Question

What is the arclength of `(t-2t^3,4t+1)` on `t in [-2,4]`?

Answer 1

Approximate arc length via a numerical method is:

`146.64` (2dp)

Explanation:

The arc length of a curve:

`vec(r) (t) = << f(t), g(t)) >>`

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

`L = int_a^b \ || vec(r) (t) || \ dt`
`\ \ = int_a^b \ sqrt(f'(t)^2 +g'(t)^2) \ dt`

So, for the given curve:

`vec(r)(t) = (t-2t^3, 4t+1) \ \ \ t in [-2,4]`

the arc length is given by:

`L = int_(-2)^4 \ sqrt( (d/dt(t-2t^3))^2 + (d/dt(4t+1))^2 ) \ dt`
`\ \ = int_(-2)^4 \ sqrt( (1-6t^2)^2 + (4)^2 ) \ dt`
`\ \ = int_(-2)^4 \ sqrt( 1-12t^2+36t^4 + 16 ) \ dt`
`\ \ = int_(-2)^4 \ sqrt( 36t^4-12t^2 + 17 ) \ dt`

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

`L = 146.64637228 ...`