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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.

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4 votes

Answer:


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

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)=(\partial)/(\partial x)V(x,y,z)\hat{i}+(\partial)/(\partial y)V(x,y,z)\hat{i}+(\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.

User Safwan Hijazi
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