How do you graph $r = 3 + 2 cos θ$ $r=3+2costheta$ ?

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Answer 1 · 2018-07-23T04:34:19

See the graph of this dimpled limacon and details.

This limacon $r = 3 + 2 cos θ \in [1, 5]$ $r = 3 + 2 cos theta in [ 1, 5 ]$ .

The pole r = 0 is outside.

As r is f ( cos theta ) = f ( cos (- theta ))#, the graph is symmetrical

about the initial line $θ = 0$ $theta = 0$ .

Use $r = \sqrt{x^{2} + y^{2}} \geq 0$ $r = sqrt ( x^2 + y^2 ) >= 0$

and $r (cos θ . sin θ) = (x, y)$ $r (cos theta. sin theta ) = ( x, y )$ , for the Cartesian form

$(x^{2} + y^{2}) - 3 \sqrt{x^{2} + y^{2}} - 2 x = 0$ $( x^2 + y^2 ) - 3 sqrt ( x^2 + y^2 ) - 2 x =0$

The Socratic graph of this dimpled limacon is immediate.
graph{ ( x^2 + y^2 ) - 3 sqrt ( x^2 + y^2 ) - 2 x =0}

How do you graph r=3+2costhetar=3+2cosθr=3+2costheta?