## Calculus 8th Edition

$10.3311$
As we are given that Given: $r(t)=\lt \cos \pi t, 2t,\sin 2 \pi t$; $0 \leq t \leq 2$ Length of the curve can be obtained by using formula, such as $L=\int_a^b |r'(t)| dt$ Now,$r'(t)=\lt -\pi \sin \pi t, 2,2 \pi \cos 2 \pi t$ ; $|r'(t)|=\sqrt {( -\pi \sin \pi t)^2+(2)^2+(2 \pi \cos 2 \pi t)^2}dt$ Since, $L=\int_{0}^2(\sqrt {( -\pi \sin \pi t)^2+(2)^2+(2 \pi \cos 2 \pi t)^2}dt) dt=10.3311$ This is the result calculated by using a calculator.