How do you graph $r = 2 cos 2 θ$ $r=2 cos 2theta$ ?

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Answer 1 · 2018-07-15T15:59:03

Here, the non-negative $r \in [0.2]$ $r in [0. 2]$ .

The period of $cos 2 θ$ $cos 2theta$ is $\frac{2 π}{2} = π$ $(2pi)/2=pi$ .

In one period $θ \in [- \frac{π}{2}, \frac{π}{2}}$ $theta in [ -pi/2, pi/2 }$ ,

$r \geq 0$ $r >= 0$ for $θ \in [- \frac{π}{4}, \frac{π}{4}]$ $theta in [ - pi/4, pi/4 ]$ only.

Using,

$(x, y) = r (cos θ, sin θ)$ $( x, y ) = r ( cos theta, sin theta )$ and $r = \sqrt{x^{2} + y^{2}} \geq 0$ $r =sqrt (x^2 + y^2 ) >=0$ ,

the Cartesian form of

$r = 2 cos 2 θ = 2 ({cos}^{2} θ - {sin}^{2} θ)$ $r = 2 cos 2theta = 2 ( cos^2theta - sin^2theta )$ is

${(x^{2} + y^{2})}^{1.5} = 2 (x^{2} - y^{2})$ $( x^2 + y^2 )^1.5 = 2 (x^2 - y^2 )$ .

The 2-loop Socratic graph is immediate.
graph{(x^2+y^2)^1.5-2(x^2-y^2)=0[-6 6 -3 3]}

Invalid ( r < 0 ) loops are shown below.
graph{(x^2+y^2)^1.5+2(x^2-y^2)=0[-6 6. -3 3]}

Valid and invalid combined graph:
graph{(x^2+y^2)^3-4(x^2-y^2)^2=0[-6 6. -3 3]}

How do you graph r=2 cos 2thetar=2cos2θr=2 cos 2theta?