Question 2 Find the arc length (s) of the curve \( \vec{r}(t)=\langle 4 \cos (t), 4 \sin (t), 3 t\rangle \) for \( -4 \leq t \leq 5 \) \( s= \)

To find the arc length \( s \) of the curve defined by \( \vec{r}(t) = \langle 4 \cos(t), 4 \sin(t), 3t \rangle \) from \( t = -4 \) to \( t = 5 \), we start by calculating the derivative \( \vec{r}'(t) \): \[ \vec{r}'(t) = \langle -4 \sin(t), 4 \cos(t), 3 \rangle. \] Next, we find the magnitude of this derivative: \[ |\vec{r}'(t)| = \sqrt{(-4 \sin(t))^2 + (4 \cos(t))^2 + 3^2} = \sqrt{16 \sin^2(t) + 16 \cos^2(t) + 9} = \sqrt{16 + 9} = \sqrt{25} = 5. \] Now, we can compute the arc length using the formula: \[ s = \int_{-4}^{5} |\vec{r}'(t)| \, dt = \int_{-4}^{5} 5 \, dt = 5 \times (5 - (-4)) = 5 \times 9 = 45. \] Thus, the arc length \( s = 45 \).