Question

Find the work done by the force field \({\rm{F}}\left( {{\rm{x,y}}} \right){\rm{ = xi + }}\left( {{\rm{y + 2}}} \right){\rm{j}}\) in moving an object along an arch of the cycloid

\({\rm{r}}\left( {\rm{t}} \right){\rm{ = }}\left( {{\rm{t - sint}}} \right){\rm{i + }}\left( {{\rm{1 - cost}}} \right){\rm{j}}\),

\({\rm{0}} \le {\rm{t}} \le {\rm{2\pi }}\)

Short answer

The work done by the force field is \({\rm{2}}{{\rm{\pi }}^{\rm{2}}}\).

Step-by-step solution

To find the work done by the force field.

\({{\rm{r}}^\prime }{\rm{(t) = (1 - cost)i + sintj}}\)

Plug in all the values and solve the line inte\({\rm{F(x,y) = xi + (y + 2)j}}\)

In moving an object along an arch of the cycloid which is parameterized by

\({\rm{r(t) = (t - sint)i + (1 - cost)j,}}\;\;\;{\rm{0}} \le {\rm{t}} \le {\rm{2\pi }}\)

Solve by using the formula

\({\rm{F}} \cdot {\rm{dr = }}_{\rm{a}}^{\rm{b}}{\rm{F(r(t))}} \cdot {{\rm{r}}^\prime }{\rm{(t)dt}}\)

\({\rm{a = 0,b = 2\pi ,x(t) = t - sint,y(t) = 1 - cost}}\)

Differentiate \({\rm{x}}\)and \({\rm{y}}\) in terms of \({\rm{t}}\)

\(\begin{aligned}&{{\rm{x}}^\prime }{\rm{(t) = (1 - cost)dt}}\\&{{\rm{y}}^\prime }(t) = \sin tdt\end{aligned}\)

Therefore,

\({{\text{r}}^{\prime }}\text{(t)=(1-cost)i+sintj}\)

Plug in all the values and solve the line integral to get the work done

\(\begin{aligned}{\rm{F}} \cdot {\rm{dr = }}&_{\rm{a}}^{\rm{b}}{\rm{F(r(t))}} \cdot {{\rm{r}}^\prime }{\rm{(t)dt}}\\&{\rm{ = }}_{\rm{0}}^{{\rm{2\pi }}}{\rm{((t - sint)i + (1 - cost + 2)j)}} \cdot {\rm{((1 - cost)i + sintj)}}\\&{\rm{ = }}\int_{{\rm{2\pi }}} {{\rm{(t - sint)}}} {\rm{(1 - cost) + (3 - cost)sintdt}}\\&{\rm{ = }}\int_{{\rm{2\pi }}} {\rm{t}} {\rm{ - tcost - sint + sintcost + 3sint - sintcostdt}}\end{aligned}\)

Evaluate the integrals\({{\rm{I}}_{\rm{1}}}\),\({{\rm{I}}_{\rm{2}}}\),\({{\rm{I}}_{\rm{3}}}\).

Solve \({{\rm{I}}_{\rm{1}}}\) using power rule

\(\begin{array}{c}{{\rm{x}}^{\rm{n}}}{\rm{ = }}\frac{{{{\rm{x}}^{{\rm{n + 1}}}}}}{{{\rm{n + 1}}}}\\ \Rightarrow \frac{{{{\rm{t}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ + C}}\end{array}\)

\({{\rm{I}}_{\rm{3}}}\) is a table integral;

\( \Rightarrow {\rm{ - cost + C}}\)

Solve\({{\rm{I}}_{\rm{2}}}\)using integration by parts;

\(\begin{align} & \text{u=t}\ \ \ \Rightarrow \ \ \ \text{du=dt} \\ & \text{dv=costdt}\ \ \ \Rightarrow \ \ \ \text{v=sint} \\ & \Rightarrow \text{tsint-}\ \ \ \text{sintdt} \\ & \text{=tsint-(-cost)+C} \\ & \text{=tsint+cost+C} \end{align}\)

Plug in solved integrals and evaluate;

\(\begin{align} & \left. \Rightarrow \frac{{{\text{t}}^{\text{2}}}}{\text{2}} \right|_{\text{0}}^{\text{2 }\!\!\pi\!\!\text{ }}\text{-tsint+}\left. \text{cost} \right|_{\text{0}}^{\text{2 }\!\!\pi\!\!\text{ }}\text{-}\left. \text{2cost} \right|_{\text{0}}^{\text{2 }\!\!\pi\!\!\text{ }} \\ & \text{=2}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}\text{-0-(2 }\!\!\pi\!\!\text{ sin2 }\!\!\pi\!\!\text{ +cos2 }\!\!\pi\!\!\text{ -0-cos0)-2cos2 }\!\!\pi\!\!\text{ +2cos0} \\ & \text{=2}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}\text{-(1-1)-2+2} \\ & \text{=2}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}} \end{align}\)

Therefore, the work done is \({\rm{2}}{{\rm{\pi }}^{\rm{2}}}\).