Answer to Question #144508 in Calculus for Hussaina

Question

Find the directional derivative of \\(Ï•=xy^2+yz^2+xyz\\) at the point (2, -1, 1) in the direction of the vector \\(A=2i+4j-k\\).

Answer 1

$f(x, y, z)=xy^2 +yz^2+xyz, M(2, -1, 1)$

$$\dfrac{\partial f}{\partial x}=y^2+yz,$$

$$\dfrac{\partial f}{\partial y}=2xy+z^2+xz,$$

$$\dfrac{\partial f}{\partial z}=2yz+xy$$

$$\dfrac{\partial f(M)}{\partial x}=(-1)^2+(-1)(1)=0,$$

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

$$\dfrac{\partial f(M)}{\partial z}=2(-1)(1)+2(-1)=-4$$

Given $\vec{u}=2\vec{i}+4\vec{j}-\vec{k}$

$$|\vec{u}|=\sqrt{(2)^2+(4)^2+(-1)^2}=\sqrt{21}$$

$$\vec{l}=\dfrac{2}{\sqrt{21}}\vec{i}+\dfrac{4}{\sqrt{21}}\vec{j}-\dfrac{1}{\sqrt{21}}\vec{k}$$

$$\dfrac{\partial f}{\partial l}=0\cdot\dfrac{2}{\sqrt{21}}+(-1)\cdot\dfrac{4}{\sqrt{21}}+(-4)\cdot(-\dfrac{1}{\sqrt{21}})=0$$

The directional derivative 

$$\dfrac{\partial f}{\partial l}=0$$