What is the arc length of f(t)=(ln(1/t) ,5-lnt) over t in [3,4] ?

1 Answer
Jun 28, 2018

L=sqrt2ln(4/3) units.

Explanation:

f(t)=(ln(1/t),5-lnt)=(-lnt,5-lnt)

f'(t)=(-1/t,-1/t)

Arc length is given by:

L=int_3^4sqrt(1/t^2+1/t^2)dt

Simplify:

L=sqrt2int_3^4 1/tdt

Integrate directly:

L=sqrt2[lnt]_3^4

Insert the limits of integration:

L=sqrt2ln(4/3)