### Work Done By Force Field on Particle (Vector Fields) Example 1

Find the work done by the force field $$F(x,y) = {x^2}\mathop i\limits^ \to + xy\mathop j\limits^ \to$$ on a particle that moves once counterclockwise around the circle governed by $${x^2} + {y^2} = 4$$.

We need to make a vector equation for "r", which will tell us the path this goes in, and we should use polar coordinates since we see we're given a circle. We're told the path is a circle of radius 2 (square root the 4).

In other words, $r(t) = \left\langle {2\cos (t),2\sin (t)} \right\rangle$ for $$0 \le t \le 2\pi$$

Why? Treat the components in r(t) as x and y, plug the x component into x squared, and plug the y component into y squared, and you get 4 = 4 as a check. We're also told once counterclockwise, so it starts from 0 to $$2\pi$$ since each quadrant is $$\frac{\pi }{2}$$.

We need to find the derivative of r(t). $r'(t) = \left\langle { - 2\sin (t),2\cos (t)} \right\rangle$

Then we find F(r(t)), which is just substituting our r(t) into F(x,y). $F\left( {r(t)} \right) = \left\langle {4{{\cos }^2}(t),4\sin (t)\cos (t)} \right\rangle$

Then all we need to find is $$\int_0^{2\pi } {\left( {F \cdot dr} \right)} dt$$. This is just the dot product between F(r(t)) and r'(t) and we use our domain (once counterclockwise means 0 to 2pi) as the integral boundaries. From there it's algebra and trigonometry in the integral.

$\begin{array}{l}\int_0^{2\pi } {\left( {4{{\cos }^2}(t)\left( { - 2\sin (t)} \right) + 4\sin (t)\cos (t) \cdot 2\cos (t)} \right)} dt\\\int_0^{2\pi } {\left( { - 8{{\cos }^2}(t)\sin (t) + 8{{\cos }^2}(t)\sin (t)} \right)} dt\\ = 0\end{array}$

So the work done by the force field is 0. Why would this happen? The vectors in the field are just perpendicular to the path r(t), it's never pushing/pulling the particular in the direction it's going in.

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) Ex2

Finding Potential of Conservative Vector Field and Work Ex1

Finding Potential of Conservative Vector Field Ex2

Finding Potential of Conservative Vector Field Ex3 