Question

How do you graph `r=2+sin theta`?

Answer 1

In `[0. pi]`, the graph crowns over the circle r = 2. In `[pi, 2pi]`, it hangs from the circle. The maximum distance from the circle, either way, is 1 unit.

Explanation:

`r(theta)` is periodic, with period `2pi` in `theta`.

The Table for one period #theta in [0, 2pi] is sufficient.

`(r, theta)`:

`(2, 0), (2+1/sqrt2, pi/4) (3, pi/2), (2+1/sqrt2, 3/4pi) (2, pi)`

`(2-1/sqrt2, 5/4pi) (1, 3/2pi) (2-1/sqrt2, 7/4pi) (2, 2pi)` #

Altogether this graph is a wave, twining around a circle.

graph{x^2 + y^2 - 2sqrt(x^2 + y^2) - y = 0}

Credit for the graphs goes to Socratic.

graph{(x^2 + y^2)^1.5 - 2(x^2 + y^2) - 2xy = 0}

The second graph is for `r = 2 + sin (2theta)`

graph{(x^2 + y^2)^2 - 2(x^2 + y^2)^1.5 - 3(x^2y-xy^2) = 0}

The third is for `r = 2 + sin (3theta)`

graph{(x^2 + y^2)^2.5 - 2(x^2 + y^2)^2 -4(x^3y-xy^3) = 0}

The fourth is for `r = 2 + sin (4theta)`