What Is the Method to Complete the Square in Algebra?

[shreddinglicks]

Homework Statement

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

Homework Equations

r = 3sin(θ)

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.

 

[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?

 

[shreddinglicks]

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?

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

 

[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 $x^2=(x-0)^2$

 

[shreddinglicks]

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 $x^2=(x-0)^2$

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.

 

[Mentallic]

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.

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.

 

[HallsofIvy]

To answer your basic question, "how do I complete the square",

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: $y^2- 3y= y^2- 3y+ 9/4- 9/4= (y- 3/2)^2- 9/4$.