What is the arclength of #(1/(1-e^t),2t)# on #t in [2,4]#?

Question

1087 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-04T14:05:33

#approx 4.003334361#

We have

#x(t)=1/(1-e^t)#
and by the power and chain rule we get

#x'(t)=-1*(1-e^(t))^(-2)(-e^t)#

#y(t)=2t#

then

#y'(t)=2#

so our integral is given by

#int_2^4sqrt((e^t/(1-e^t))^2+4)dt#
by a numerical method we get

#approx 4.003334361#