Question

Find the work done by the force field F(x, y) = xi + (y + 6)j in moving an object along an arch of the cycloid r(t) = (t − sin(t))i + (1 − cos(t))j, 0 ≤ t ≤ 2π. Step 1 We know that the work done by the force field F in moving an object along the path C which is parameterized by the vector function r(t) is found by the following equation.

Answer 1

The work done by $\vec F$ along $\vec r(t)$ is given by

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r$

where $C$ denotes the given path.

Since we already have the parameterization, all we need to do is compute the differential/length element:

$\mathrm d\vec r=\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt=(1-\cos t)\,\vec\imath+\sin t\,\vec\jmath$

Then the work is

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r$

$\displaystyle=\int_0^{2\pi}((t-\sin t)\,\vec\imath+(7-\cos t)\,\vec\jmath)\cdot((1-\cos t)\,\vec\imath+\sin t\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}(t-t\cos t+6\sin t)\,\mathrm dt=\boxed{2\pi^2}$