How do you graph $r = 4 + 6 cos θ$ $r=4+6costheta$ ?

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Answer 1 · 2017-02-19T16:18:19

See Socratic graph for the Cartesian form of the equation.

This limacon passes through the pole.

The pole is a dimple.

As $r = f (cos θ) = f (cos (- θ)$ $r = f(costheta)=f(cos(-theta)$ , the graph is symmetrical about

$θ = 0$ $theta = 0$ .

Maximum $r = f (0) = f (2 π) = 10$ $r = f(0)=f(2pi)=10$ .

I have used the Cartesian form of the polar equation.

$x^{2} + y^{2} - 4 \sqrt{x^{2} + y^{2}} - 6 x = 0$ $x^2+y^2-4sqrt(x^2+y^2)-6x=0$ .

graph{x^2+y^2-4sqrt(x^2+y^2)-6x=0 [-20, 20, -10, 10]}

How do you graph r=4+6cosθr=4+6costheta?