Question

What is the arclength of `f(t) = (t^3-t^2+t,t-t^2)` on `t in [0,1]`?

Answer 1

Arc Length `=1.1404` (4dp)

Explanation:

The Arc Length for a Parametric Curve is given by

`L=int_alpha^beta sqrt((dx/dt)^2+(dy/dt)^2) \ dt`

So in this problem we have

`x=t^3-t^2+t => dx/dt = 3t^2-2t+1`
`y= t-t^2 \ \ \ \ \ \ \ \ => dy/dt = 1-2t`

So the Arc Length is;

`L=int_0^1 sqrt((3t^2-2t+1)^2+(1-2t)^2) \ dt`
`\ \ \= int_0^1 sqrt((9t^4 - 12t^3 + 10t^2 - 4t + 1) + (4t^2 - 4 t + 1)) \ dt`
`\ \ \= int_0^1 sqrt(9t^4 - 12t^3 + 14t^2 - 8t + 2) \ dt`

Which we can evaluate using Numerical Techniques to get:

`L= 1.14038999 ...`