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

1 Answer
A. S. Adikesavan
Oct 21, 2016

I strictly adhere to length r = sqrt(x^2+y^2)>=0r=√x2+y2≥0

So, despite that r is periodic with period 2pi2π, I have to exclude

theta in (pi-sin^(-1)(-1/4), 2pi+sin ^(-1) (-1/4))θ∈(π−sin−1(−14),2π+sin−1(−14)) for which r <0r<0
.
Use this Table for making the graph in one period that excludes the

above subinterval.

(r, theta)(r,θ):

(1, 0) (3, pi/3) (5, pi/2) (3, 2/3pi) (1, pi) (0, pi-sin^(-1)(-1/4)(1,0)(3,π3)(5,π2)(3,23π)(1,π)(0,π−sin−1(−14)

After discontinuity with respect to thetaθ,

(0, 2pi+sin^(-1)(-1/4)) (1, 2pi)(0,2π+sin−1(−14))(1,2π)

The graph looks like the cardioid r=1+sin thetar=1+sinθ that is elongated

in the theta= pi/2θ=π2 direction.

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