Chapter H.1, Problem 63E

To determine

To graph:

Relation of $r=f\left(\theta -\alpha \right)$ to the graph $r=f\left(\theta \right)$ .

Explanation of Solution

Given information:

  $r=1+\sin \left(\theta -\pi /6\right)$

  $r=1+\sin \left(\theta -\pi /3\right)$

Graph:

The graphs of $r=1+\sin \left(\theta \right),r=1+\sin \left(\theta +\frac{\pi }{6}\right)$ and $r=1+\sin \left(\theta -\frac{\pi }{3}\right)$ are plotted as follows using maple,

  

The graphs of $r=1+\sin \left(\theta +\frac{\pi }{6}\right)$ and $r=1+\sin \left(\theta -\frac{\pi }{6}\right)$ is get from $r=1+\sin \left(\theta \right)$ by rotating counter clock wise the graph of $r=1+\sin \left(\theta \right)$ through $\frac{\pi }{6}$ and $\frac{\pi }{3}$ radians respectively.

About the origin, in general graph of $r=f\left(\theta -\alpha \right)$ is get from $r=f\left(\theta \right)$ rotating the graph $r=f\left(\theta \right)$ counterclockwise through $\alpha$ radians about the origin.