How do you graph $r = 2 + sin θ$ $r=2+sin theta$ ?

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Answer 1 · 2018-07-28T01:09:34

See the dimpled upright limacon.

$0 \leq r = 2 + sin θ \in [2 - 1, 2 + 1] = [1, 3]$ $0 <= r = 2 + sin theta in [2-1, 2 + 1 ] = [ 1, 3 ]$

The period = period of $sin θ = 2 π$ $sin theta = 2 pi$ .

Using

$r = \sqrt{x^{2} + y^{2}} and (x, y) = r (cos θ, sin θ)$ $r = sqrt ( x^2 + y^2 ) and ( x, y )= r ( cos theta, sin theta )$ ,

the Cartesian equation is obtained as

$x^{2} + y^{2} = 2 \sqrt{x^{2} + y^{2}} + y$ $x^2 + y^2 = 2 sqrt ( x^2 + y^2 ) + y$ and the dimpled graph of

this limacon id immediate.

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

Note that the general equation of limacons is

$r = a + b cos (θ - α)$ $r = a + b cos ( theta - alpha )$ , where $α$ $alpha$ is for anticlockwise

rotation, about the pole $r = 0$ $r = 0$ .

Here, a = 2, b = 1 and $α = \frac{π}{2}$ $alpha = pi/2$ .

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