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Tangent will not meet the curve again

  1. May 2, 2014 #1
    1. The problem statement, all variables and given/known data

    can anyone give me hint on how to show the tangent will not meet the curve again?

    2. Relevant equations



    3. The attempt at a solution
     

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  3. May 2, 2014 #2

    mfb

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    Just plug your coordinates in the equation for the tangent and look for all solutions?
     
  4. May 2, 2014 #3

    Ray Vickson

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    You can try to check the convexity (or concavity) properties of the curve ##y = Y(x)##, by checking the second derivative ##d^2 y / dx^2##.
     
  5. May 5, 2014 #4
    what's the relationship between checking the concavity and show the curve will not meet the tangent again?
     
  6. May 6, 2014 #5

    Curious3141

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    Have you tried sketching the curve?
     
  7. May 6, 2014 #6

    Ray Vickson

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    Google "convex function" or "concave function". See, eg., http://ece.tamu.edu/~cui/ECEN629/lecture2.pdf [Broken] slide 4.
     
    Last edited by a moderator: May 6, 2017
  8. May 6, 2014 #7

    HallsofIvy

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    It might be simplest to eliminate the parameter and write a Cartesian equation for this curve:
    [itex]x= ln(cos(\theta)[/itex] so [itex]e^x= cos(\theta)[/itex] and [itex]y= ln(sin(\theta)[/itex] so [itex]e^y= sin(\theta)[/itex]. Then [itex]e^{2x}+ e^{2y}= cos^2(\theta)+ sin^2(\theta)= 1[/itex]. Of course, for [itex]0< \theta< \pi/2[/itex], [itex]cos(\theta)[/itex] goes from 1 to 0 so x goes from 0 to [itex]-\infty[/itex] and y goes from [itex]-\infty[/itex] to 0. The graph is in the third quadrant.

    At [itex]\theta= \pi/4[/itex], [itex]y= x= -(1/2)ln(2)[/itex] so that [itex]e^{2y}= e^{2x}= 2^{-1/2}= 1/2[/itex]. Further, differentiating [itex]e^{2x}+ e^{2y}= 1[/itex], [itex]2e^{2x}+ 2e^{2y}y'= 0[/itex] so, at [itex](1/2, 1/2)[/itex], [itex]y'= -1[/itex]. The tangent line is [itex]y= -1(x+ (1/2)ln(2))- (1/2)ln(2)= -x- ln(2)[/itex] and the question becomes solving [itex]e^{2x}+ e^{2(-x- ln(2)}= e^{2x}- (1/2)e^{-2x}= 1[/itex].
     
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