# Integration of a square root

• Jan 20th 2010, 02:29 PM
DonnMega
Integration of a square root
While working on finding the arc length of the parametrized curve $\displaystyle g(t)=(logt,2t,t^2)$ where $\displaystyle t$ is in $\displaystyle [1,e]$. i got to an integral that i couldn't manage to integrate. Here is the work so far leading up to the problematic integral so that it may be checked for errors.

arc length $\displaystyle =\int_{a}^{b} |g'(t)| dt$ by definition

$\displaystyle =\int_{1}^{e} |( 1/t,2,2t)| dt$ by taking the partials of g

$\displaystyle =\int_{1}^{e} \sqrt{(1/t)^2+2^2+(2t)^2} dt$ by taking the length

$\displaystyle =\int_{1}^{e} \sqrt{1/t^2+4+4t^2} dt$ by simplification

$\displaystyle =$ ...? by integration

I'm thinking it has something to do with simplifying the inside of the sqrt some more, or maybe a trig substitution (ugh!)

By the way, the answer is $\displaystyle e^2$

(Appologies if this is posted in the wrong thread as it is coming from an analysis course, but it really is only the integral that i am needing help with, so i see it fit that i place this thread here. Thank you)
• Jan 20th 2010, 02:39 PM
skeeter
Quote:

Originally Posted by DonnMega
While working on finding the arc length of the parametrized curve $\displaystyle g(t)=(logt,2t,t^2)$ where $\displaystyle t$ is in $\displaystyle [1,e]$. i got to an integral that i couldn't manage to integrate. Here is the work so far leading up to the problematic integral so that it may be checked for errors.

arc length $\displaystyle =\int_{a}^{b} |g'(t)| dt$ by definition

$\displaystyle =\int_{1}^{e} |( 1/t,2,2t)| dt$ by taking the partials of g

$\displaystyle =\int_{1}^{e} \sqrt{(1/t)^2+2^2+(2t)^2} dt$ by taking the length

$\displaystyle =\int_{1}^{e} \sqrt{1/t^2+4+4t^2} dt$ by simplification

$\displaystyle =$ ...? by integration

I'm thinking it has something to do with simplifying the inside of the sqrt some more, or maybe a trig substitution (ugh!)

By the way, the answer is $\displaystyle e^2$

does this help?

$\displaystyle \frac{1}{t^2} + 4 + 4t^2 = \left(\frac{1}{t} + 2t\right)^2$
• Jan 20th 2010, 02:45 PM
DonnMega
That does....a lot. As they always say, calculus is easy, it's the algebra that'll get ya......