Results 1 to 5 of 5

Thread: Tangent to hyperbolic function

  1. #1
    Newbie
    Joined
    Jun 2009
    Posts
    19

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    let $\displaystyle y = \sinh(x) = \frac {e^{x}-e^{-x}}{2} $

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

    $\displaystyle 2y = e^{x}-e^{-x} $

    multlipy both sides by $\displaystyle e^{x} $

    $\displaystyle 2ye^{x} = e^{2x} -1 $

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

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

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

    and then take the ln of both sides to arrive at the answer
    Last edited by Random Variable; Jun 12th 2009 at 11:16 AM. Reason: correcting errors
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,797
    Thanks
    2828
    Awards
    1
    Solve this equation: $\displaystyle e^x - e^{-x}=2$.
    That is the value of arcsinh(1).
    You should get $\displaystyle e^x=1+\sqrt(2)$ so $\displaystyle x=?$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2009
    Posts
    19
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    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
    $\displaystyle (e^x)^2 - 2 e^x - 1 = 0$

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

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

    Note that $\displaystyle \sqrt{8} = 2 \sqrt{2}$, divide by the common factor of 2, reject the -ve root etc.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: Sep 2nd 2010, 11:54 AM
  2. Replies: 5
    Last Post: Apr 6th 2010, 11:53 PM
  3. Hyperbolic Tangent
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Oct 29th 2009, 07:33 PM
  4. hyperbolic tangent
    Posted in the Geometry Forum
    Replies: 3
    Last Post: Sep 26th 2009, 08:47 AM
  5. hyperbolic tangent issue
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Dec 1st 2008, 06:30 AM

Search Tags


/mathhelpforum @mathhelpforum