Q. 34 - Calculus

Question

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

$\mathbf{F}(x,y)=(x^2+y^2,\cos y)$

Step-by-Step Solution

Answer

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

Step 1. Given Information

We have to show that the vector fields in the given exercise is not conservative.
$\mathbf{F}(x,y)=(x^2+y^2,\cos y)$

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 ∂ y ≠ ∂ F 2 ∂ x .

For the vector field $\mathbf{F}(x,y)=(x^2+y^2,\cos y)$

$$\begin{gathered} \frac{\partial F_1}{\partial y}=\frac{\partial }{\partial y}(x^2+y^2) \\ \frac{\partial F_1}{\partial y}=\frac{\partial }{\partial y}{x}^2+\frac{\partial }{\partial y}{y}^2 \\ \frac{\partial F_1}{\partial y}=0+2y \\ \frac{\partial F_1}{\partial y}=2y \end{gathered}$$

Step 3. Now finding ∂ F 2 ∂ y

$$\begin{gathered} \frac{\partial F_2}{\partial x}=\frac{\partial }{\partial x}\cos y \\ \frac{\partial F_2}{\partial x}=0 \end{gathered}$$

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