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Writing x^2 + y^2 = 1 + sin^2(xy) in polar form

  • Thread starter Rasmus
  • Start date
  • #1
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Homework Statement


Write the equation

[tex]x^2 + y^2 = 1 + sin^2(xy)[/tex]

in polar form assuming

[tex]x = rcos(\phi)[/tex]
[tex]y = rsin(\phi)[/tex]

[tex]0<r, 0<= \phi < 2pi[/tex]

solve for r as a function of [itex]\phi[/itex]

The Attempt at a Solution




[tex](rcos(\phi))^2 + (rsin(\phi))^2 = 1 + sin^2(r^2cos(\phi)sin(\phi))[/tex]

[tex]r^2(cos^2(\phi) + sin^2(\phi)) = 1 + sin^2(r^2cos(\phi)sin(\phi))[/tex]

[tex]r^2 = 1 + sin^2(r^2cos(\phi)sin(\phi))[/tex]

At this point I'm feeling pretty lost, since I have no idea how to get the all r:s alone on one side of the equation. More specifically I don't understand how to get them out of the trig function.
 

Answers and Replies

  • #2
1,225
75
Hello.
I continued your calculation. it should be
φ = 1/2 * arcsin [2 arcsin{sqrt(r^2 - 1)} / r^2 ]
Regards.
 
  • #3
8
0
Thanks for the help, but I'm afraid I don't really understand exactly what you have done.

It looks like you applied arcsin twice as well as some other operations to solve for [itex]\phi[/itex] rather than r.

Wouldn't arcsin be a problem since that isn't defined for [0 to 2pi]?
 
  • #4
1,225
75
Hi.
The function stands for both φ and -φ, i.e. even function or symmetric for x axis.
The function stands for both φ and π-φ, i.e. even function or symmetric for y axis.
The function stands for both φ and π/2-φ, i.e. symmetric for y=x
So you can narrow the range of φ to [0,π/4] in your estimation.
Regards.
 

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