Question

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

Answer 1

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 ...`