What is the arclength of (t/(t+5),t) on t in [-1,1]?

1 Answer
May 25, 2018

L=2+4sum_(n=1)^oo((1/2),(n))1/(4n-1)(5/16)^(2n)(1-(2/3)^(4n-1)) units.

Explanation:

f(t)=(t/(t+5),t)=(1-5/(t+5),t)

f'(t)=(5/(t+5)^2,1)

Arclength is given by:

L=int_-1^1sqrt(25/(t+5)^4+1)dt

Apply the substitution t+5=u:

L=int_4^6sqrt(1+25/u^4)du

For u in [4,6], 25/u^4<1. Take the series expansion of the square root:

L=int_4^6sum_(n=0)^oo((1/2),(n))(25/u^4)^ndu

Isolate the n=0 term:

L=int_4^6du+sum_(n=1)^oo((1/2),(n))25^nint_4^6u^(-4n)du

Integrate directly:

L=2+sum_(n=1)^oo((1/2),(n))25^n/(1-4n)[u^(1-4n)]_4^6

Hence

L=2+4sum_(n=1)^oo((1/2),(n))1/(4n-1)(5/16)^(2n)(1-(2/3)^(4n-1))