Calculate the arc length of the vector function

[jaydnul]

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]

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.

 

[jaydnul]

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.

 

[jaydnul]

Nevermind, got it. multiply t^2

 

[jaydnul]

Thanks for the help!

 

[Mark44]

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.