Question

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

Answer 1

Graph is inserted.

Explanation:

The period for the graph is `2pi`

The range for r is [0, 6]>

As`r >= 0, cos theta >=-1/2`

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

`theta in [-2/3pi, 2/3pi].`.

For `theta in (-2/3pi, 2/3pi), r < 0`.

`r = sqrt(x^2+y^2)>=0 and cos theta = x/r`.

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

`x^2+y^2-2sqrt(x^2+y^2)-4x=0`. And the graph is inserted.

`r= f(cos theta)=f(cos(-theta))`. 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+- 4 cos theta` and

`r = 2+- 4 sin theta` follows. To get this, rotate the given one

about the pole, through `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]}