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Chapter 6: Problem 41

Test for symmetry and then graph each polar equation. $$r=\sin \theta \cos ^{2} \theta$$

Short Answer

Expert verified
The polar equation \(r = \sin (\theta)\cos^2 (\theta)\) does not show symmetry about the origin. The graph of this equation offers a visualization of how \(\sin(\theta)\cos^2 (\theta)\) changes in a polar coordinate system over a range of 0 to \(2\pi\).

Step by step solution

01

Test Symmetry about the Origin

Replace \(r\) with \(-r\) and \(\theta\) with \(-\theta\) in the equation of \(r = \sin(\theta)\cos^2(\theta)\). If the equation remains the same, then it has symmetry about the origin. \(-r = \sin(-\theta)\cos^2(-\theta)\) simplifies to \(-r = -\sin(\theta)\cos^2(\theta)\). Thus, the equation is not symmetric about the origin because the original equation is not obtained.
02

Plot the Graph

After doing the symmetry test, the next step is to plot the polar equation. Since \(\theta\) varies from 0 to \(2\pi\), plot values within this range for \(r\) and \(\theta\) and connect these points to form a smooth curve. You may also use a graphing calculator. To acquire this plot, one must understand the effects of the trigonometric functions \(\sin(\theta)\) and \(\cos^2(\theta)\) on the value of \(r\).
03

Interpreting the Graph

This graph illustrates the variations in \(r\) as per the value of \(\theta\). Note that while it does not show symmetry about the origin, the graph gives a visualization of \(\sin(\theta)\cos^2(\theta)\)'s effect on a polar coordinate system.

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