How do you graph the polar equation #r=1.5theta#?

Polar equations vary #r# as #theta# cycles counterclockwise through from #0# to #2pi#, #2pi - 4pi#, etc.

#r = 1.5theta#

This equation has #r# increase by #1.5# for every #1# radian that we pass through (#(180/pi)^@#). If #r = theta#, we would have gotten a spiral.

Here, with #r = 1.5theta# in #[0,8pi]#, we just have a bigger spiral:

For example, if you look at #theta = pi/2#, you should see

#r = 1.5 xx pi/2 ~~ ul2.35# on the vertical axis.

If we pass through #pi/2# radians, we move to

#r = 1.5 xx (pi/2 + pi/2) => ul(-4.71)# on the horizontal axis

If we pass through #(3pi)/2# radians, we move to

#r = 1.5 xx (pi/2 + pi) => ul(-7.07)# on the vertical axis.

If we pass through #2pi# radians, we move up to

#r = 1.5 xx (pi/2 + (3pi)/2) ~~ ul9.42# on the horizontal axis.

Then, if we pass through #(5pi)/2# radians, we move up to

#r = 1.5 xx (pi/2 + 2pi) ~~ ul11.78# on the vertical axis.

Once you have those major points, connect them in a spiral.