How do you graph #r=4costheta+2#?

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Answer 1 · 2016-12-20T12:59:41

Graph is inserted.

The period for the graph is #2pi#

The range for r is [0, 6]>

As# r >= 0, cos theta >=-1/2#

So, the loop of the limacon ( with dimple at the pole ) is drawn for

#theta in [-2/3pi, 2/3pi].#.

For #theta in (-2/3pi, 2/3pi), r < 0#.

#r = sqrt(x^2+y^2)>=0 and cos theta = x/r#.

So, the the cartesian form for #r = 4 cos theta + 2# is

#x^2+y^2-2sqrt(x^2+y^2)-4x=0#. And the graph is inserted.

#r= f(cos theta)=f(cos(-theta))#. So, the graph is symmetrical about

the initial line #theta = 0#.

graph{x^2+y^2-2sqrt(x^2+y^2)-4x=0 [-10, 10, -5, 5]}

The combined graph of four limacons #r = 2+- 4 cos theta # and

#r = 2+- 4 sin theta # follows. To get this, rotate the given one

about the pole, through #pi/2#, three times in succession.

graph{(x^2+y^2-2sqrt(x^2+y^2)-4x)(x^2+y^2-2sqrt(x^2+y^2)+4x)(x^2+y^2-2sqrt(x^2+y^2)-4y)(x^2+y^2-2sqrt(x^2+y^2)+4y)=0 [-20, 20, -10, 10]}