What is the arc length of teh curve given by $r (t) = (2 t, 2 e^{- t}, e^{t})$ $r(t)=(2t , 2e^(-t) , e^(t) )$ in the interval $t \in [0, 3]$ $t in [0,3]$ ?

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Answer 1 · 2018-07-10T18:30:26

1+e^3-2e^-3#

The parametric equation for the curve is

$x(t) = 2t qquadimplies dx = 2dt$
$y(t) = 2e^-t implies dy = -2e^-t dt$
$z(t) = e^t \ quad implies dz = e^t dt$

Thus

$ds = sqrt{dx^2+dy^2+dz^2}$
$qquad = sqrt{4+4e^{-2t}+e^{2t}}dt$
$qquad = sqrt{(e^t+2e^-t)^2}dt$
$qquad = (e^t+2e^-t)dt$

Hence, the required length is

$int ds = int_0^3 (e^t+2e^-t)dt$
$qquadquad =[e^t-2e^-t]_0^3$
$quadqquad = (e^3-2e^-3)-(e^0-2e^0)$
$qquadquad =1+e^3-2e^-3$

What is the arc length of teh curve given by r(t)=(2t , 2e^(-t) , e^(t) )r(t)=(2t,2e−t,et)r(t)=(2t , 2e^(-t) , e^(t) ) in the interval t in [0,3]t∈[0,3]t in [0,3]?