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February 12th, 2017, 03:14 PM   #1
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Question about Arch Length

Find the arc length of r(t) = < t^2, sin(t), cos(t) >, 0 <= t <= 1

Answer: sqrt(5) / 2 + ln( sqrt(5) + 2 ) / 4

I use the standard fomula: Integral( |r'(t)|) between 0 and 1,
but I keep getting the following incorrect result:

( 2 * sqrt(5) - ln ( sqrt(5) - 2)) / 4


I would be much appreciated if you could point me at the right direction...
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February 12th, 2017, 04:39 PM   #2
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$\displaystyle \int_0^1 \sqrt{4t^2+1} \, dt$

$2t = \tan{\theta}$

$2 \, dt = \sec^2{\theta} \, d\theta$

$\displaystyle \dfrac{1}{2}\int_0^{\phi} \sec^3{\theta} \, d\theta$, where $\phi = \arctan(2)$

using integration by parts, note ...

$\displaystyle \int \sec^3{\theta} \, d\theta = \sec{\theta}\tan{\theta} - \int \sec{\theta}\tan^2{\theta} \, d\theta$

$\displaystyle \int \sec^3{\theta} \, d\theta = \sec{\theta}\tan{\theta} - \int \sec{\theta}(\sec^2{\theta}-1) \, d\theta$

$\displaystyle \int \sec^3{\theta} \, d\theta = \sec{\theta}\tan{\theta} - \int \sec^3{\theta}-\sec{\theta} \, d\theta$

$\displaystyle \int \sec^3{\theta} \, d\theta = \sec{\theta}\tan{\theta} - \int \sec^3{\theta}\, d\theta + \int \sec{\theta} \, d\theta$

$\displaystyle 2\int \sec^3{\theta} \, d\theta = \sec{\theta}\tan{\theta} + \int \sec{\theta} \, d\theta$

$\displaystyle \int \sec^3{\theta} \, d\theta = \dfrac{1}{2}\bigg[\sec{\theta}\tan{\theta} + \int \sec{\theta} \, d\theta\bigg]$

therefore ...

$\displaystyle \dfrac{1}{2}\int_0^{\phi} \sec^3{\theta} \, d\theta = \dfrac{1}{4}\bigg[\sec{\theta}\tan{\theta} + \ln|\sec{\theta}+\tan{\theta}|\bigg]_0^{\phi}$

$\displaystyle \dfrac{1}{2}\int_0^{\phi} \sec^3{\theta} \, d\theta = \dfrac{1}{4}\bigg[2\sqrt{5} + \ln(\sqrt{5}+2)\bigg]$
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February 13th, 2017, 03:27 PM   #3
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May I ask what gives you the clue for u-substitution 2t = tan(x)
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February 13th, 2017, 05:26 PM   #4
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Quote:
Originally Posted by zollen View Post
May I ask what gives you the clue for u-substitution 2t = tan(x)
note $4x^2 + 1 =(2x)^2 + 1$

compare $(2x)^2+1$ to $\tan^2{x} + 1$ ...
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