Question

Find the arc length of the curves defined by the vector-valued functions on the specified intervals in Exercises 22–27.

$\mathbf{r}\left(t\right)=⟨t-\sin t,1-\cos t⟩,[0,2\pi ]$

Short answer

The arc length of curve$r\left(t\right)=⟨t-\sin t,1-\cos t⟩\text{on}[0,2\pi ]\text{is}8.$

Step-by-step solution

Step 1. Given information.

The given curve is$r\left(t\right)=⟨t-\sin t,1-\cos t⟩\text{on}[0,2\pi .$

Step 2. Arc length.

The arc-length is given by,

$$\begin{gathered} l(a,b)={\int }_a^b||r^{\text{'}}\left(t\right)||dt \\ ||r^{\text{'}}\left(t\right)||=\sqrt{(1-\cos t{)}^2+(\sin t{)}^2} \\ =\sqrt{1+{\cos }^{2}t-2\cos t+{\sin }^{2}t} \\ =\sqrt{2}\sqrt{1-\cos t} \\ =\sqrt{2}\sqrt{2{\sin }^2\frac{t}{2}} \\ =2\sin \frac{t}{2} \\ \left(l\right)={\int }_a^b||r^{\text{'}}\left(t\right)||dt \\ ={\int }_0^{2\pi }2\sin \frac{t}{2}dt \\ {\left.=\frac{-2\cos \frac{t}{2}}{\frac{1}{2}}\right]}_0^{2\pi } \\ =-4\left[\cos \frac{2\pi }{2}-\cos 0\right] \\ =-4(-1-1) \\ =8 \end{gathered}$$