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Tangent to Hyperbolic functions graph

  1. Oct 19, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that the tangent to ##x^2-y^2=1## at points ##x_1=\cosh (u)## and ##y_1=\sinh(u)## cuts the x axis at ##{\rm sech(u)}## and the y axis at ##{\rm -csch(u)}##.

    2. Relevant equations
    Hyperbolic sine: ##\sinh (u)=\frac{1}{2}(e^u-e^{-u})##
    Hyperbolic cosine: ##\cosh (u)=\frac{1}{2}(e^u+e^{-u})##

    3. The attempt at a solution
    $$2x-2yy'=0~\rightarrow~\frac{x}{y}=y'=\frac{\cosh (u)}{\sinh (u)}=\frac{e^u+e^{-u}}{e^u-e^{-u}}$$
    The equation of the tangent is ##y=y'x+b## and it's intersection with the y axis is at ##y=\sinh (u)##:
    $$y=y'x+b~~\rightarrow~~\sinh (u)=\frac{e^u+e^{-u}}{e^u-e^{-u}}(e^u+e^{-u})+b~~\rightarrow~~b=4$$
    But b cant be 4 because it is a general expression, it should depend on u
     
  2. jcsd
  3. Oct 19, 2016 #2

    stevendaryl

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    Staff Emeritus
    Science Advisor

    How are you getting 4 out of that? There is no reason to go back and forth between the hyperbolic trig functions and exponentials. In terms of the hyperbolic trig functions, you have:

    [itex]y = \frac{cosh(u)}{sinh(u)} x + b[/itex]

    So plugging in [itex]y=sinh(u), x = cosh(u)[/itex] gives:

    [itex]sinh(u) = \frac{cosh(u)}{sinh(u)} cosh(u) + b[/itex]

    So what do you get for [itex]b[/itex]? (Remember the identity: [itex]cosh^2(u) - sinh^2(u) = 1[/itex])
     
  4. Oct 19, 2016 #3
    Thank you stevendaryl
     
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