Q. 37 - Calculus

Question

Show that the vector fields in Exercises 33–40 are not conservative.

$\mathbf{F}(x,y,z)=2\mathbf{i}-z\mathbf{j}+{\mathrm{e}}^{\mathrm{yz}}\mathbf{k}$

Step-by-Step Solution

Answer

The vector fields is not conservative because $\frac{\partial F_1}{\partial z}\ne \frac{\partial F_2}{\partial y}$. 

Step 1. Given Information

We have to show that the vector fields in the given exercise is not conservative.
$\mathbf{F}(x,y)=2\mathbf{i}-z\mathbf{j}+{\mathrm{e}}^{\mathrm{yz}}\mathbf{k}$

Step 2. A vector field F ( x , y ) = ( F 1 ( x , y ) , F 2 ( x , y ) ) is not conservative if and only if ∂ F 1 ∂ z ≠ ∂ F 2 ∂ y .

For the vector field $\mathbf{F}(x,y)=2\mathbf{i}-z\mathbf{j}+{\mathrm{e}}^{\mathrm{yz}}\mathbf{k}$

$$\begin{gathered} \frac{\partial F_1}{\partial z}=\frac{\partial }{\partial z}(-z) \\ \frac{\partial F_1}{\partial z}=-1 \end{gathered}$$

Step 3. Now finding ∂ F 2 ∂ y

$$\begin{gathered} \frac{\partial F_2}{\partial y}=\frac{\partial }{\partial y}{e}^{yz} \\ \frac{\partial F_2}{\partial y}=z{e}^{yz} \end{gathered}$$

Hence, $\frac{\partial F_1}{\partial z}\ne \frac{\partial F_2}{\partial y}.$