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

1 Answer
Aug 3, 2017

Approximate arc length via a numerical method is:

# 0.79 # (2dp)

Explanation:

The arc length of a curve:

# vec(r) (t) = << x(t), y(t) >> #

Over an interval #[a,b]# is given by:

# L = int_a^b \ || vec(r) (t) || \ dt #
# \ \ = int_a^b \ sqrt(x'(t)^2 +y'(t)^2) \ dt #

So, for the given curve:

# vec( r )(t) = << lnt/t,lnt >> \ \ \ t in [1,2] #

Differentiating the components wrt #t# we get:

# x'(t) = ( (t)(d/dt lnt) - (lnt)(d/dt t) ) / (t^2) #
# " " = ( (t)(1/t) - (lnt)(1) ) / (t^2) #
# " " = (1-lnt)/t^2 #

# y'(t) = 1/t #

So, the arc length is given by:

# L = int_1^2 \ sqrt( x'(t)^2+y'(t)^2 ) \ dt #
# \ \ = int_1^2 \ sqrt( ((1-lnt)/t^2)^2 + (1/t)^2 ) \ dt #
# \ \ = int_1^2 \ sqrt( ((1-lnt)/t^2)^2 + 1/t^2 ) \ dt #

The integral does not have an elementary antiderivative,and so we evaluate the definite integral al numerically:

# L = 0.7890476183035367 ... #