# Math Help - Arc Length

1. ## Arc Length

x = 1 + 3t^2
y = 4 + 2t^3
0<= t >= 1

k so i got that it is the integral from 0 to 1 of the sqrt((6t)^2+(6t^2)^2)
then i mess up somewhere when im solving that

2. Originally Posted by amiv4
x = 1 + 3t^2
y = 4 + 2t^3
0<= t >= 1

k so i got that it is the integral from 0 to 1 of the sqrt((6t)^2+(6t^2)^2)
then i mess up somewhere when im solving that
$L = \int_{0}^{1} 6t \sqrt{1 + t^2} \, dt$.

Make the substitution $u = 1 + t^2$.