How would you change this to a polar equation?

Question

x^2+y^2-3y=0
Please explain you method.

Answer 1

Remember these basic trig relationships and identities:

x = r cos Θ ----> x² = r² cos² Θ
y = r sin Θ ----> y² = r² sin² Θ
cos² Θ + sin² Θ = 1

If we substitute r cos Θ for x and r sin Θ for y in the original equation we get this:

x² + y² - 3y = 0 ----> ** r² cos² Θ + r² sin² Θ - 3 (r sin Θ) = 0 **.

r² cos² Θ + r² sin² Θ = r² (cos² Θ + sin² Θ).

Since cos² Θ + sin² Θ = 1, then the starred equation above collapses to r² (1) - 3 (r sin Θ) = r² - 3r (sin Θ) = 0.

We can simplify the above equation even further:

r² - 3r (sin Θ) = 0 ----> r² = 3r (sin Θ).

Now divide both sides by r:

(r²) / r = (3 r) / r (sin Θ) ----> r = 3 sin Θ.

So the given rectangular coordinate equation is equivalent to the polar equation:

r = 3 sin Θ.

Answer 2

x^2+y^2=r^2 and y= rsin(theta)
so i guess it's r^2+rsin(theta)=0