What is the arc length of #f(t)=(ln(1/t) ,5-lnt) # over #t in [3,4] #?
1 Answer
Jun 28, 2018
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)#