How do I complete the square?

1. Sep 5, 2014

shreddinglicks

1. The problem statement, all variables and given/known data

I identify the curve by finding a Cartesian equation for the curve.

2. Relevant equations

r = 3sin(θ)

3. The attempt at a solution

r^2 = 3r(y/r)

x^2+y^2 = 3y

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

Now I'm stuck, I don't remember how to complete the square as I haven't done it in ages.

2. Sep 5, 2014

Mentallic

Notice that the only x in the equation is a single x2 term so that's already a square. We just need to turn y2-3y into a complete square now.

The binomial expansion for the following square value is

$$(y-a)^2 = y^2-2ay+a^2$$

So we want to choose the value of 'a' such that we have turned the right hand side into the $y^2-3y$ expression.

So essentially since we want $y^2-3y=y^2-2ay$ then comparing coefficients of y clearly we want -3=-2a, so a=3/2.

But we also need the a2 term in there in order to turn it back into a perfect square. If we add a2 to both sides of the equation then we keep everything balanced and we get

$$x^2+y^2-3y+(3/2)^2=(3/2)^2$$

Can you take it from here?

3. Sep 5, 2014

shreddinglicks

I see, so I have radius r = 3/2. How do I get the center?

4. Sep 5, 2014

Mentallic

Have you completed the square on y? The general form for the circle is

$$(x-a)^2+(y-b)^2=r^2$$

And the centre is located at (a,b). Note that $x^2=(x-0)^2$

5. Sep 5, 2014

shreddinglicks

That is interesting. My textbook has me making a table using radian angles and plugging into the equation to find values of r and plotting them on a graph.

6. Sep 6, 2014

Mentallic

Well the thing about parametric curves is that you can't always find a nice cartesian equation to represent them. In this case it's simply a circle and so doing what you did is best to find what the parametric curve actually is.
Also, it would be good practise to do what your textbook is expecting so that you can find the rough shape of curves that normally wouldn't be as nice.

7. Sep 6, 2014

HallsofIvy

Staff Emeritus
you have $-3y+ y^2$. You should recall, or be able to calculate, that $(y- a)^2= y^2- 2ay+ a^2$. Comparing that with $y^2- 3y$, you see that your "2a" is 3. Since 2a= 3, a= 3/2 and $a^2= 9/4$. You need to add (and subtract so you won't actually change the value of the expression) 9/4: $y^2- 3y= y^2- 3y+ 9/4- 9/4= (y- 3/2)^2- 9/4$.