Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Test - Page 800: 9

Answer

See below:
1571562236

Work Step by Step

To determine the graph, we will check the symmetry across the polar axis, the line $\theta =\frac{\pi }{2}$ and across the pole. To find the symmetry across the polar axis, we will replace $\theta $ with $-\theta $. Then, $\begin{align} & r=1+3\cos \left( -\theta \right) \\ & =1+3\cos \theta \end{align}$ So, it is symmetrical across the polar axis because $ r=1+3\cos \theta $. Now we will check across the line $\theta =\frac{\pi }{2}$ then replace $\left( r,\theta \right)=\left( -r,-\theta \right)$, $\begin{align} & -r=1+3\cos \left( -\theta \right) \\ & -r=1+3\cos \theta \\ & r=-1-3\cos \theta \end{align}$ So, it is not symmetrical across the line $\theta =\frac{\pi }{2}$ because $ r\ne 1+3\cos \theta $. Now we will check the symmetry across the pole by replacing $ r\,\,\text{with}\,\,-r $, $\begin{align} & -r=1+3\cos \theta \\ & =-1-3\cos \theta \end{align}$ So, it is not symmetrical across the pole because $ r\ne 1+3\cos \theta $. So, from the above, we have symmetry across the polar axis but not across the line $\theta =\frac{\pi }{2}$ and not across the pole. Now to plot the graph in the range taken between $\left[ 0,\pi \right]$, use the symmetry across the polar axis. Put $\theta =0{}^\circ $ in $ r=1+3\cos \theta $, Then, $\begin{align} & r=1+3\cos 0{}^\circ \\ & =1+3 \\ & =4 \end{align}$ So at $\theta =0$, $ r=4$. Similarly, put the values between $0\,\,\text{to}\,\,\pi $ and find the values of r seen below, $\theta $ 0 $\frac{\pi }{6}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\frac{5\pi }{6}$ $\pi $ $ r $ 4 3.6 2.5 1 -0.5 -1.6 -2
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