What is the arclength of (t/(t+5),t) on t in [-1,1]?
1 Answer
May 25, 2018
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
L=int_4^6sqrt(1+25/u^4)du
For
L=int_4^6sum_(n=0)^oo((1/2),(n))(25/u^4)^ndu
Isolate the
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))