Find the derivative of the scalar field: $\Omega=x^2yz+4xz^2$ at the direction of the vector: $(2, -1, -1)$ at the point: $(1, -2, -2)$

Question

Find the derivative of the scalar field: $$\Omega=x^2yz+4xz^2$$ at the direction of the vector: $(2, -1, -1)$ at the point: $P(1, -2, -2)$

Hint: Write the unit vector $\hat n$ at the beginning, qhich has the same direction that the given vector, and evaluate $n \cdot \nabla \Omega$ at $P$.

Answer 1

$\nabla \Omega=(2xyz+4z^2, x^2z, x^2y+8xz)$

$\nabla \Omega(P)=(2.1.(-2).(-2)+4(-2)^2, 1^2.(-2), 1^2.(-2)+8.1.(-2))=(8, -4, -16)$ $\mathbf{n}=(\frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}, \frac{-1}{\sqrt{6}})$

$\mathbf{n}.\nabla \Omega(P)=8.\frac{2}{\sqrt{6}}-4.\frac{-1}{\sqrt{6}}-16.\frac{-1}{\sqrt{6}}=\frac{16+4+16}{\sqrt{6}}=6\sqrt{6}$