Question

How are the graphs of \(r = 1 + \sin \left( {\theta - \frac{\pi }{6}} \right)\) and \(r = 1 + \sin \left( {\theta - \frac{\pi }{3}} \right)\) related to the graph of \(r = 1 + \sin \theta \) ? In general, how is the graph of \(r = f\left( {\theta - \alpha } \right)\) related to the graph of \(r = f\left( \theta \right)\) ?

Short answer

Both equation \(r = 1 + \sin \left( {\theta - \frac{\pi }{6}} \right)\) and \(r = 1 + \sin \theta \) has the same shape. The only difference is the first equation has been rotated counterclockwise by an amount of \(\frac{\pi }{6}\) . Similarly, the equation \(r = 1 + \sin \left( {\theta - \frac{\pi }{3}} \right)\) and \(r = 1 + \sin \theta \) has the same shape and has been rotated counterclockwise by an amount of \(\frac{\pi }{3}\) . In general if we have \(r = f\left( {\theta - \alpha } \right)\) then we have a rotation counterclockwise of \(\alpha \) from the original graph.

Step-by-step solution

Step 1:

Consider the polar equations,

\(\begin{aligned}{l}r &= 1 + \sin \left( {\theta - \frac{\pi }{6}} \right)\\r &= 1 + \sin \left( {\theta - \frac{\pi }{3}} \right)\\r &= 1 + \sin \theta \end{aligned}\)

Both equation \(r = 1 + \sin \left( {\theta - \frac{\pi }{6}} \right)\) and \(r = 1 + \sin \theta \) has the same shape. The only difference is the first equation has been rotated counterclockwise by an amount of \(\frac{\pi }{6}\) . Similarly, the equation \(r = 1 + \sin \left( {\theta - \frac{\pi }{3}} \right)\) and \(r = 1 + \sin \theta \) has the same shape and has been rotated counterclockwise by an amount of \(\frac{\pi }{3}\) . In general if we have \(r = f\left( {\theta - \alpha } \right)\) then we have a rotation counterclockwise of \(\alpha \) from the original graph.

Step 2:

Sketch the curves with polar equations.