How would you write r=2+sinθ as a rectangular equation?

Answer 1

Use the following rules to convert from polar to rectangular equation:

1) r²=x²+y²

2) x=r*cos(θ)

3) y=r*sin(θ)

4) tan(θ)=y/x

Starting with your polar equation:

r=2+sin(θ)

First thing I notice is that if we multiply everything by "r", then we will have an "r*sin(θ)" which equals y, and an r² which equals x²+y². Let's start by doing that:

r=2+sin(θ)

Multiplying everything by r:

r²=2r+r*sin(θ)

Substituting in from the converting rules:

x²+y²=2r+y

Now that we are left with an "r" in the equation, let's use rule #1 but take the square root:

r²=x²+y²

r=(x²+y²)^(1/2)

So, plugging this in for r, the equation becomes:

x²+y²=2*(x²+y²)^(1/2)+y

I don't see a way to simplify further, so this would be the answer. Hope this helps! :)

Answer 2

y = rsin(theta), so sin(theta) = y/r.

r = 2 + y/r;

r^2 = 2r+y; multiply both sides by r

x^2+y^2 = 2*sqrt(x^2+y^2)+y; use fact that r^2 = x^2+y^2.

Does this help?

Answer 3

r-2=sin θ

y=r(r-2) = r² - 2r = x² + y² - 2sqrt(x² + y²)

This is an ugly implicit expression but it is graphically correct.

I don't know if there is a simpler way to do it.