Arc length

[Guest]

Hello to you all. I need help with a math problem, and yehh its given me a headeach :/
Can anyone please help me? Im in serious troubl

The question

Show that the length s of the catenary between x=-b and x=b is decided by the expressions L=2 asinh(b/a), where sinhx is a the function sinus hyperbolicus x=e^x-e^(-x)/2

I know that the the arclength of \sinh(x) between a and b is given by http://www.mathwords.com/a/arc_length_of_a_curve.htm

But I have no idea what to do about x=e^x-e^(-x)/2 and L=2 asinh(b/a

[Guest]

Use the fact that [tex]\cosh^2 x - \sinh^2 x = 1[/tex] (for most trigonometry identities there is a hyperbolic version usually with an additional minus sign wherever there is an explicit or implicit [tex]\sin^2 x[/tex] term). Also remember that [tex]\sinh x[/tex] differentiates to [tex]\cosh x[/tex] and [tex]\cosh x[/tex] differentiates to [tex]\sinh x[/tex].

The equation of a Catenary is [tex]y = a\ \cosh(x/a)[/tex]. So
Arc length [tex]= \int_{-b}^b \sqrt{1+(dy/dx)^2}\quad dx[/tex]
[tex]= \int_{-b}^b \sqrt{1+\sinh^2 (x/a)}\quad dx[/tex]
[tex]= \int_{-b}^b \sqrt{\cosh^2 (x/a)}\quad dx[/tex]
[tex]= \int_{-b}^b \cosh (x/a)\quad dx[/tex]
[tex]= [a\ \sinh(x/a)]_{-b}^b[/tex]
[tex]= a\ \sinh(b/a) - a\ \sinh(-b/a)[/tex]
[tex]= 2a\ \sinh(b/a)[/tex]
as desired.

Hope this helped,

R. Baber.