Question

V(x, y, z) = 5x2 − 3xy + xyz (a) Find the rate of change of the potential at P(6, 6, 5) in the direction of the vector v = i + j − k.

Answer 1

Answer:

$D_{\vec{v}}V(6,6,5)=48$

Step-by-step explanation:

You have the following potential function:

$V(x,y,z)=5x^2-3xy+xyz$           (1)

To find the rate of change of the potential at the point P(6,6,5) in the direction of v = i + j - k, you use the following formula:

$D_{\vec{v}}V(x,y,z)=\bigtriangledown V(x,y,z)\cdot \vec{v}$           (2)

First, you calculate the gradient of V:

$\bigtriangledown V(x,y,z)=\frac{\partial}{\partial x}V(x,y,z)\hat{i}+\frac{\partial}{\partial y}V(x,y,z)\hat{i}+\frac{\partial}{\partial z}V(x,y,z)\hat{i}\\\\\bigtriangledown V(x,y,z)=(10x-3y+yz)\hat{i}+(-3x+xz)\hat{j}+(xy)\hat{k}\\\\\bigtriangledown V(6,6,5)=(10(6)-3(6)+(6)(5))\hat{i}+(-3(6)+(6)(5))\hat{j}+((6)(6))\hat{k}\\\\\bigtriangledown V(6,6,5)=72\hat{i}+12\hat{j}+36\hat{k}$

Next, you replace in the equation (2):

$D_{\vec{v}}V(6,6,5)=(72\hat{i}+12\hat{j}+36\hat{k})\cdot(\hat{i}+\hat{j}-\hat{k})\\\\D_{\vec{v}}V(6,6,5)=48$

Then, the rate of change of the potential at the point P(6,6,5) in the direction of v, is 48.