What is the arc length of the curve given by #r(t)= (1,t,t^2)# on # t in [0, 1]#?

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Answer 1 · 2018-06-20T10:36:19

#L=1/2sqrt5+1/4ln(2+sqrt5)# units.

#r(t)=(1,t,t^2)#

#r'(t)=(0,1,2t)#

Arc length is given by:

#L=int_0^1sqrt(0^2+1^2+(2t)^2)dt#

Simplify:

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

Apply the substitution #2t=tantheta#:

#L=1/2intsec^3thetad theta#

This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals:

#L=1/4[secthetatantheta+ln|sectheta+tantheta|]#

Reverse the substitution:

#L=1/4[2tsqrt(1+4t^2)+ln|2t+sqrt(1+4t^2)|]_0^1#

Insert the limits of integration:

#L=1/2sqrt5+1/4ln(2+sqrt5)#