Using the arc length formula I get to the integral of 16sec(t) evaluated at the given limits, but I cant seem to get the correct answer.

Arc length = [(dx/dt) (dy/dt) (dz/dt)] dt

= [16sin(4t) 16cos(4t) 16*tan(t)] dt eval. from 0 to /4

= (16 16*(sec(t) 1) dt eval. from 0 to /4

= 4*sec(t) dt eval. from 0 to /4

= 4*ln|sec(t) tan(t)| eval. from 0 to /4

= 4ln(2 1) 4ln(1)

= 4*ln(2 1)

This post is last updated on hrtanswers.com at Date : 1st of September – 2022