How do you graph #r=5-2sintheta#?

Question

4147 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-03T15:24:29

See explanation.

#r(theta)=5-2 sin theta# is periodic in #theta#, with period #2pi#.

This graph is oscillatory about the circle #r = 5#, with relative

amplitude 2 and period #2pi#. A short Table for one period #[0, 2pi] #

gives the whole graph that is simply repeat of this graph for one

period.

#(r, theta)#:

#(5, 0) (4, pi/6) (5-sqrt 3, pi/3) (3, pi/2)#

In #[pi/2, pi]#, use symmetry about #theta=pi/2#.

Then, in #[pi. 2pi]#, use #sin (pi+theta)=-sin theta#.

The whole curve is laid between the circles r = 3 and r = 7, touching

them at #theta=pi/2 and 3/2pi#, respectively.