Question

What is the arc length of `f(t)=((4-t^2)e^t,te^t)` over `t in [3,4]`?

Answer 1

Approximate arc length via a numerical method is:

`577.06` (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) = << (4-t^2)e^t, te^t >> \ \ \ t in [3,4]`

Differentiating the components wrt `t` we get:

`x'(t) = (4-t^2)(d/dte^t) + (d/dt(4-t^2))(e^t)`
`" " = (4-t^2)(e^t) + (-2t)(e^t)`
`" " = -(t^2+2t-4)e^t`

`y'(t) = (t)(d/dte^t) + (d/dt t)(e^t)`
`" " = te^t+e^t`
`" " = (t+1)e^t`

So, the arc length is given by:

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

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

`L = 577.055737868 ...`