# Calculate the arc length of the vector function

## Homework Statement

Calculate the arc length of $<2t,t^2,lnt>$ from $1=<t=<e$

## Homework Equations

Arc length=$∫√{(x')^2 + (y')^2 + (z')^2}$

## The Attempt at a Solution

So I have gotten to this point:
$∫√{4 + 4t^2 + \frac{1}{t^2}}$

Am I on the right track, and if so, how do I integrate that?

Mark44
Mentor

## Homework Statement

Calculate the arc length of $<2t,t^2,lnt>$ from $1=<t=<e$

## Homework Equations

Arc length=$∫√{(x')^2 + (y')^2 + (z')^2}$
Missing dt in the integrand.

Also, if you use \sqrt instead of √, it will look better.
$$\int \sqrt{x'^2 + y'^2 + z'^2}dt$$

## The Attempt at a Solution

So I have gotten to this point:
$∫√{4 + 4t^2 + \frac{1}{t^2}}$

Am I on the right track, and if so, how do I integrate that?
Write the quantity in the radical as 4t2 + 4 + 1/t2, and then factor it.

Thanks. Good advice so far. I think I might just be an idiot by how do you factor that? Haven't had to do that for a while.

Nevermind, got it. multiply t^2

Thanks for the help!

Mark44
Mentor
Nevermind, got it. multiply t^2
You can't do that, but you can multiply by t2 over itself. OTOH, the expression can be factored directly, to (2t + 1/t)2.