# Tricky integral from calc 3 Arc Length question

1. Feb 19, 2015

### RJLiberator

1. The problem statement, all variables and given/known data
Find the arc length of:

r(t)=<e^t, e^(-t),sqrt(2)*t>
from 0 to ln(2)

2. Relevant equations
L=integral from a to b of the magnitude of r'(t)

3. The attempt at a solution

Okay, this was an Exam question, the one exam question that I could not get on our Calc 3 exam. This breaks down into an integral. The answer is 3/2 and there seems to be an obvious trick that I am missing.

r'(t)=<e^t, -e^(-t), sqrt(2)>
Take the magnitude
sqrt((e^t)^2+(-e^-t)^2+2)
Take the integral of that from 0 to ln(2).

There is some trick here that I simply could not get. I spent an hour on this problem during the exam :p.

2. Feb 19, 2015

### LCKurtz

Write that as$$\sqrt{e^{2t} + 2 +e^{-2t}}$$and note that the quantity under the square root sign is a perfect square.

3. Feb 19, 2015

### RJLiberator

I'm still not getting it. I've searched google for 'perfect square' and I see what it means, but how is it visible in this instance?

4. Feb 19, 2015

### Dick

Guess a factorization. If that's equal to (a+b)^2 what do you think a and b might be? Here's a hint. (e^x)^2=e^(2x). After you've guessed, check it.

Last edited: Feb 19, 2015
5. Feb 20, 2015

### RJLiberator

Ugh, I sat here and thought about it for 30 minutes, but something isn't sticking out to me.

I know it must be in form (a+b)^2 so that the square root will cancel and the problem will be easily solvable.
Hm.
so we have: (e^(2x)+2+e^(-2x)).
The negative sign is throwing me off greatly.

6. Feb 20, 2015

### HallsofIvy

Staff Emeritus
$e^{2x}= (e^x)^2$. $e^{-2x}$ is equal to $(e^{-x})^2$.

7. Feb 20, 2015

### RJLiberator

Oh... I just got it, it clicked to perfection:
(e^x+e^(-x))^2

This way the e^(-x)*e^x cancel out to create 1+1=2.
A perfect square.

And then taking out the square root you simply are left with the integral of e^x+e^(-x).