Question

What is the arclength of `(1/(t+te^t),-t)` on `t in [3,4]`?

Answer 1

Approximate arc length via a numerical method is:

`1.00` (2dp)

Explanation:

The arc length of a curve:

`bb(vec(r)) (t) = << x(t), y(t) >>`

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

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

So, for the given curve:

`bb(vec( r ))(t) = << 1/(t+te^t),-t >> \ \ \ t in [3,4]`

Differentiating the components wrt `t` we get:

`x'(t) = -(t+te^t)^(-2)d/dt(t+te^t)`
`" " = -(t+te^t)^(-2)(1+(t)(e^t)+(1)(e^t))`
`" " = -((1+e^t+te^t))/(t+te^t)^(2)`

`y'(t) = -1`

So, the arc length is given by:

`L = int_3^4 \ sqrt( x'(t)^2+y'(t)^2 ) \ dt`
`\ \ = int_3^4 \ sqrt( (-((1+e^t+te^t))/(t+te^t)^(2))^2+(-1)^2 ) \ dt`
`\ \ = int_3^4 \ sqrt( 1+ (1+e^t+te^t)^2/(t+te^t)^4) \ dt`

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

`L = 1.000072796710095 ...`