Help converting triangular to polar

  • Thread starter Thread starter dare2dream
  • Start date Start date
  • Tags Tags
    Polar
Click For Summary

Homework Help Overview

The discussion revolves around converting the equation \(x^2 + 4y^2 = 4\) into polar coordinates. Participants are exploring the relationships between Cartesian and polar forms, particularly focusing on how to express the given equation using polar variables.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial substitution of \(x\) and \(y\) with their polar equivalents, \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). There is uncertainty about how to handle the coefficient of 4 in the equation and how to simplify the resulting expressions. Some participants question the implications of the trigonometric identities involved.

Discussion Status

The discussion is active, with participants providing guidance on expanding and factoring the equation. There is a focus on simplifying the expression obtained after substitution, and while some participants express confusion, others are attempting to clarify the relationships between the variables.

Contextual Notes

Participants mention that this problem is part of a take-home test, indicating a potential constraint on resources or time for problem-solving. There is also a sense of frustration expressed by some regarding their progress on the problem.

dare2dream
Messages
14
Reaction score
0

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.
 
Last edited:
Physics news on Phys.org
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 [tex]\theta[/tex] to x and y.
 
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...
 
Last edited:
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
 
Okay...well I don't know how to do that..>.>
 
You have this (rcos theta)^2 + 4(rsin theta)^2 = 4. So expand it: [tex]r^2\cos^2\theta+4r^2\sin^2\theta=4[/tex].

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

I don't know how to simplify that.
 
Well, I presume you know that sin^2(x)+cos^2(x)=1. Can you use this here?
 
Yeah, I remember that, but I don't know what to do with the 4 then.
 
  • #10
4sin^2(x)=sin^2(x)+3sin^2(x)
 
  • #11
so r^2 (1 + 3sin^2 theta) = 4?
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
11
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K