Question

Obtain the gradient of the following scalar field :
U(X,Y,Z)=X^2Zcos2Y

Answer 1

Question#38737, Math, Differential Calculus

Obtain the gradient of the following scalar field :

$$U(X,Y,Z) = X^2 Z \cos 2Y$$

Solution

By the definition, $\operatorname{grad} U = \frac{\partial U}{\partial X} \mathbf{i} + \frac{\partial U}{\partial Y} \mathbf{j} + \frac{\partial U}{\partial Z} \mathbf{k}$.

All we have to do then is to find the partial derivatives.

1. If the brackets were omitted, then the partial derivatives are

$$\frac{\partial U}{\partial X} = 2XZ \cos(2Y), \quad \frac{\partial U}{\partial Y} = -2X^2Z \sin(2Y), \quad \frac{\partial U}{\partial Z} = X^2 \cos(2Y).$$

Then gradient of the scalar field is the following vector

$$\operatorname{grad} U = 2XZ \cos(2Y) \mathbf{i} - 2X^2Z \sin(2Y) \mathbf{j} + X^2 \cos(2Y) \mathbf{k}.$$

Answer

$$\operatorname{grad} U = 2XZ \cos(2Y) \mathbf{i} - 2X^2Z \sin(2Y) \mathbf{j} + X^2 \cos(2Y) \mathbf{k}$$

2. If the brackets were not omitted, then the partial derivatives are equal to

$$\frac{\partial U}{\partial X} = 2XZY \cos 2, \quad \frac{\partial U}{\partial Y} = X^2Z \cos 2, \quad \frac{\partial U}{\partial Z} = X^2Y \cos 2.$$

Then gradient of the scalar field is equal to the following vector

$$\operatorname{grad} U = (2ZY\mathbf{i} + XZ\mathbf{j} + XY\mathbf{k})X \cos 2.$$

Answer

$$\operatorname{grad} U = (2ZY\mathbf{i} + XZ\mathbf{j} + XY\mathbf{k})X \cos 2$$