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Math Help - hyperbolic angle = ln(x) ?

  1. #1
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    hyperbolic angle = ln(x) ?

    Hello,

    It is often said that a hyperbolic angle \phi is the area of the hyperbolic sector (the hyperbola being y=1/x) swept out by a ray drawn from the origin to the point (x, 1/x), where x>1. The starting line of the "sweep" is the ray y= x. And it is said that the hyperbolic angle is therefore equal to ln(x).

    \phi=\ln{x}

    (For example here: Hyperbolic angle - Wikipedia, the free encyclopedia)

    But then you also see that the hyperbolic angle is the hyperbolic sector (in this case the hyperbola being y=\sqrt{x^2-1}) swept out by the ray drawn from the origin to the point (x, \sqrt{x^2-1}). The starting ray of the sweep is the x axis in this case.

    (For example here: Hyperbolic function - Wikipedia, the free encyclopedia)

    How can \phi be the area of both of these hyperbolic sectors, since an increase in one hyperbolic sector means the decrease of the other?

    Thanks


    PS- I'm never sure where to put questions regarding hyperbolic functions. The calculus forum seems like a good place because hyperbolic functions are important for working through difficult integrals, as in my case.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by rainer View Post
    Hello,

    It is often said that a hyperbolic angle \phi is the area of the hyperbolic sector (the hyperbola being y=1/x) swept out by a ray drawn from the origin to the point (x, 1/x), where x>1. The starting line of the "sweep" is the ray y= x. And it is said that the hyperbolic angle is therefore equal to ln(x).

    \phi=\ln{x}

    (For example here: Hyperbolic angle - Wikipedia, the free encyclopedia)

    But then you also see that the hyperbolic angle is the hyperbolic sector (in this case the hyperbola being y=\sqrt{x^2-1}) swept out by the ray drawn from the origin to the point (x, \sqrt{x^2-1}). The starting ray of the sweep is the x axis in this case.

    (For example here: Hyperbolic function - Wikipedia, the free encyclopedia)

    How can \phi be the area of both of these hyperbolic sectors, since an increase in one hyperbolic sector means the decrease of the other?

    Thanks


    PS- I'm never sure where to put questions regarding hyperbolic functions. The calculus forum seems like a good place because hyperbolic functions are important for working through difficult integrals, as in my case.
    They are the same as both curves are after a suitable scalling the same thing after rotation through \pi/4 (and are dealing with the magnitudes here when we are considering the areas)

    CB
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