Polar to Rectangular conversions

  • Thread starter Thread starter Eng67
  • Start date Start date
  • Tags Tags
    Polar Rectangular
Eng67
Messages
21
Reaction score
0
I am at a standstill with the solution to this problem.

I need to convert r^2=2cos(2 theta) to rectangular form.

I know that x = rcos(theta) and y = rsin(theta)

so far I have r = (2cos(2theta))/r

then I substitute for r

sqrt(x^2+y^2)= (2cos(2theta))/sqrt(x^2+y^2)

Then I hit a brick wall.

please help me knock down this wall.

Thanks
 
Physics news on Phys.org
Replace r² by x²+y² and theta = arctan(y/x).
 
So I then would have

sqrt(x^2+y^2)= (2cos(2arctan y/x))/sqrt(x^2+y^2)

I am still stuck.
 
Eng67 said:
I am at a standstill with the solution to this problem.

I need to convert r^2=2cos(2 theta) to rectangular form.

I know that x = rcos(theta) and y = rsin(theta)

so far I have r = (2cos(2theta))/r

then I substitute for r

sqrt(x^2+y^2)= (2cos(2theta))/sqrt(x^2+y^2)

Then I hit a brick wall.

please help me knock down this wall.

Thanks

Back up a little! You have r^2= 2 cos(2\theta) so first note that cos(2\theta)= cos^2(\theta)- sin^2(\theta) so that
r^2= 2(cos^2(\theta)- sin^2(\theta))
Now multiply on both sides by r2 to get
(r^2)^2= 2(r^2cos^2(\theta)- r^2sin^2(\theta))
I'll bet you can convert that to rectangular coordinates!
 
Thanks!

This is now so simple.
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top