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

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Answer 1 · 2018-05-25T06:00:19

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

#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))#