}$ is $D_{\mathbf{p}}\Phi =\nabla \Phi \cdot \hat{\mathbf{p}}$, where $\hat{\mathbf{p}}=\frac{\mathbf{p}}{|\mathbf{p}|}$ is the unit vector along $\mathbf{p}$.

2. Partial derivatives of $\Phi =x{y}^2+y{z}^2+z{x}^2$: $\frac{\partial \Phi }{\partial x}=y^2+2zx$, $\frac{\partial \Phi }{\partial y}=2xy+z^2$, $\frac{\partial \Phi }{\partial z}=2yz+x^2$.

3. At $(2,-1,1)$: $\frac{\partial \Phi }{\partial x}=(-1{)}^2+2(1)(2)=1+4=5$, $\frac{\partial \Phi }{\partial y}=2(2)(-1)+1^2=-4+1=-3$, $\frac{\partial \Phi }{\partial z}=2(-1)(1)+2^2=-2+4=2$, so $\nabla \Phi =5\mathbf{i}-3\mathbf{j}+2\mathbf{k}$.

4. Unit vector: $|\mathbf{p}|=\sqrt{1^2+2^2+2^2}=\sqrt{9}=3$, so $\hat{\mathbf{p}}=\frac{1}{3}(\mathbf{i}+2\mathbf{j}+2\mathbf{k})$.

5. Dot product: $$D_{\mathbf{p}}\Phi =\frac{(5)(1)+(-3)(2)+(2)(2)}{3}=\frac{5-6+4}{3}=\frac{3}{3}=1$$

6. The directional derivative is $1$, which is option (A).