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Parametric expression

  1. Apr 22, 2005 #1
    I have x(t)=t(1-t) and y(t)=t(1-t^2). As t goes from 0 to 1 in forms a loop and I need to know the point where the tangent line is vertical. I know this must be easy but I'm clueless right now. Any help?
  2. jcsd
  3. Apr 22, 2005 #2


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    U need the derivative in that point to be infinite.That is the slope must be infinite...

    [tex] \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} [/tex]

  4. Apr 23, 2005 #3
    So basically you need to set [tex]\frac{dx}{dt}[/tex] equal to zero.

  5. Apr 23, 2005 #4
    solving [itex]dx/dt=0[/itex] is not quite enough in general. Dexter's method is fine.

    Consider [itex]x(t)=t, \ y(t)=t[/itex].
  6. Apr 24, 2005 #5
    Data: but x = t, and y = t doesnt have a vertical asymptote?
  7. Apr 24, 2005 #6


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    It was a counterexample to the statement of analyzing [itex] \frac{dx}{dt} [/itex] instead of [tex] \frac{\displaystyle{\frac{dy}{dt}}}{\displaystyle{\frac{dx}{dt}}} [/tex]

    Last edited: Apr 24, 2005
  8. Apr 24, 2005 #7
    I still don't see how that's an example where simply setting dx/dt equal to zero will not yield the vertical asymptotes.

  9. Apr 24, 2005 #8
    This was not, because if x(t)=t, dx/dt=1

    However, if you take the dumb example : x(t)=y(t)=sin(t). Then obviously, y=x, so this is just the diagonal, so there are no vertical tangents.

    However, there exist some t such that dx/dt=0...but this is precisely where dy/dt=0, so that both cancel....you can imagine example where this gives other value than 1...

    In fact to solve the question why you have to look at (dy/dt)/(dy/dt)...it's because you could have dy/dt=infty and dx/dt finite...or other limits such as dx/dt->0, and dy/dt->0...(indefinite)...
  10. Apr 24, 2005 #9
    oops, yeah, i needed to use [itex]x(t)=y(t)=t^2[/itex]!
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