Question

Find the work done by the force field F(x, y) = xi + (y + 5)j in moving an object along an arch of the cycloid r(t) = (t − sin(t))i + (1 − cos(t))j, 0 ≤ t ≤ 2π.

Answer 1

Integrate the force field along the given path (call it C):

$W=\displaystyle\int_C\mathbf F(x,y)\cdot\mathrm d\mathbf r=\int_0^{2\pi}\mathbf F(x(t),y(t))\cdot\frac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}\bigg((t-\sin t)\,\mathbf i+(6-\cos t)\,\mathbf j\bigg)\cdot\bigg((1-\cos t)\,\mathbf i+\sin t\,\mathbf j\bigg)\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}(t-t\cos t+5\sin t)\,\mathrm dt=\boxed{2\pi^2}$

Answer 2

By direct calculation we will find that the work done is equal to 2π²

The formula to compute the work done is given by:

$W = \int\limits^a_b {F(x(t), y(t))\cdot\frac{dr(t)}{dt} } \, dt$

Here we have:

$r(t) = (t - sin(t))i + (1 - cos(t))j$

This means that:

$x(t) = (t - sin(t))\\y(t) = (1 - cos(t))\\\\\frac{dr(t)}{dt} = (1-cos(t))i + sin(t)j = (1-cos(t), sin(t))$

And we know that 0 ≤ t ≤ 2π, so b = 0 and a = 2π

Replacing that in the work integral we get:

$W = \int\limits^{2\pi}_0 {(t - sin(t), 1 - cos(t) + 5)\cdot(1-cos(t), sin(t))} \, dt \\\\W = \int\limits^{2\pi}_0 {(t - sin(t), 6 - cos(t))\cdot(1-cos(t), sin(t))} \, dt\\\\W = \int\limits^{2\pi}_0 {(-t*cos(t) +t-sin(t)+ cos(t)*sin(t)+ 6*sin(t) - cos(t)*sin(t) )} \, dt\\\\W = \int\limits^{2\pi}_0 {(-cos(t)*t + 5*sin(t) + t)} \, dt$

the sin(t) integral can be removed because it is equal to zero, so we get:

$W = \int\limits^{2\pi}_0 {(-cos(t)*t + t)} \, dtW = [(-t*sin(t) - cos(t)) + \frac{t^2}{2} ]^{2\pi}_0\\\\W = -2\pi*sin(2\pi) - cos(2\pi) + 0*sin(0) + cos(0) + \frac{(2\pi)^2}{2} - \frac{(0)^2}{2}\\\\W = 2\pi^2$

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