Convert the equation to polar form x^2+y^2-3y=0?

Question

x^2+y^2-3y=0

 

how does this get converted into polar form?

Answer 1

From cartesian to polar coordinates we use x=ρcosθ and y=ρsinθ. Then

x²+y²-3y

=(ρcosθ)²+(ρsinθ)²-3ρ sinθ

=ρ²(cos²θ+sin²θ)-3ρ sinθ= ρ²-3ρcosθ =0

 

So the equation in polar coordinates is

ρ=3cosθ.

Answer 2

Hi Avas.

The trick to converting to polar coordinates is to substutute the following:

 

EVERY "x" gets replaced with "cos(theta)".

EVERY "y" gets replaced with "sin(theta)".

 

So what you have is x² is replaced with [cos(theta)]²

y² is replaced with [sin(theta)]², and 3y is replaced with 

3 sin(theta).

 

So now we have 

 

cos²(theta) + sin²(theta) - 3[sin(theta)] = 0

 

Remember that sin²x+cos²x=1,

 

So the given equation becomes

 

1 - 3sin(theta) = 0