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Could someone help me with cylindrical and rectangular conversions?

  1. Mar 24, 2004 #1
    I have the following cylindrical equation:

    z = r^2 cos(2theta)

    I am suppose to convert it into a rectangular equation. I'm stumped.
  2. jcsd
  3. Mar 24, 2004 #2
    would i be correct to assume the trig property that cos(2theta) = cos^2 theta - sin^2 theta ?

    then i'd end up with

    r^2 cos^2 theta - r^2 sin^2 theta = z

    then substitute x^2 and y^2 to get z = x^2 - y^2 ?
  4. Mar 24, 2004 #3
    no if someone could help me convert this rectangular equation to cylindirical i'd be eternally greatful:

    z^2 (x^2 - y^2) = 4xy
  5. Mar 24, 2004 #4

    matt grime

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    You seem to be doing well enough on your own:

    you know what x^2-y^2 is from the previous example.

    What other trig identities do you know? say, sin(2theta)?
  6. Mar 24, 2004 #5
    yeah i realized i could use cos(2theta) from the previous problem but im still stuck since i still have the 4xy on the other side of the equation :(
  7. Mar 24, 2004 #6

    matt grime

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    Hope this doesn't appear too RTFM, but you have formulae:

    so use them as you did in the previous question.
  8. Mar 24, 2004 #7
    i know what youre hinting at that i could divide both sides of the equation by 2 and then sub x and y for r cos(theta) and r sin(theta) and then cancel the r^2 on both sides and have sin(2theta) left.. but that would leave me with

    (z^2 cos(2theta)) / 2 = sin(2theta) is that correct though? doesnt seem right.
    Last edited: Mar 24, 2004
  9. Mar 24, 2004 #8
    (z^2) / 2 = tan(2theta)

    if sin(2theta) / cos(2theta) = tan(2theta) is valid... is it?
    Last edited: Mar 24, 2004
  10. Mar 24, 2004 #9
    im an idiot.. of course its valid.. its just an angle. thanks guys. i guess this thread was pretty pointless ;)
  11. Mar 24, 2004 #10


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    Since I am not particularly bright and have great difficulty remembering trig identities, I would probably change
    z^2 (x^2 - y^2) = 4xy into polar coordinates in the obvious way:

    Since x= r sinθ and y= r cosθ, x^2= r^2 sin^2θ, y^2= r^2 cos^2θ so x^2- y^2= r^2(cos^2θ- sin^2θ), and 4xy= 4r^2 cosθ sinθ. Now the "r^2" terms cancel leaving z^2(cos^2θ-sin^2θ)= 4sinθcosθ or

    z^2= (4sinθcosθ)/(cos^2θ- sin^2θ).

    Now IF I were smart I might remember (or look up!) both
    "cos(2θ)= cos^2θ- sin^2θ" and
    "sin(2θ)= 2sinθcosθ to write that equation as

    z^2= 2sin(2θ)/cos(2θ)
  12. Mar 24, 2004 #11
    yep thats what i got. thanks
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