Yahoo Answers is shutting down on 4 May 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
help with an integral over a hyperbola?
please can you help with this question:
the integral with respect to s of (3x^2+3y^2)^0.5 over the hyperbola x^2-y^2=1 from (1,0) to (cosh2,sinh2).
Any help is much appreciated, I don't get why it is with respect to s.
1 Answer
- chauncyLv 71 decade agoFavourite answer
It's an integral with respect to arc length, the symbol for which is s. I think the following is correct:
Let x = t
Then y^2 = x^2 - 1
= t^2 - 1
y = sqrt(t^2 - 1)
The integral of (3x^2+3y^2)^0.5 w.r.t. arc length
= integral (3x^2+3y^2)^0.5 ds
= integral (3x^2+3y^2)^0.5 (ds/dt) dt
ds/dt = sqrt[(dx/dt)^2 + (dy/dt)^2]
dx/dt = 1
dy/dt = t/sqrt(t^2 - 1)
so ds/dt = sqrt[1 + t^2/(t^2 - 1)]
= sqrt[(2t^2 - 1)/(t^2 - 1)]
(3x^2 + 3y^2)^(0.5)
= (sqrt 3)(x^2 + y^2)^(0.5)
= (sqrt 3)(2t^2 - 1)^(0.5)
So the integral (3x^2+3y^2)^0.5 (ds/dt) dt
= integral (t = 1 ---> cosh 2) (sqrt 3)sqrt(2t^2 - 1)sqrt[(2t^2 - 1)/(t^2 - 1)] dt
= (sqrt 3) integral (t = 1 ---> cosh 2) (2t^2 - 1)/sqrt (t^2 - 1) dt
I'll leave that for you to evaluate, since it's just ordinary calculus.
