What is the arclength of $(9t^3-2t^2+15,-2t^2-12t-1)$ on $t in [-2,3]$ ?

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Answer 1 · 2016-08-28T13:35:07

$approx 325$

Arc length is the time integral of speed, so $s = int_(-2)^3 dot s dt$

and speed $dot s = sqrt(vec v * vec v)$ where $vec v$ is the velocity vector

$vec v = d/dt vec r$

$= d/dt ((9t^3 - 2t^2 + 15),(-2t^2 - 12t - 1))$

$= ((27t^2 - 4t ),(-4t - 12))$

So $s = int_(-2)^3 sqrt(vec v * vec v) dt$

$= int_(-2)^3 sqrt(((27t^2 - 4t ),(-4t - 12)) * ((27t^2 - 4t ),(-4t - 12))) dt$

$= int_(-2)^3 sqrt( (27t^2 - 4t )^2 + (4t + 12)^2) dt$

$approx 325$ units according to computer solution

What is the arclength of (9t^3-2t^2+15,-2t^2-12t-1)(9t^3-2t^2+15,-2t^2-12t-1) on t in [-2,3]t in [-2,3]?