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

1 Answer
Jun 20, 2018

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

Explanation:

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)