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

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Answer 1 · 2017-01-22T14:41:29

The Socratic graph for the cartesian equation
$r^{2} = (x^{2} + y^{2}) = 6 r cos θ = 6 x$ $r^2=(x^2+y^2)=6r cos theta=6x$ is inserted.

This is the equation of the circle through the pole O( r = 0 ), with

center C at ( 3, 0 ). If P is any point ( r. theta ) and OA the diameter

through O.

$O P = r = O A cos θ = 6 cos θ .$ $OP = r = OA cos theta = 6 cos theta.$ .

graph{(x^2+y^2-6x)((x-3)^2+y^2-.005)=0 [0, 14, -3.5, 3.5]}

This circle ia a member of the family of curves

{ $r = k cos (n θ)$ $r = k cos (ntheta)$ }, where k and n scale factors for r and

$θ$ $theta$ .

I let readers to see the grandeur in the graphs of the members

$r = cos (\frac{θ}{3}) and r = cos (\frac{θ}{5})$ $r = cos (theta/3) and r = cos (theta/5)$ .

Graph of $r = cos (\frac{θ}{3})$ $r = cos (theta/3)$ :
graph{x - (x^2+y^2)(4(x^2+y^2)-3)=0[-2 2 -1 1]}

Graph of $r = cos (\frac{θ}{5})$ $r = cos (theta/5)$ :
graph{x - (x^2+y^2)(16(x^2+y^2)^2-20(x^2+y^2)+5)=0[-2 2 -1 1]}

How do you graph r=6cosθr=6costheta?