A particle travels in the xy-plane so that the position is given by x(t) = sin(t) and y(t) = e-3t - 1. What is the total distance traveled by the particle for 0 ≤ t ≤ 3?

Question

anonymous · Answer

i think it use integral

anonymous · Answer

yeah i know the formula is total distance is teh integral of velocity from 0 to 3 but idk what to do next

anonymous · Answer

$$\Delta x= \int\limits\limits_{0}^{3}\sin (t)$$
$$=\left[ -\cos(t) \right] $$
=-cos(3)-(-cos(0))

anonymous · Answer

$$\Delta y=\int\limits_{0}^{3}e-3t-1$$
$$=et-3/2 t ^{2}-t$$ 
={e*3-(3/2)*3^2-3}-{0}

anonymous · Answer

e=2.718

anonymous · Answer

$$
\int_0^3 \sqrt{ x'(t)^2 + y'[t]^2} dt
$$

anonymous · Answer

Is 
$$ y(t) =e ^{-3t} -1
$$
or $$
y(t) =e ^{-3t  -1}
$$

anonymous · Answer

In either case you end up with a messy integral that needs some numerical approximation.