# How do you graph r=4costheta+2?

Dec 20, 2016

Graph is inserted.

#### Explanation:

The period for the graph is $2 \pi$

The range for r is [0, 6]>

As$r \ge 0 , \cos \theta \ge - \frac{1}{2}$

So, the loop of the limacon ( with dimple at the pole ) is drawn for

$\theta \in \left[- \frac{2}{3} \pi , \frac{2}{3} \pi\right] .$.

For $\theta \in \left(- \frac{2}{3} \pi , \frac{2}{3} \pi\right) , r < 0$.

$r = \sqrt{{x}^{2} + {y}^{2}} \ge 0 \mathmr{and} \cos \theta = \frac{x}{r}$.

So, the the cartesian form for $r = 4 \cos \theta + 2$ is

${x}^{2} + {y}^{2} - 2 \sqrt{{x}^{2} + {y}^{2}} - 4 x = 0$. And the graph is inserted.

$r = f \left(\cos \theta\right) = f \left(\cos \left(- \theta\right)\right)$. So, the graph is symmetrical about

the initial line $\theta = 0$.

graph{x^2+y^2-2sqrt(x^2+y^2)-4x=0 [-10, 10, -5, 5]}

The combined graph of four limacons $r = 2 \pm 4 \cos \theta$ and

$r = 2 \pm 4 \sin \theta$ follows. To get this, rotate the given one

about the pole, through $\frac{\pi}{2}$, three times in succession.

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