Question

In Problems 8 and 9, test the polar equation for symmetry with respect to the pole, the polar axis, and the line $\theta =\frac{\pi }{2}$.

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

Short answer

The polar equation is symmetric about the line $\theta =\frac{\pi }{2}$. The tests for symmetry about the pole and the polar axis fail, so the graph of polar equation may or may not be symmetric about the pole or polar axis.

Step-by-step solution

Step 1. Given information  

We are given a polar equation,

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

We have to check the polar equation for symmetry with respect to the pole, the polar axis, and the line $\theta =\frac{\pi }{2}$.

Step 2. Checking the symmetry about polar axis 

We have to replace $\theta$with $-\theta$to check the symmetry,

$$\begin{gathered} r=5\sin (-\theta ){\cos }^2(-\theta ) \\ r=-5\sin \theta {\cos }^2\theta \end{gathered}$$

Since, the substitution does not yield an equivalent equation, the polar equation may or may not be symmetric about the polar axis.

Step 3. Checking the symmetry about pole 

We have to replace r with $-r$to check the symmetry,

$$\begin{gathered} -r=5\sin \theta {\cos }^2\theta \\ r=-5\sin \theta {\cos }^2\theta \end{gathered}$$

Since, the substitution does not yield an equivalent equation, the polar equation may or may not be symmetric about the pole.

Step 4. Checking the symmetry about the line θ=π2

The graph of a polar equation is symmetric with respect to the line $\theta =\frac{\pi }{2}$if the substitution $\theta =(\pi -\theta )$yields an equivalent equation,

role="math" $$\begin{gathered} r=5\sin (\pi -\theta ){\cos }^2(\pi -\theta ) \\ =5\sin \theta {\cos }^2(-\theta ) \\ =5\sin \theta {\cos }^2\theta \end{gathered}$$

Hence, the substitution yields an equivalent equation, the polar equation is symmetric with respect to the line $\theta =\frac{\pi }{2}$.