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)#