Question

What is the arclength of `f(t) = (lnt/t,ln(t+2))` on `t in [1,e]`?

Answer 1

Not a full solution, but...

Explanation:

This function is a parametric curve `f(t)=(x(t),y(t))`, where
`x(t)=lnt/t` and `y(t)=ln(t+2)`.

To measure its length we sould first use Pythagorean theorem to find an element of lenght `ds` (a curve is approximated with a series of small, infinitesimal segments)
`ds=sqrt(dx^2+dy^2)`

Now because `x` and `y` are functions of `t` we use chain rule
`ds=sqrt((dx/dt)^2 dt^2+(dy/dt)^2 dt^2)=sqrt((dx/dt)^2+(dy/dt)^2)dt`

The total lenght for `t in [1,e]` is an integral

`s=int_"curve"ds=int_1^e sqrt((dx/dt)^2+(dy/dt)^2)dt`

Let's find `dx/dt` and `dy/dt` first.

`x(t)=lnt*1/t`

By product rule
`dx/dt=1/t*1/t+lnt*1/t^2=(lnt+1)/t^2`

`y(t)=ln(t+2)t`

By chain rule
`dy/dt=1/(t+2)*1`

By plugging this in, we get

`s=int_1^e sqrt(((lnt+1)/t^2)^2+(1/(t+2))^2)dt`
`s=int_1^e sqrt((lnt+1)^2/t^4+1/(t+2)^2)dt`
Common denominator
`s=int_1^e sqrt((t+2)^2(lnt+1)^2+t^4)/(t^2(t+2))dt`
`s=int_1^e sqrt((t+2)^2ln^2(et)+t^4)/(t^2(t+2))dt`