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dy/dx = 1/(x.sqrt(x^2-1)) |
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| Apr1-05, 06:38 PM | #1 |
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dy/dx = 1/(x.sqrt(x^2-1))
Hi,
Can anyone help me here! dy/dx = 1/(x.sqrt(x^2-1)) , x>0 Note y = pi when x = 2 Here is what I did: Let u = x^2 - 1 x^2 = u + 1 Substituting this in the above yields: dy/dx = 1/x.sqrt(u) Differentiating: u = x^2-1 du = 2x.dx x = du/2.dx Then, dy/dx= 2/sqrt(u).du I don't know if this is rite. Plz help me! |
| Apr1-05, 06:41 PM | #2 |
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What are u trying to do?Find the antiderivative of [itex] \frac{1}{x\sqrt{x^{2}-1}} [/tex] ...?If so,then use a substitution involving [itex] \cosh u [/itex]...
Daniel. |
| Apr1-05, 06:45 PM | #3 |
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Why is it that everyone has been taught about hyperbolic trig functions when we skipped it in high school and calc 1-3?
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| Apr1-05, 06:47 PM | #4 |
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dy/dx = 1/(x.sqrt(x^2-1))
I was trying to simplify the Right Hand side, before, I could integrate , by making a substitution.
Anyway what would be the hyperbolic sub, for this involving coshu. I don't have my text book with me. I am at work, and I am trying to finish my homework. |
| Apr1-05, 06:48 PM | #5 |
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Need i say that your initial approach is incorrect...?
It also works with [itex] \sec u [/itex]... Daniel. |
| Apr1-05, 06:51 PM | #6 |
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Daniel. |
| Apr1-05, 07:04 PM | #7 |
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Ok, then,
Let x = secu dx/du = secu.tanu sec^2u-tan^2u = 1 dy/dx = 1/secu.tanu Then what, to do! |
| Apr1-05, 07:07 PM | #8 |
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[tex] \int \frac{\sec u \ \tan u \ du}{\sec u \ \tan u} =\int du = u+C = \mbox{arcsec} \ x +C [/tex]
Daniel. |
| Apr1-05, 07:21 PM | #9 |
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Ok, I applied the initial conditions:
y = pi when x = 2 and I got, y = sec-1(x) + pi - sec-1(2) can we write this as: y = sec-1 (x/2) + pi |
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