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Math Help - Tangent to hyperbolic function

  1. #1
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    Tangent to hyperbolic function

    Question:
    At what point of the curve y=cosh (x) does the tangent have slope 1?

    It seems so simple but I'm having a hard time working through this one. I know I need to find X when sinh (x)=1. I looked at the answer however and it is;

    (ln(1+sqr(2)), sqr(2))

    The X term in this point is similar to the arcsinh (x)= ln(x+sqr(x^2 + 1)). I don't understand how it got there? Any help walking through this one??

    Thanks all,
    cheers
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  2. #2
    Super Member Random Variable's Avatar
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    let  y = \sinh(x) = \frac {e^{x}-e^{-x}}{2}

    If we can find x in terms of y, then we have found  \sinh^{-1}x

     2y = e^{x}-e^{-x}

    multlipy both sides by e^{x}

     2ye^{x} = e^{2x} -1

    so  {(e^{x})}^{2} - 2y(e^{x}) - 1 = 0

    now apply the quadratic forumula to find  e^{x}

    we only need the positive root because  e^{x} is always positive

    and then take the ln of both sides to arrive at the answer
    Last edited by Random Variable; June 12th 2009 at 12:16 PM. Reason: correcting errors
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  3. #3
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    Solve this equation: e^x - e^{-x}=2.
    That is the value of arcsinh(1).
    You should get e^x=1+\sqrt(2) so x=?
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  4. #4
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    Thank you all,

    Plato - I'm still getting stuck after e^x - e^-x = 2. I multiplied both sides by e^x, then used the quadratic function but I'm gettin x= ln( (-2+sqr(8))/2 )....

    cheers
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  5. #5
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    Quote Originally Posted by klooless View Post
    Thank you all,

    Plato - I'm still getting stuck after e^x - e^-x = 2. I multiplied both sides by e^x, then used the quadratic function but I'm gettin x= ln( (-2+sqr(8))/2 )....

    cheers
    (e^x)^2 - 2 e^x - 1 = 0

    a = 1, b = -2, c = -1:

    e^x = \frac{2 \pm \sqrt{4 + 4}}{2}.

    Note that \sqrt{8} = 2 \sqrt{2}, divide by the common factor of 2, reject the -ve root etc.
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