Answer to Question #144509 in Calculus for Hussaina

Question

. If \\(θ(x,y,z)=3x^2 y-y^3 z^2\\),find \\( ˆ‡Î¸\\) (grad θ) at the point (1, -2, -1).

Answer 1

$f(x, y, z)=3x^2y -y^3z^2, M(1, -2, -1)$

$$\dfrac{\partial f}{\partial x}=6xy,$$

$$\dfrac{\partial f}{\partial y}=3x^2-3y^2z^2,$$

$$\dfrac{\partial f}{\partial z}=-2y^3z$$

$$\dfrac{\partial f(M)}{\partial x}=6(1)(-2)=-12,$$

$$\dfrac{\partial f(M)}{\partial y}=3(1)^2-3(-2)^2(-1)^2=-9,$$

$$\dfrac{\partial f(M)}{\partial z}=-2(-2)^3(-1)=-16$$

$$\text{grad}f(M)=-12\vec{i}-9\vec{j}-16\vec{k}$$