# Thread: Find the length of the curve ⃗r(t) = ⟨ 1/3*t^3, t^2, 1⟩, 0 ≤ t ≤ 2.

1. ## Find the length of the curve ⃗r(t) = ⟨ 1/3*t^3, t^2, 1⟩, 0 ≤ t ≤ 2.

r'(t) = <t^2, 2t,0>

|r'(t)| = sqrt( t^4 + 4t^2)

Arc length is integral of |r'(t)|, but how can i simplify it so i can integrate it?

2. ## Re: Find the length of the curve ⃗r(t) = ⟨ 1/3*t^3, t^2, 1⟩, 0 ≤ t ≤ 2.

Originally Posted by linalg123
r'(t) = <t^2, 2t,0>

|r'(t)| = sqrt( t^4 + 4t^2)

Arc length is integral of |r'(t)|, but how can i simplify it so i can integrate it?
Integral , click "show steps"

3. ## Re: Find the length of the curve ⃗r(t) = ⟨ 1/3*t^3, t^2, 1⟩, 0 ≤ t ≤ 2.

$\int_0^2 \sqrt{t^4+4t^2}dt = \int_0^2 t \sqrt{t^2+4} dt$

make the substitution $u = t^2 + 4$
and the integral should be easy.