How do you graph $r = 4 cos θ + 2$ $r=4costheta+2$ ?

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Answer 1 · 2016-12-20T12:59:41

Graph is inserted.

The period for the graph is $2 π$ $2pi$

The range for r is [0, 6]>

As $r \geq 0, cos θ \geq - \frac{1}{2}$ $r >= 0, cos theta >=-1/2$

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

$θ \in [- \frac{2}{3} π, \frac{2}{3} π] .$ $theta in [-2/3pi, 2/3pi].$ .

For $θ \in (- \frac{2}{3} π, \frac{2}{3} π), r < 0$ $theta in (-2/3pi, 2/3pi), r < 0$ .

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

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

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

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

the initial line $θ = 0$ $theta = 0$ .

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

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

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

about the pole, through $\frac{π}{2}$ $pi/2$ , 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]}

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