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Calculus 8th Eighth Edition 10.3 #22

  1. Feb 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the equations of the tangent lines at the point where the curves crosses itself.

    x=2-piCos(t), y=2t-piSin(t)

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 6, 2008 #2
    What have you tried so far?
  4. Feb 6, 2008 #3
    well, all i need are the points and after that its easy. But the PROBLEM is the points thats is where i am stuck
  5. Feb 6, 2008 #4
    Points? The problem only mentions one. Anyway, there wasn't any trick to it rather than realizing that the textbook wouldn't make a calculus problem very difficult in precalculus terms. I'd observe where the curve intercepts y=0. Any time you see such a strange formulation of the x and y coordinates, you know something is bound to simplify dramatically, so start plugging in things!
  6. Feb 6, 2008 #5


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    At the point where the graph crosses itself, you must have the point [itex]2- \pi cos(t)= 2- \pi sin(s)[/itex] for some s and t and [itex]2t- \pi sin(t)= 2s-\pi sin(s)[/itex]. Solve those equations for s and t.
  7. Feb 6, 2008 #6
    sorry, but still confusion.. can you take it step by step please
  8. Feb 6, 2008 #7
    [itex]2- \pi cos(t)= 2- \pi sin(s)[/itex] and is this right? or a typo because I thought its was suppose to be [itex]2- \pi sin(t)= 2- \pi sin(s)[/itex]
  9. Feb 7, 2008 #8


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    Sorry that was a typo. It should be [itex]2- \pi cos(t)= 2- \pi cos(s)[/itex] which says that the x values for s and t are the same. Remember that [itex]x= 2- \pi cos(t)[/itex].
    The other equation is [itex]2t- \pi sin(t)= 2s- \pi sin(s)[itex] which says that the y values for s and t are the same- from [itex]y= 2t- \pi cos(t)[/itex].

    It should be obvious that [itex]2- \pi cos(t)= 2- \pi cos(x)[/itex] gives cos(t)= cos(s) which does not mean t= s but rather that [itex]t= 2\pi- s[/itex].
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