How do you graph #r=4sintheta-2#?

Question

2017 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-12T09:59:57

See explanation.

#r=4 sin theta - 2 >=0 to sin theta >=1/2 to theta in (pi/6, 5/6pi)#

A Table for making one half of the graph:

#(r, theta): (0, pi/5) (2(sqrt2-1), pi/4) ((2(sqrt3-1), pi/3) (2, pi/2)#

The graph is symmetrical about the vertical #theta=pi/2#.

I have inserted graph for the cartesian double

#4y -(x^2+y^2)=+-2sqrt(x^2+y^2)# that is supposed to be the

equivalent.

As #r=sqrt(x^2+y^2)>=0#, my graph chooses + sign only.

For the given polar equation, the graph would be the inner

loop only. Please feel such nuances while making graphs, when you

make conversions..

graph{x^4+y^4-8y^3+2x^2y^2-8x^2y+12y^2-4x^2=0 [-10, 10, -5, 5]}