What is the arc length of #r(t)=(t^2,2t,4-t)# on #tin [0,5]#?
1 Answer
May 24, 2018
Explanation:
#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
#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)#