What is the arclength of #(t/(5-t),8/t)# on #t in [1,4]#?

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Answer 1 · 2018-07-06T08:56:11

#approx 8.248414706#

we have
#x(t)=t/(5-t)# so #x'(t)=5/(t-5)^2#

#y(t)=8/t#

#y'(t)=-8/t^2#

so we have to calculate

#int_1^4 sqrt((5/(t-5)^2)^2+(-8/t^2)^2)dt#
this leads to an elliptic integral

with a numerical approach we get

#8.248414706#