At time \(t\geq 0\), a particle moving in the \(xy\)-plane has velocity vector given by \(v(t)= \Big\langle 4e^{-t}, \sin{(1+\sqrt{t})} \Big\rangle \). What is the total distance the particle travels between \(t=1\) and \(t=3\) ?

Summary Submit

Recall that when given parametric equations, arc length can be calculated with the following:

$$ L = \int\limits_a^b \sqrt{[x'(t)]^2 + [y'(t)]^2}   dt $$

Since we are given the velocity vector \((x'(t) \text{ and } y'(t))\), we can use our calculator to evaluate the following integral.

$$ L = \int\limits_1^3 \sqrt{(4e^{-t})^2 + (\sin{(1+\sqrt{t})})^2 }   dt $$ $$ L = \int\limits_1^3 \sqrt{16e^{-2t} + \sin^2{(1+\sqrt{t})} }   dt $$ $$ = \boxed{1.861} $$