February 12th, 2017, 03:14 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3  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... 
February 12th, 2017, 04:39 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,949 Thanks: 1555 
$\displaystyle \int_0^1 \sqrt{4t^2+1} \, dt$ $2t = \tan{\theta}$ $2 \, dt = \sec^2{\theta} \, d\theta$ $\displaystyle \dfrac{1}{2}\int_0^{\phi} \sec^3{\theta} \, d\theta$, 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 ... $\displaystyle \dfrac{1}{2}\int_0^{\phi} \sec^3{\theta} \, d\theta = \dfrac{1}{4}\bigg[\sec{\theta}\tan{\theta} + \ln\sec{\theta}+\tan{\theta}\bigg]_0^{\phi}$ $\displaystyle \dfrac{1}{2}\int_0^{\phi} \sec^3{\theta} \, d\theta = \dfrac{1}{4}\bigg[2\sqrt{5} + \ln(\sqrt{5}+2)\bigg]$ 
February 13th, 2017, 03:27 PM  #3 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 
May I ask what gives you the clue for usubstitution 2t = tan(x)

February 13th, 2017, 05:26 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,949 Thanks: 1555  

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