## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.3 Polar Coordinates - Exercises - Page 618: 25

#### Answer

Point $A$: ${\ \ \ }$ $\theta=0$, $\theta=\pi$ Point $B$: ${\ \ \ }$ $\theta = \frac{\pi }{4}$, $\theta = \frac{{5\pi }}{4}$ Point $C$: ${\ \ \ }$ $\theta = \frac{\pi }{2}$, $\theta = \frac{{3\pi }}{2}$ Point $D$: ${\ \ \ }$ $\theta = \frac{{3\pi }}{4}$, $\theta = \frac{{7\pi }}{4}$ (a) $0 \le \theta \le \frac{\pi }{2}$ The curve traces out the upper half-circle. (b) $\frac{\pi }{2} \le \theta \le \pi$ The curve traces out the lower half-circle. (c) $\pi \le \theta \le \frac{{3\pi }}{2}$ The curve traces out the upper half-circle.

#### Work Step by Step

The points $A$, $B$, $C$, $D$ in Figure 19 have the following rectangular coordinates: $A = \left( {4,0} \right)$, ${\ }$ $B = \left( {2,2} \right)$, ${\ }$ $C = \left( {0,0} \right)$, ${\ }$ $D = \left( {2, - 2} \right)$. Consider the interval $0 \le \theta \le 2\pi$. The coordinates of a point on the circle is given by $\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right) = \left( {4{{\cos }^2}\theta ,4\cos \theta \sin \theta } \right)$. For $A = \left( {4,0} \right)$, we have $\left( {4,0} \right) = \left( {4{{\cos }^2}\theta ,4\cos \theta \sin \theta } \right)$. So, ${\cos ^2}\theta = 1$ ${\ \ \ }$ and ${\ \ \ }$ $\cos \theta \sin \theta = 0$. The solutions are $\theta=0$, $\theta=\pi$. For $B = \left( {2,2} \right)$, we have $\left( {2,2} \right) = \left( {4{{\cos }^2}\theta ,4\cos \theta \sin \theta } \right)$. So, ${\cos ^2}\theta = \frac{1}{2}$ ${\ \ \ }$ and ${\ \ \ }$ $\cos \theta \sin \theta = \frac{1}{2}$. The solutions are $\theta = \frac{\pi }{4}$, $\theta = \frac{{5\pi }}{4}$. For $C = \left( {0,0} \right)$, we have $\left( {0,0} \right) = \left( {4{{\cos }^2}\theta ,4\cos \theta \sin \theta } \right)$. So, ${\cos ^2}\theta = 0$ ${\ \ \ }$ and ${\ \ \ }$ $\cos \theta \sin \theta = 0$. The solutions are $\theta = \frac{\pi }{2}$, $\theta = \frac{{3\pi }}{2}$. For $D = \left( {2, - 2} \right)$, we have $\left( {2, - 2} \right) = \left( {4{{\cos }^2}\theta ,4\cos \theta \sin \theta } \right)$. So, ${\cos ^2}\theta = \frac{1}{2}$ ${\ \ \ }$ and ${\ \ \ }$ $\cos \theta \sin \theta = - \frac{1}{2}$. The solutions are $\theta = \frac{{3\pi }}{4}$, $\theta = \frac{{7\pi }}{4}$. (a) $0 \le \theta \le \frac{\pi }{2}$ For this interval, the curve passes through the points $A$ (at $\theta=0$), B (at $\theta = \frac{\pi }{4}$) and C (at $\theta = \frac{\pi }{2}$), so it traces out the upper half-circle. (b) $\frac{\pi }{2} \le \theta \le \pi$ For this interval, the curve passes through the points $C$ (at $\theta = \frac{\pi }{2}$), $D$ (at $\theta = \frac{{3\pi }}{4}$) and $A$ (at $\theta = \pi$), so it traces out the lower half-circle. (c) $\pi \le \theta \le \frac{{3\pi }}{2}$ For this interval, the curve is back again to the top as it traces out the upper half-circle passing through the points $A$ (at $\theta = \pi$), $B$ (at $\theta = \frac{{5\pi }}{4}$) and $C$ (at $\theta = \frac{{3\pi }}{2}$).

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