## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.2 Arc Length and Speed - Exercises - Page 611: 11

#### Answer

The length of one arch of a cycloid generated by a circle of radius $R$ is $8R$.

#### Work Step by Step

The parametrization of a cycloid generated by a circle of radius $R$ is given by (Exercise 97 of Section 12.1) $c\left( t \right) = \left( {Rt - R\sin t,R - R\cos t} \right)$. So, we have $x\left( t \right) = Rt - R\sin t$, ${\ \ \ }$ $x'\left( t \right) = R - R\cos t$, $y\left( t \right) = R - R\cos t$, ${\ \ \ }$ $y'\left( t \right) = R\sin t$. The $x$-interval for one arch of the cycloid is $0 \le x \le 2\pi R$ as is shown in Figure 25 of Exercise 97 of Section 12.1. The corresponding $t$-interval is $0 \le t \le 2\pi$. Using Eq. (3) of Theorem 1, the arc length is $s = \mathop \smallint \limits_0^{2\pi } \sqrt {{{\left( {R - R\cos t} \right)}^2} + {{\left( {R\sin t} \right)}^2}} {\rm{d}}t = \mathop \smallint \limits_0^{2\pi } \sqrt {2{R^2}\left( {1 - \cos t} \right)} {\rm{d}}t$ $s = \sqrt 2 R\mathop \smallint \limits_0^{2\pi } \sqrt {1 - \cos t} {\rm{d}}t$ Since $2{\sin ^2}\frac{t}{2} = 1 - \cos t$, we get $s = 2R\mathop \smallint \limits_0^{2\pi } \sin \frac{t}{2}{\rm{d}}t = - 4R\cos \frac{t}{2}|_0^{2\pi } = 8R$.

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