Help converting triangular to polar

[dare2dream]

Homework Statement

Find an equivalent equation in polar coordinates.

Homework Equations

x^2 + 4y^2 = 4

Anyone know how to do this? I don't remember how when it's an equation...=S

My best guess is (rcos theta)^2 + 4(rsin theta)^2 = 4...but that's as far as I can get.

 

[mjsd]

how do you work out the x, y projections of a vector in the 1st quadrant? think about the unit circle and how you define sin and cos... you should be able to work out the rules that relate r and \theta to x and y.

 

[dare2dream]

I know that r^2 = x^2 + y^2 but I don't know how to use that because of the 4 in the equation. I've thought about this problem for a week and this is as far as I've gotten. -.- (This is an equation off a take-home test we received to do over spring break.)

So r^2 = x^2 + y^2, x = rcos (theta), y = rsin (theta), and tan theta = y/x

However, knowing these, I still don't get how to figure out the problem...

 

[mjsd]

My best guess is (rcos theta)^2 + 4(rsin theta)^2 = 4...but that's as far as I can get.

you know the answer, so all you need is rearrange (if you want) and get r on one side while the rest on the other... I don't understand your dilemma

 

[dare2dream]

Okay...well I don't know how to do that..>.>

 

[cristo]

You have this (rcos theta)^2 + 4(rsin theta)^2 = 4. So expand it: r^2\cos^2\theta+4r^2\sin^2\theta=4.

Now, spot that both terms on the left contain r², so factor this out. What do you obtain on the left hand side? Is there any way you can think of simplifying this expression you obtain?

 

[dare2dream]

So then you get r^2 (cos^2 theta + 4 sin^2 theta) = 4

I don't know how to simplify that.

 

[cristo]

Well, I presume you know that sin^2(x)+cos^2(x)=1. Can you use this here?

 

[dare2dream]

Yeah, I remember that, but I don't know what to do with the 4 then.

 

[cristo]

4sin^2(x)=sin^2(x)+3sin^2(x)

 

[dare2dream]

so r^2 (1 + 3sin^2 theta) = 4?