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

Question

3744 views around the world

You can reuse this answer
Creative Commons License

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.

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