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Simple hyperbolic calculus

  1. Nov 9, 2013 #1
    q: http://gyazo.com/297417b9665206ae8e38cb8b5d930a83

    I'm stuck trying to find the value of x when TN is a minimum

    here's what I've tried so far:

    Let T be the point (a,0) and N be the point (b,0)

    line of tangent through P:

    ## y = sinh(x)(x-a) ##
    line of normal through P ## y = \dfrac{-1}{sinh(x)}(x-b) ##

    my plan was to rearrange for a and b and find b-a and try differentiate that and set the derivative to be 0 and solve for x:

    ## a = x - \dfrac{y}{sinh(x)} ##
    ## b = y(sinh(x)) + x ##
    ## b - a = y(sinh(x) + \dfrac{1}{sinh(x)}) ##
    ## \dfrac{d(b-a)}{dx} = \dfrac{dy}{dx} (sinh(x) + \dfrac{1}{sinh(x)}) + y(cosh(x) - coth(x)cosech(x)) = sinh^2(x) + 1 + y(cosh(x) - coth(x)cosech(x)) = 0## but I can't seem to solve that

    any ideas on where I went wrong?
     
  2. jcsd
  3. Nov 9, 2013 #2

    tiny-tim

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    hi synkk! :smile:
    erm :redface:

    y isn't a variable, y is coshx ! :wink:
     
  4. Nov 9, 2013 #3
    oh thank you, but is my method correct?

    following on from this I get ## x = arsinh(\dfrac{1}{\sqrt{2}}) ## is this correct?
     
  5. Nov 9, 2013 #4

    tiny-tim

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  6. Nov 10, 2013 #5

    FeDeX_LaTeX

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    You could have saved yourself differentiating ##\sinh(x) + \frac{1}{\sinh(x)}## by instead adding the two terms and noting the hyperbolic trig identity -- otherwise, all correct.
     
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