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 Eccentric Integral (Posted on 2010-02-18)
The respective eccentricities of the hyperbola x2/ a2 - y2/ b2 = 1 and its conjugate y2/ b2 - x2/ a2 = 1 are denoted by e and f(e).

Evaluate:

√2
(f(e) + f(f(e))) de
1

 No Solution Yet Submitted by K Sengupta No Rating

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 analytical solution Comment 2 of 2 |
I agree with Charlie that
f(e) = sqrt(1+1/(e^2-1)) and f(f(e)) = e
further we can simplify f(e) to
sqrt(e^2/(e^2-1)) = e/sqrt(e^2-1)
so we want
int e + e/sqrt(e^2-1) de so split it into 2 integrals

int e de = e^2/2 from 1 to sqrt(2) equals 2/2 - 1/2 = 1/2

int e/sqrt(e^2-1) de
use substitution with u = e^2-1 thus du =2e de thus we have
int 0.5*u^(-1/2) du =
(1/2)*2*u^(1/2)  =
sqrt(x^2-1)
thus from 1 to sqrt(2) we get
1 - 0 = 1

thus the total integral is equal to
(1/2) + 1 = 3/2
which agrees with charlie's numerical calculation

 Posted by Daniel on 2010-02-19 12:15:26

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