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From: Jonathan Doolin on 18 Jul 2010 14:09
I am considering the right-side rectangular hyperbola, and I want the
path integral

x=cosh(t)
y=sinh(t)

-infinity < t < infinity

If I am not mistaken, the path length from (1,0) to (x(t0), y(t0))

dr = integral from 0 to t0 sqrt( (dx/dt)^2 + (dy/dt)^2 ) dt

= integral from 0 to t0 sqrt ( (sinh(t))^2 + (cosh(t))^2 ) dt

Now sinh^2 + cosh^2 has several representations

= 2 sinh(t)^2 + 1
= 2 cosh(t)^2 -1
= cosh(2t)

Also, if it helps, if you let v=tanh(t), then

cosh(t) = 1/sqrt(1-v^2), and

sinh(t)=v/sqrt(1-v^2)

Thanks for any help you can provide.

Jonathan Doolin
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