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

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Answer 1 · 2016-01-15T09:54:22

$s=2.86658$ units

Given $x(t)=sqrt(e^(t^2-t) +1)$ and $y(t)=t^3$

$dx/dt=((e^(t^2-t))(2t-1))/(2 sqrt(e^(t^2-t) +1)$

$dy/dt = 3 t^2$

$s = int_-1^1 sqrt((dx/dt)^2 + (dy/dt)^2) dt$

$s = int_-1^1 sqrt( (( ( e^(t^2-t) )(2t-1))/(2 sqrt(e^(t^2-t) +1)))^2 + (3 t^2)^2) dt$

the integral is complicated so Simpson's Rule is used:

so that $s=2.86658$

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