How do you graph #r=2+sin theta#?

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Answer 1 · 2018-07-28T01:09:34

See the dimpled upright limacon.

#0 <= r = 2 + sin theta in [2-1, 2 + 1 ] = [ 1, 3 ]#

The period = period of #sin theta = 2 pi#.

Using

#r = sqrt ( x^2 + y^2 ) and ( x, y )= r ( cos theta, sin theta )#,

the Cartesian equation is obtained as

#x^2 + y^2 = 2 sqrt ( x^2 + y^2 ) + y# and the dimpled graph of

this limacon id immediate.

graph{x^2 + y^2 - 2 sqrt ( x^2 + y^2 ) - y = 0}

Note that the general equation of limacons is

# r = a + b cos ( theta - alpha )#, where #alpha# is for anticlockwise

rotation, about the pole #r = 0#.

Here, a = 2, b = 1 and #alpha = pi/2 #.