Find the Potential of a Conservative Vector Field and Find the Work Example 1

Find \(\mathop f\limits^ \to \) such that \(\mathop F\limits^ \to = \nabla \mathop f\limits^ \to \) and find \(\int\limits_C^{} {\mathop F\limits^ \to \cdot dr} \) for a curve C. We're given that \(F(x,y) = x{y^2}\mathop i\limits^ \to + {x^2}y\mathop j\limits^ \to \) and 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 \(0 \le t \le 1\).

Basically we're asked to find a function f, where if you take the gradient of it (each partial derivative on each component) you get back F. And we're asked to find that integral, but that's really just work.

You can find f sometimes by just playing around and making an equation, 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.

\[f = \int {x{y^2}dx} \]

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

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. So,

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

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!

\[x = t + \sin \left( {\frac{{\pi t}}{2}} \right)\]

\[x(t = 1) = 2\]

\[x(t = 0) = 0\]

\[y = t + \cos \left( {\frac{{\pi t}}{2}} \right)\]

\[y(t = 1) = 1\]

\[y(t = 0) = 1\]

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.

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

\[f\left( {r(t = 1)} \right) = 2 + K\]

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

\[f\left( {r(t = 0)} \right) = K\]

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.

\[\int\limits_C^{} {\mathop F\limits^ \to \cdot dr} = f\left( {r(t = 1)} \right) - f\left( {r(t = 0)} \right)\]

\[\int\limits_C^{} {\mathop F\limits^ \to \cdot dr} = 2 + K - K\]

\[\int\limits_C^{} {\mathop F\limits^ \to \cdot dr} = 2\]

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.

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 Ex2

Finding Potential of Conservative Vector Field Ex3

Alternatively return to the vector fields hub page:

Conservative Vector Fields and Work

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