How long is the path ?

Calculus Level 1

Let $p : [0, 1] \to \mathbb{R}^3$ be a path given by

$p(t) = (2 t^3, 6 t^2, 12 t )$

Determine the length of this path.

2 solutions

Hosam Hajjir
Dec 24, 2020

The length of the path is given by

$L = \displaystyle \int_0^1 | p'(t) | dt$

where $p'(t) = \dfrac{dp}{dt} = (6 t^2 , 12 t, 12 )$

so that $| p'(t) | = \sqrt{ (6 t^2)^2 + (12 t)^2 + 12^2 } = 6 \sqrt{ t^4 + 4 t^2 + 4 } = 6 \sqrt{ (t^2 + 2)^2 } = 6 ( t^2 + 2 )$

Therefore, $L =\displaystyle \int_0^1 6 (t^2 + 2 ) dt = 6 ( \dfrac{1}{3} + 2 ) = 2 + 12 = \boxed{14}$

Ron Gallagher
Dec 23, 2020

dp/dt = (6 t^2, 12 t, 12) so that

norm(dp/dt) = (36 t^4 + 144 t^2 + 144)^.5 = (36 (t^2 + 2)^2)^.5 = 6 (t^2 + 2)

We get the length by integrating this last quantity from t = 0 to t = 1, giving length = 14

You're missing a ^2 in the factor of (t^2 +2) in the third expression. Should be (36(t^2 + 2)^2)^.5. The final expression is correct.

Richard Desper - 5 months, 3 weeks ago

Thanks, Richard - I'll fix it now.

Ron Gallagher - 5 months, 3 weeks ago