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

Question

1404 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-28T05:22:35

#L=sqrt2ln(4/3)# units.

#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)#