How do you graph $r = 8 cos 2 θ$ $r=8cos2theta$ on a graphing utility?

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Answer 1 · 2016-12-10T16:14:14

Use the Cartesian form.

The period for the graph is $\frac{2 π}{2} = π$ $(2pi)/2=pi$ .

In one half period $θ \in [- \frac{π}{4}, \frac{π}{4}], r \geq 0; r < 0$ $theta in [-pi/4, pi/4], r>=0; r<0$ . for the other half

$θ \in (\frac{π}{4}, \frac{π}{2}]$ $theta in (pi/4, pi/2]$ . In the double period #theta in [0, 2pi], two

loops are created.

The cartesian form of the given equation is

$(x^{2} + y^{2}) \sqrt{x^{2} + y^{2}} = 9 (x^{2} - y^{2})$ $(x^2+y^2)sqrt(x^2+y^2)=9(x^2-y^2)$ that befits the graphic utility

that is readily available here.

graph{(x^2+y^2)^1.5-8(x^2-y^2)=0 [-10, 10, -5, 5]}

How do you graph r=8cos2thetar=8cos2θr=8cos2theta on a graphing utility?