Area Sector of Hyperbola: Arcsinh & Arcosh Explained w/ Example

  • Context: Graduate 
  • Thread starter Thread starter Point Conception
  • Start date Start date
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 5K views
Point Conception
Gold Member
Messages
1,166
Reaction score
1,923
Could someone explain with a numerical example showing how the inverse hyperbolic
functions, arcsinh and arcosh in log form ,can compute the area sector of a unit hyperbola
x^2-y^2=1
If possible with a graph, thanks
 
Physics news on Phys.org
This picture's worth a thousand area calculations of a sector...
 

Attachments

  • Hyperbolic_functions.png
    Hyperbolic_functions.png
    1.8 KB · Views: 526
benorin said:
This picture's worth a thousand area calculations of a sector...
Hello that picture from wikipedia is on the hyperbolic functions, sinhx =e^x-e^-x/2
coshx=e^x+e^-x/2
If you substitute a = area for x in the equations above then you can obtain the points on the hyperbola with the sinh(a) as the y cooridinate and cosh(a) as the x cooridinate.
As the ray in picture, that passes through the point cosh(a),sinh(a), sweeps down you can see how the values of area, sinhx,coshx change.
My question is on the inverse hyperbolic functions , arcsinhx= ln(x+sqrt(x^2+1)
See Wikipedia for definitions. Then from the first two lines of that page ,wiki inv hyp func.
continue to 'area sector of unit hyperbola'
Now this is where my question is: what values of x would be used for arcsinh x and arccosh x
to obtain that area sector
 
Last edited: