Thread: hyperbolic angle = ln(x) ?

1. hyperbolic angle = ln(x) ?

Hello,

It is often said that a hyperbolic angle $\displaystyle \phi$ 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).

$\displaystyle \phi=\ln{x}$

(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 $\displaystyle y=\sqrt{x^2-1}$) swept out by the ray drawn from the origin to the point $\displaystyle (x, \sqrt{x^2-1})$. The starting ray of the sweep is the x axis in this case.

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

How can $\displaystyle \phi$ 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.

2. Originally Posted by rainer
Hello,

It is often said that a hyperbolic angle $\displaystyle \phi$ 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).

$\displaystyle \phi=\ln{x}$

(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 $\displaystyle y=\sqrt{x^2-1}$) swept out by the ray drawn from the origin to the point $\displaystyle (x, \sqrt{x^2-1})$. The starting ray of the sweep is the x axis in this case.

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

How can $\displaystyle \phi$ 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 $\displaystyle \pi/4$ (and are dealing with the magnitudes here when we are considering the areas)

CB