# Arc length of vector function curve

## Homework Statement

1. Find the length of the curve from t=0 to t=1.
r(t) = <2t, t^2, (1/3)t^3>

2. Reparametrize the curve with respect to arc length measured from the point where t=0 in the direction of increasing t.
r(t) = <e^(2t)cos2t, 2, e^(2t)sin2t>

## Homework Equations

$$S = \int{r'(t)} dt$$

## The Attempt at a Solution

1. I take the definite integral of the absolute value of r'(t).

r'(t) = <2, 2t, t^2>
S = integral[sqrt(4 + 4t^2 + t^4)dt] from [0,1]
This is a pretty complicated integral to solve. It's not that I can't solve it but I think I'm doing something wrong because this is only the second practice question of the section so I don't think it should be that complicated. Our prof has only show us the case for a helix, so the sine and cosine simplify to some number with no "t" terms in the integrand.

2. Basically, the same problem as number 1. I need to use the arc length function to find s in term of t. But I'm having trouble evaluating the integral for the arc length.
r'(t) becomes a pretty complicated function by itself but when you square the components, square root them and then take the integral of that from [0, t], it becomes very messy. I don't think I can solve this one. So again, I'm thinking I'm doing something wrong conceptually.

Last edited:

lanedance
Homework Helper
note that (t^2+2)^2 = ...

lanedance
Homework Helper
for the 2nd, consider some trig inequalities to simpilfy

lanedance
Homework Helper
note just to clarify, though its looks like you're already doing it, but you need the absolute value to show you're intergating a scalar function:
$$s = \int |r'(t)|dt$$

Alright, I guess I was too tired. I didn't realize the obvious factor for the 1st question. And the 2nd question isn't hard to integrate at all.

Thanks.