Question

How to rewrite r=3sin(theta) as a Cartesian equation?

Answer 1

I understand parts of it but then I get stuck. Perhaps outlining my thought process will make someone more likely to help haha.

I started out with

r=3sin(theta), then multiplied both sides by r
r^2=3rsin(theta) , at which point I changed forms into
x^2 + y^2 = 3y

Answer 2

Looks good to me.

Answer 3

ignore what i wrote

Answer 4

I have notes from doing nearly the same problem but looking back I don't understand my process at all.

Answer 5

Observe the folllowing:\[\bf y=rsin(\theta) \implies \sin(\theta)=\frac{y}{r}\]Now substitute this in to the given equation:\[\bf \implies r= 3\frac{ y }{ r } \implies r^2=3y\]Now note the following identity:\[\bf r^2=x^2+y^2 \]Now let's make this substitution:\[\bf \implies x^2+y^2=3y \implies x^2+y^2-3y=0\]Now "complete the square":\[\bf \implies x^2+\left( y -\frac{ 3 }{ 2 }\right)^2-\frac{ 9 }{ 4 }=0\]Now bring the constant over to the other side and you get:\[\bf \implies x^2+\left( y -\frac{ 3 }{ 2 }\right)^2=\left( \frac{ 3 }{ 2 } \right)^2\]Which is the equation of a circle with radius 3/2

Answer 6

Could you explain to be how to complete the square? It's a concept I haven't reviewed for ages.

Answer 7

*me

Answer 8

Basically, you are trying to find a binomial that when squared, gives us the part that we need in in \(\bf y^2-3y\). I know when i square \(\bf y-3/2\), i get \(\bf y^2-3y+9/4\) which has the y^2 - 3y but we don't need the 9/4 so we subtract it. Now bring the 9/4 to the other side and we're done.

@daliamichelle