What is the arc length of $r(t)=(t^2,2t,4-t)$ on $tin [0,5]$ ?

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Answer 1 · 2018-05-24T07:19:15

$L=5sqrt105+5/2ln(2sqrt5+sqrt21)$

$r(t)=(t^2,2t,4−t)$

$r'(t)=(2t,2,−1)$

Arc length is given by:

$L=int_0^5sqrt(4t^2+4+1)dt$

Simplify:

$L=int_0^5sqrt(4t^2+5)dt$

Apply the substitution $2t=sqrt5tantheta$ :

$L=5/2intsec^3thetad theta$

This is a known integral:

$L=5/2[secthetatantheta+ln|sectheta+tantheta|]$

Reverse the substitution:

$L=[tsqrt(4t^2+5)+5/2ln|2t+sqrt(4t^2+5)|]_0^5$

Hence

$L=5sqrt105+5/2ln(2sqrt5+sqrt21)$

What is the arc length of r(t)=(t^2,2t,4-t)r(t)=(t^2,2t,4-t) on tin [0,5]tin [0,5]?