## Integration problem regarding parabolic length

Limits, differentiation, related rates, integration, trig integrals, etc.
mightytosave
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Joined: Tue Jun 03, 2014 11:55 pm
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### Integration problem regarding parabolic length

$\mbox{Hence, the total length of the cable can be}$

$\mbox{determined by integration. Using Eqn. 4, we have}$

. . . . .$\mathscr{L}\, =\, \displaystyle{\int\, ds\, =\, 2\, \int_{0}^{L/2}\, \sqrt{1\, +\, \left(\frac{8h}{L^2}x\right)^2\,}\, dx}$

$\mbox{Integrating yields}$

. . . . .$\mathscr{L}\, =\, \displaystyle{\frac{L}{2}\, \left[\,\sqrt{1\, +\, \left(\frac{4h}{L}\right)^2\, }\, +\, \frac{L}{4h}\sinh^{-1}\left(\frac{4h}{L}\right)\,\right]}$
How did the author arrive at his answer? I've been racking my brain around this for days to no avail... :(

stapel_eliz
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Joined: Mon Dec 08, 2008 4:22 pm
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$\mbox{Hence, the total length of the cable can be}$

$\mbox{determined by integration. Using Eqn. 4, we have}$

. . . . .$\mathscr{L}\, =\, \displaystyle{\int\, ds\, =\, 2\, \int_{0}^{L/2}\, \sqrt{1\, +\, \left(\frac{8h}{L^2}x\right)^2\,}\, dx}$

$\mbox{Integrating yields}$

. . . . .$\mathscr{L}\, =\, \displaystyle{\frac{L}{2}\, \left[\,\sqrt{1\, +\, \left(\frac{4h}{L}\right)^2\, }\, +\, \frac{L}{4h}\sinh^{-1}\left(\frac{4h}{L}\right)\,\right]}$
How did the author arrive at his answer? I've been racking my brain around this for days to no avail... :(
What is "Eqn. 4"? What preceded the exercise (such as "the cable" in question)?

Thank you!

mightytosave
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Joined: Tue Jun 03, 2014 11:55 pm
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### Re: Integration problem regarding parabolic length

Thanks for the reply! Eqn. 4 is 8hx/L^2.

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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Eqn. 4 is 8hx/L^2.
Lacking an "equals" sign, this is just an expression, not an equation. What was on the other side of the "equals" sign? How are the various variables defined? Thank you!

mightytosave
Posts: 3
Joined: Tue Jun 03, 2014 11:55 pm
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### Re: Integration problem regarding parabolic length

Here is the whole problem. I got lost when integrating eq. (7) on page 2. Thank you for replies! really appreciate them.