What is the arclength of #(ln(t+3),lnt/t)# on #t in [1,2]#?

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Answer 1 · 2018-07-09T09:46:49

#approx 0.4287742274#

We have

#x(t)=ln(t+3)# then

#x'(t)=1/(t+3)#

#y(t)=ln(t)/t#

then

#y'(t)=(1/t*t-ln(t))/t^2#

#y'(t)=(1-ln(t))/t^2#
and we have to integrate
#int_1^2sqrt((1/(t+3))^2+((1-ln(t))/t^2)^2)dt#
by a numerical method we get

#approx 0.4287742274#