What is the arclength of f(t) = (-(t+3)^2,3t-4)f(t)=((t+3)2,3t4) on t in [0,1]t[0,1]?

1 Answer
Jun 15, 2018

L=2sqrt73-9/2sqrt5+9/4ln((8+sqrt73)/(6+3sqrt5))L=273925+94ln(8+736+35) units.

Explanation:

f(t)=(−(t+3)2,3t−4)f(t)=((t+3)2,3t4)

f'(t)=(−2(t+3),3)

Arclength is given by:

L=int_0^1sqrt(4(t+3)^2+9)dt

Apply the substitution 2(t+3)=3tantheta:

L=9/2intsec^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=9/4[secthetatantheta+ln|sectheta+tantheta|]

Reverse the substitution:

L=[1/2(t+3)sqrt(4(t+3)^2+9)+9/4ln|2(t+3)+sqrt(4(t+3)^2+9)|]_0^1

Insert the limits of integration:

L=2sqrt73-9/2sqrt5+9/4ln((8+sqrt73)/(6+3sqrt5))