Question

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

Answer 1

Approximately 68.8711...

Explanation:

Arclength is defined by the integral

`int_a^bsqrt(((dx)/(dt))^2+((dy)/(dt))^2)dt`

In this case, `x=t-t^3`, `y=t+1`, `a=-2`, and `b=4`. Plugging these values into the formula gives

`int_-2^4sqrt((1-3t^2)^2+(1)^2)dt`

Now simplifying gives

`int_-2^4sqrt(9t^4-6t^2+2)dt ~~68.8711...`