How do you graph #r=1+4sintheta#?

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Answer 1 · 2016-10-21T06:51:25

I strictly adhere to length #r = sqrt(x^2+y^2)>=0#

So, despite that r is periodic with period #2pi#, I have to exclude

#theta in (pi-sin^(-1)(-1/4), 2pi+sin ^(-1) (-1/4))# for which #r <0#
.
Use this Table for making the graph in one period that excludes the

above subinterval.

#(r, theta)#:

#(1, 0) (3, pi/3) (5, pi/2) (3, 2/3pi) (1, pi) (0, pi-sin^(-1)(-1/4)#

After discontinuity with respect to #theta#,

#(0, 2pi+sin^(-1)(-1/4)) (1, 2pi)#

The graph looks like the cardioid #r=1+sin theta# that is elongated

in the #theta= pi/2# direction.