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Parametric Tangent Problem driving me insane

  1. Oct 6, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex] x = e^{t} , y = (t-1)^{2} , (1,1) [/tex]
    Find an equation of the tangent to the curve at a given point by two methods. Without eliminating the parameter and by first eliminating the parameter.

    The answer in the book says [tex] y = -2x + 3 [/tex] and I cannot see how you get it.

    So I can't do it either way.


    2. Relevant equations

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


    3. The attempt at a solution

    [tex] \frac{dy}{dt} = 2(t-1)[/tex]

    [tex]\frac{dx}{dt} = e^{t}[/tex]

    [tex]\frac{dy}{dx} = \frac{2(t-1)}{e^{t}}[/tex]



    okay so now I need to substitute in (1,1)

    Rearranging x and y in terms of t does no good:

    [tex]x = e^{t}[/tex]

    so[tex] t = lnx ...(1)[/tex]

    [tex]\sqrt{y} = t - 1[/tex]

    so [tex]t = \sqrt{y} + 1 ... (2)[/tex]

    At point (1,1):

    t = 0 by equation (1)

    t = 2 by equation (2)

    So which t to use?

    Lets try a different approach:

    [tex] \frac{dy}{dx} = \frac{2(\sqrt{y} + 1 - 1)}{e^{lnx}}[/tex]

    so [tex]\frac{dy}{dx} = \frac{2\sqrt{y}}{x}[/tex]

    Substituting (1,1)

    [tex]\frac{dy}{dx} = 2 = m[/tex]

    so [tex]y - y_{1} = m(x - x_{1})[/tex]

    [tex]y = 2x - 1[/tex]

    I really can't see what I'm doing wrong. Any help appreciated.
     
  2. jcsd
  3. Oct 6, 2008 #2

    Ben Niehoff

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    Science Advisor
    Gold Member

    Remember every positive number has two real square roots. Try using the other one.
     
  4. Oct 6, 2008 #3
    Ah of course. That sorts it out:

    [tex]\frac{dy}{dx} = \frac{2(0-1)}{e^{0}}[/tex]

    [tex]\frac{dy}{dx} = -2[/tex]

    [tex] y - 1 = -2x + 2[/tex]
    [tex] y = -2x + 3[/tex]

    Thanks!
     
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