Chapter 11.1, Problem 35E

(a)

To determine

To calculate: The length of the semicircle with the different parameterizations.

Answer to Problem 35E

The required length of the semicircle is $\pi$ .

Explanation of Solution

Given information:

The parametric equations: $x=\cos 2t$ , $y=\sin 2t$ , $0\le t\le \frac{\pi }{2}$

Calculation:

The parametric equations are $x=\cos 2t$ , $y=\sin 2t$ .

Differentiate the above equations with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left[\cos 2t\right]\\ =-2\sin 2t\end{array}$

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left[2\cos t\right]\\ =2\cos 2t\end{array}$

Consider that L be the length of the parametric curve.

So, $L={\displaystyle \underset{t_2}{\overset{t_2}{\int }}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}}dt$

Write the length of the path traveled by the projectile as:

  $\begin{array}{c}L={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{{\left(-2\sin 2t\right)}^2+\left(2\cos 2t\right)}dt}\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}2dt}\\ ={2t|}_0^{\frac{\pi }{2}}\\ =2\left(\frac{\pi }{2}\right)\\ =\pi \end{array}$

Hence, the required length of the semicircle is $\pi$ .

(b)

To determine

To calculate: The length of the semicircle with the different parameterizations.

Answer to Problem 35E

The required length of the semicircle is $\pi$ .

Explanation of Solution

Given information:

The parametric equations: $x=\sin \pi t$ , $y=\cos \pi t$ , $-\frac{1}{2}\le t\le \frac{1}{2}$

Calculation:

The parametric equations are $x=\sin \pi t$ , $y=\cos \pi t$ .

Differentiate the above equations with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left[\sin \pi t\right]\\ =\pi \cos \pi t\end{array}$

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left[2\cos t\right]\\ =-\pi \sin \pi t\end{array}$

Consider that L be the length of the parametric curve.

So, $L={\displaystyle \underset{t_2}{\overset{t_2}{\int }}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}}dt$

Write the length of the path traveled by the projectile as:

  $\begin{array}{c}L={\displaystyle \underset{-\frac{1}{2}}{\overset{\frac{1}{2}}{\int }}\sqrt{{\left(\pi \cos \pi t\right)}^2+{\left(-\pi \sin \pi t\right)}^2}dt}\\ ={\displaystyle \underset{-\frac{1}{2}}{\overset{\frac{1}{2}}{\int }}\pi dt}\\ =\pi \left[\frac{1}{2}-\left(-\frac{1}{2}\right)\right]\\ =\pi \end{array}$

Hence, the required length of the semicircle is $\pi$ .