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

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Answer 1 · 2015-12-25T19:56:18

#f(t) = (e^tln(t))/t#

#g(t) = t/(e^tln(t))#

#(df)/dt = (e^t((t-1)ln(t)+1))/t^2#

#(dg)/dt = (e^(-t)(ln(t)(1-t)-1))/ln^2(t)#

#Gamma(t) = int_2^4sqrt(((df)/dt)^2+((dg)/dt)^2)dt#

#Gamma(t) = int_2^4sqrt(((e^t((t-1)ln(t)+1))/t^2)^2 + ((e^(-t)(ln(t)(1-t)-1))/ln^2(t))^2)#

Which is #~~ 16.371300234854#

Note : Of course integral is not defined