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

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Answer 1 · 2016-05-31T04:55:20

Approximately 68.8711...

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

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

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