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How to calculate parametric representation of a circle?

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]y^2 + 4y + z^2 = 5, x = 3[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I know that the calculated coordinates must satisfy the above equation, but I don't know how to go about solving for those coordinates. The best I could do was to equate [tex]z = \sqrt{(-y + 5)(y + 1)}[/tex].
     
  2. jcsd
  3. Sep 18, 2010 #2

    CompuChip

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    Hmm,
    [tex]y^2 + 4y + z^2 = 5[/tex]
    doesn't really look like a circle, does it?
    I mean, the general equation for a circle (in the y-z plane) is
    [tex](y - a)^2 + (z - b)^2 = r^2[/tex]

    Can you maybe start by rewriting it to this form?

    For the next step, remember the usual parametrisation for a circle [itex]y^2 + z^2 = r^2[/itex] is
    [tex]y(t) = r \sin t, z(t) = r \cos t[/tex]
     
  4. Sep 18, 2010 #3
    Oh! By completing the square, I get

    [tex](y + 2)^2 + (z)^2 = 3^2[/tex]

    [tex]r(t) = (3, 3 \sin t, 3 \cos t)[/tex]

    Is this correct?
     
    Last edited: Sep 18, 2010
  5. Sep 18, 2010 #4

    CompuChip

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    Almost!
    The first line is right. Just think very carefully about what you call sin(t).
    When in doubt, calculate
    [tex](3 \sin t)^2 + 4 (3 \sin t) + (3 \cos t)^2 [/tex]
    and check if it gives 5.
     
  6. Sep 18, 2010 #5
    Is it [tex]r(t) = (3, 3 \sin t - 2, 3 \cos t)[/tex]?

    Why isn't it 3 sin t + 2? I had to change the sign to get the answer to fit.
     
  7. Sep 18, 2010 #6

    CompuChip

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    Very well done.

    The sign is there, because you want 3 (y + 2/3) to be 3 cos(t).
     
  8. Sep 18, 2010 #7
    Thank you very much. I have another (somewhat) related question, but I'd rather not create a new thread. I certainly hope you don't mind.

    1. The problem statement, all variables and given/known data
    Given a curve C: r(t), find a tangent vector r'(t), a unit tangent vector u'(t), and the tangent of C at P.

    [tex]r(t) = (\cosh t, \sinh t), P: (\frac{5}{3}, \frac{4}{3})[/tex]

    2. Relevant equations

    3. The attempt at a solution
    [tex]r'(t) = (\sinh t, \cosh t)[/tex]
    [tex]u'(t) = (\cosh 2x)^{-1/2}(\sinh t, \cosh t)[/tex]

    The last part stumps me.
     
  9. Sep 21, 2010 #8

    CompuChip

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    You mean about u'(t)?
    It is simply a unit vector along r'(t), so you will have to divide r'(t) by its length.
    To get the cosh(2x) there is just some (hyperbolic) identities magic using the formulas analagous to [tex]\cos^2 t + \sin^2 t = 1[/tex] and [tex]\cos^2 t - \sin^2 t = \sin 2t[/tex].
     
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