Question

Obtain the directional derivative for a scalar field (x,y,z)=3x^2y-y^3z^2 at the point (1,-2,-1) in the direction i+j+k

Answer 1

Find the gradient at the point $(1,-2,-1)$

$\frac{\partial\Phi}{\partial x}=6xy=6\cdot(-2)\cdot1=-12$

$\frac{\partial\Phi}{\partial y}=3x^2-3y^2z^2=3\cdot 1^2-3 \cdot(-2)^2 \cdot (-1)^2=-9$

$\frac{\partial\Phi}{\partial z}=-2y^3z=-2\cdot(-2)^3\cdot(-1)=16$

$\nabla \Phi(1,-2,-1)=-12\overrightarrow{i}-9\overrightarrow{j}+16\overrightarrow{k}=(-12,-9,16)$

Let $\overrightarrow{u}=u_1\overrightarrow{i}+u_2\overrightarrow{j}+u_3\overrightarrow{k}$ be a unit vector. The directional derivative at $(1,-2,-1)$ in the direction of $\overrightarrow{u}$ is

$D_u\Phi(1,-2,-1)=\nabla\Phi(1,-2,-1)\cdot \overrightarrow{u}=$

$=(-12\overrightarrow{i}-9\overrightarrow{j}+16\overrightarrow{k})(u_1\overrightarrow{i}+u_2\overrightarrow{j}+u_3\overrightarrow{k})=$

$=-12u_1-9u_2+16u_3$

To find the directional derivative in the direction of the vector $(1,1,1)$ , we need to find a unit vector in the direction of the vector $(1,1,1)$ . We simply divide by the magnitude of $(1,1,1)$.

$\overrightarrow{u}=\frac{(1,1,1)}{\lvert\lvert(1,1,1)\rvert\rvert}=\frac{(1,1,1)}{\sqrt{1^2+1^2+1^2}}=\frac{(1,1,1)}{\sqrt{3}}=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$

So,

$D_u\Phi(1,-2,-1)=-12u_1-9u_2+16u_3=$

$=-12\frac{1}{\sqrt{3}}-9\frac{1}{\sqrt{3}}+16\frac{1}{\sqrt{3}}=-\frac{5}{\sqrt{3}}$ Answer