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Tangent to Hyperbolic function graph

  1. Oct 19, 2016 #1
    • Moved from a technical forum, so homework template missing
    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)}##.
    $$2x-2yy'=0~\rightarrow~\frac{x}{y}=y'=\frac{\cosh (u)}{\sinh (u)}=\frac{e^u+e^{-u}}{e^u-e^{-u}}$$
    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

    mfb

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    2016 Award

    Staff: Mentor

    Why do you expect the intersection with the y-axis to be at sinh(u)? That's the y-value where the tangent touches the hyperbola, it won't be identical with the y-axis intersection in general.

    I don't understand the second part of your calculation at all.
     
  4. Oct 19, 2016 #3

    stevendaryl

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    Staff Emeritus
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    I don't see the point in going back and forth from exponentials to hyperbolic functions. If you stick to hyperbolic functions, then your equation for the tangent is:

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

    If you plug [itex]y=sinh(u), x = cosh(u)[/itex] in, what do you get for [itex]b[/itex]? Certainly not [itex]4[/itex].
     
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