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

Question

4548 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.

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