Q. 17 - Calculus

Question

In Exercises 17–24, find a potential function for the given vector field.

$\mathbf{F}(x, y)=(3x^2\cos y,-x^3\sin y)$

Step-by-Step Solution

Answer

A potential function for the given vector field is $f(x,y)=x^3\cos y$.

Step 1. Given Information

In given exercises we have to find a potential function for the given vector field. 

$\mathbf{F}(x, y)=(3x^2\cos y,-x^3\sin y)$

Step 2. Since F ( x ,   y ) = ( 3 x 2 cos y , − x 3 sin y )

$$\begin{gathered} f(x, y)=3\cos y\int x^{2}dx+\alpha +B \\ f(x, y)=3\cos y\left[\frac{x^{2+1}}{3}\right]+\alpha +B \\ f(x, y)=3\cos y·\frac{x^3}{3}+\alpha +B \\ f(x, y)=x^3\cos y+\alpha +B \end{gathered}$$

where $\alpha$ is an arbitrary constant and B is the integral with respect to y of the terms in $F_2(x,y)$ in which the factor x does not appear.

Step 3. In this case, that is all of F 2 ( x , y ) , so

$$\begin{gathered} B=\int (-x^3\sin y)dy \\ B=-x^3\int \sin ydy \\ B=-x^3(-\cos y)+\beta \\ B=x^3\cos y+\beta \end{gathered}$$

where β is an arbitrary constant.

Step 4. Setting the constants equal to zero since they do not affect the gradient of f ( x , y )

We have,

$f(x,y)=x^3\cos y$