### Determine if Vector Field is Conservative and Find the Potential of a Conservative Vector Field Example 3

Determine if F is a conservative vector field and find $$\mathop f\limits^ \to$$ such that $$\nabla \mathop f\limits^ \to = \mathop F\limits^ \to$$. We're given $$F(x,y) = {e^x}\sin (y)\mathop i\limits^ \to + {e^x}\cos (y)\mathop j\limits^ \to$$.

Let the i component be represented by P, and the j component represented by Q. Just find the partial derivatives of each component, i needs to be derived with respect to y, and j derived with respect to x.

$\frac{{\partial Q}}{{\partial x}} = {e^x}\cos (y)$

$\frac{{\partial P}}{{\partial y}} = {e^x}\cos (y)$

$\frac{{\partial Q}}{{\partial x}} = \frac{{\partial P}}{{\partial y}}$

Since these partial derivatives are equal to each other, the vector field F provided is indeed a conservative vector field.

To find the potential f, take the i component and integrate with respect to x, then derive that with respect to y.

$f = \int {({e^x}\sin (y)} )dx$

$f = {e^x}\sin (y) + g(y)$

Where g(y) is any function of y. Now derive this result with respect to y.

$\frac{{\partial f}}{{\partial y}} = {e^x}c\cos (y) + g'(y)$

This should look like the j component in F. It does, with the exception of g'(y). So, g'(y) = 0. But integrating that, g(y) = K, where K is a real-number constant. So the potential f is,

$f(x,y) = {e^x}\sin (y) + K$

And you can check it yourself, find the partial derivative with respect to x and you'll get the i component in F. If find the partial derivative with respect to y and you'll get the j component in F.

We have examples on determining on whether a vector field is conservative or not, finding the potential of a vector field, and finding the work involving particles and vector fields below:

Work Done By Force Field on Particle (Vector Fields) Ex1

Work Done By Force Field on Particle (Vector Fields) Ex2

Finding Potential of Conservative Vector Field and Work Ex1

Finding Potential of Conservative Vector Field Ex2