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Spherical equation to Rectangular

  1. Jan 29, 2016 #1
    1. The problem statement, all variables and given/known data

    An equation is given in spherical coordinates

    ## \rho sin(\phi) = 8cos(\theta) ##

    Express the equation in rectangular coordinates

    a) ## (x-4)^2 + y^2 = 16 ##
    b) ## x^2 + y^2 + z^2 = 16 ##
    c) ## x^2 + (y-4)^2 = 16 ##

    2. Relevant equations

    ## x = \rho sin \phi cos \theta, y = \rho sin \phi sin \theta, z = \rho cos \phi ##

    ## \rho = \sqrt {x^2 + y ^2 + z^2}, tan \theta = \frac y x, cos \phi = \frac {z} {\sqrt {x^2 + y^2 + z^2} } ##

    3. The attempt at a solution

    Everytime I try to do this problem, I cannot get any of the clean answers is has in the multiple choice.

    In the relevent equations above, I plug in the ##\rho## for the ##\rho##, solve for ##\theta## and ##\phi## and plug those in as well, and I get this giant mess of an equation that has no means of simplifying down further.

    I've tried using trigonometric identities and such to simplify, but I don't think I understand exactly what it wants me to do.
     
    Last edited by a moderator: Jan 29, 2016
  2. jcsd
  3. Jan 29, 2016 #2

    ehild

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    Do the opposite, substitute the equation ##\rho sin(\phi) = 8cos(\theta)## into the expression for x and y and z and then try which equation from a, b, c is valid.
     
  4. Jan 30, 2016 #3
    Okay, I get what you mean, but I'm not seeing how to plug the equation into the x, y, and z expressions, there doesn't seem any common substitution grounds between the whole equation and one of the x,y, or z's.
     
  5. Jan 30, 2016 #4

    ehild

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    You wrote that ##x = \rho sin \phi cos \theta, y = \rho sin \phi sin \theta, z = \rho cos \phi##. The problem states that ##\rho sin(\phi) = 8cos(\theta)##. You can substitute ##8 \cos(θ)## for ## ρ\sin(\phi)##, so ##x=8 \cos^2(θ)## and ##y=8 \cos(θ)\sin(θ)##. Is ##(x-4)^2+y^2=16## true?
     
  6. Jan 30, 2016 #5
    v

    After doing the substitution I determined that equation wasn't true, I ended up getting ## 4sin(2\theta) - 64cos(\theta)^2 + 64cos(\theta)^4 = 16 ## and I'm not seeing much of anyway to reduce that any further.

    My question in response to this, is from getting ## x = 8cos^2(\theta)## and ## y = 8cos(\theta)sin(\theta) ## how would I deduce the formula in x, y, z form, without having to look at those multiple choice answers?
     
  7. Jan 30, 2016 #6

    ehild

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    You did something wrong.
    Well, let's go back to the direct way.
    ##\rho = \sqrt {x^2 + y ^2 + z^2}, tan \theta = \frac {y} {x}, cos \phi = \frac {z} {\sqrt {x^2 + y^2 + z^2} }##. Write cos(θ) in terms of x, y.
     
  8. Jan 30, 2016 #7
    ## cos(\theta) = \frac {xsin(\theta)} {y} ## ?
     
  9. Jan 30, 2016 #8

    ehild

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    Last edited: Jan 30, 2016
  10. Jan 30, 2016 #9
    Oh okay... Do you mean ##cos(\theta) = \frac {x} {\sqrt {x^2 + y^2}} ##
     
  11. Jan 30, 2016 #10

    ehild

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    Correct. Now, substitute the expressions for ρ,sinΦ and cosθ into the equation ρsin(Φ)=8cos(θ).
     
  12. Jan 30, 2016 #11
    Okay, so when I did this I found an expression for ##sin(\phi)## as well before substituting, because I didn't know how else to do it.

    ## psin(\phi) = 8cos(\theta) ##
    ## \sqrt{x^2 + y^2 + z^2} \frac {\sqrt{x^2 + y^2}} {\sqrt{x^2 + y^2 + z^2}} = 8 \frac {x} {\sqrt{x^2 + y^2}}##
    ## \frac {x^2 + y^2} {x} = 8 ##

    not really sure what else I can do here besides divide the x through the left hand side.
     
  13. Jan 30, 2016 #12

    ehild

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    Multiply the equation by x, and arrange it so as the right hand side is zero. Complete square for x.
     
  14. Jan 30, 2016 #13
    Okay, I've got it!

    After completing the square I've arrived at ##(x-4)^2 + y^2 = 16##

    Thank you a lot for your time, I was really frustrated with this problem, and wasn't following it. I appreciate it.
     
  15. Jan 30, 2016 #14

    ehild

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    Well done!

    The other method would have been a bit easier. Substituting ##x = 8\cos^2(\theta)## and ##y=8 \cos(θ)\sin(θ)## into the expression ##(x-4)^2 +y^2## : ## (8\cos^2(\theta)-4)^2+(8 \cos(θ)\sin(θ))^2 = 64\cos^4(\theta)-64\cos^2(\theta)+16 + 64 \cos^2(\theta)\sin^2(\theta) = 64 \cos^2(\theta)(\cos^2(\theta)+\sin^2(\theta)) -64\cos^2(\theta) +16 = 64 \cos^2(\theta) -64\cos^2(\theta) +16 =16 ## , that is, a) proved to be true.
     
  16. Jan 31, 2016 #15

    WWGD

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    If it is a matter of choosing, you can throw away the middle choice ## x^2+y^2+z^2=16 ## , which would have equation ##\rho=4 ##. But this is the easy part, ehild took care of the hard 99%.
     
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