Chapter 9, Problem 9CT

To determine

The symmetry of the polar equation $r=5\sin \theta {\cos }^2\theta$ with respect to pole, the polar axis, and the line $\theta =\frac{\pi }{2}$ .

Answer to Problem 9CT

Solution:

The polar equation $r=5\sin \theta {\cos }^2\theta$may or may not be symmetric with respect to pole and polar axis, and it is symmetric with respect to line$\theta =\frac{\pi }{2}$ .

Explanation of Solution

Given information:

The polar equation is $r=5\sin \theta {\cos }^2\theta$ .

Use the test of symmetry with respect to polar axis, pole and line $\theta =\frac{\pi }{2}$ .

Polar axis: Replace $\theta$ with $-\theta$ .

  $r=5\sin \left(-\theta \right){\cos }^2\left(-\theta \right)$ .

Using $\sin \left(-\theta \right)=-\sin \theta$ and $\cos \left(-\theta \right)=\cos \theta$ ,

  $\Rightarrow r=-5\sin \theta {\cos }^2\theta$

The resulting equation is not same as original, so the graph may or may not be symmetric with respect to polar axis.

The line $\theta =\frac{\pi }{2}$ : Replace $\theta$ with $\pi -\theta$ .

  $r=5\sin \left(\pi -\theta \right){\cos }^2\left(\pi -\theta \right)$

  $\Rightarrow r=5\left(\sin \pi \cos \theta -\cos \pi \sin \theta \right){\left(\cos \pi \cos \theta +\sin \pi \sin \theta \right)}^2$

  $\Rightarrow r=5\left(\left(0\right)\cos \theta -\left(-1\right)\sin \theta \right){\left(\left(-1\right)\cos \theta +\left(0\right)\sin \theta \right)}^2$

  $r=5\sin \theta {\cos }^2\theta$

The resulting equation is same as original, so the graph is symmetric with respect to line $\theta =\frac{\pi }{2}$ .

Pole: Replace $r$ with $-r$ ,

  $-r=5\sin \theta {\cos }^2\theta$

  $r=-5\sin \theta {\cos }^2\theta$ .

The resulting equation is not same as original, so the graph may or may not be symmetric with respect to respect to pole.

Therefore, the polar equation $r=5\sin \theta {\cos }^2\theta$ may or may not be symmetric with respect to pole and polar axis, and it is symmetric with respect to line $\theta =\frac{\pi }{2}$ .