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# How to find the Potential of a Conservative Vector Field and Find the Work Example 1

You may be asked to find the **potential of a conservative vector field** and **work** for a given F(x,y) and r(t).

We know that (memorize these!):

\[\mathop F\limits^ \to = \nabla \mathop f\limits^ \to \]

\[\int\limits_C^{} {\mathop F\limits^ \to \cdot dr} \]

**Example problem:** Find the potential and work, given that \(F(x,y) = x{y^2}\mathop i\limits^ \to + {x^2}y\mathop j\limits^ \to \) and given a function for curve C, \(r(t) = \left\langle {t + \sin \left( {\frac{1}{2}\pi t} \right),t + \cos \left( {\frac{1}{2}\pi t} \right)} \right\rangle \) for the domain of \(0 \le t \le 1\)

In easier terms, we’re asked to find a function f, where if you take the gradient of it (which is each partial derivative on each component) you get back F.

And we’re asked to find that integral, which is the “work”.

You can find f sometimes by just playing around and making an equation (many people can just eyeball it), but a more systematic way to do it is to take the i component, integrate it with respect to x, then take the partial derivative of with respect to y.

\[\begin{array}{l}f = \int {x{y^2}dx} \\f = \frac{{{x^2}{y^2}}}{2} + g(y)\end{array}\]

Where g(y) can be any function of y.

Now take the partial derivative with respect to y.

\[\frac{{\partial f}}{{\partial y}} = {x^2}y + g'(y)\]

Now, this needs to look like what’s before the j component in F.

It looks exactly the same, but g'(y) must be 0.

If that’s the case, integrate that, so then we know g(y) = K, where K is just a real-number constant.

\[f(x,y) = \frac{{{x^2}{y^2}}}{2} + K\]

The next step is to substitute r(t) into f.

However, we’re told that t goes from 0 to 1, but we need to figure out what x and y values that corresponds to!

\[\begin{array}{l}x = t + \sin \left( {\frac{{\pi t}}{2}} \right)\\x(t = 1) = 2\\x(t = 0) = 0\\y = t + \cos \left( {\frac{{\pi t}}{2}} \right)\\y(t = 1) = 1\\y(t = 0) = 1\end{array}\]

From this, t = 1 corresponds to (2,1), and t = 0 corresponds to (0,1).

Now substitute in r(t = 0) and r(t = 1) into f for two cases.

\[\begin{array}{l}f\left( {r(t = 1)} \right) = f(x = 2,y = 1) = \frac{1}{2}{\left( 2 \right)^2}{\left( 1 \right)^2} + K\\f\left( {r(t = 1)} \right) = 2 + K\\f\left( {r(t = 0)} \right) = f(x = 0,y = 1) = \frac{1}{2}{\left( 0 \right)^2}{\left( 1 \right)^2} + K\\f\left( {r(t = 0)} \right) = K\end{array}\]

And the integral asked in the question (the work) is just the difference between these two points.

Remember, the curve goes from t = 0 to t = 1, so you do endpoint – first point, or (t = 1) – (t = 0).

It’s just 2 + K – (K) = 2

So the work is just 2.

If you’re given units in the question, the work will usually be in units of Joules (J).

Notice the “K” constant cancelled out in this problem, which it is guaranteed to, it’s just an intermediate we used.

Click here to try another example where you find potential and work of a conservative vector field

Try another topic, such as using Lagrange multipliers for optimization of multivariate problems.