Convert x^2+y^2=4y-2x to Polar Equation

  • Thread starter Thread starter Calcgeek123
  • Start date Start date
  • Tags Tags
    Cartesian Polar
Click For Summary
To convert the equation x^2 + y^2 = 4y - 2x to polar coordinates, the substitution x = rcos(θ) and y = rsin(θ) is applied, leading to the equation r^2 = 4rsin(θ) - 2rcos(θ). It is noted that an "r" was missing in the cosine term, which should be corrected to r^2 = 4rsin(θ) - 2rcos(θ). Dividing both sides by r is permissible since the curve does not pass through the origin, avoiding issues with r being zero. Taking the square root of both sides is discouraged in this context. The discussion emphasizes careful handling of polar coordinate transformations.
Calcgeek123
Messages
19
Reaction score
0

Homework Statement


Establish an equation in polar coordinates for the curve x^2+y^2=4y-2x


Homework Equations



n/a

The Attempt at a Solution


I know that x^2+y^2=r^2 so I used substitution, and now have r^2=4y-2x. Now this next part, I'm really not sure if I'm allowed to do this... i know that x=rcos@ (let @=theta) and y=rsin@. So i simply substituted to get r^2=4rsin@-2cos@. Is that allowed? Also, i feel like my answer should just look like r=... So should I take the sqaure root of both sides?
Thank you, very appreciated.
 
Physics news on Phys.org
You should have r^2 = 4rsin@ - 2rcos@. You lost an "r" on your cosine term. You can simplify by dividing both sides by r, but you don't want to take the square root of both sides. You normally have to be careful of dividing by r, in case r happens to be equal to zero. That's not a problem in this case, since the curve doesn't go through the origin.
 
Yeah, the missing r is a type. Dividing makes sense though, duh. thanks so much!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 20 ·
Replies
20
Views
2K
Replies
7
Views
2K
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K