Find the distance traveled by a particle moving on the curve given by $\vec { r } ( t ) = \langle \sin ( 3 t ) , \cos ( 3 t ) , t \rangle$ as $t$ goes from $t = 0$ to $t = \pi$
A) $3 \pi$
B) $\pi \sqrt { 10 }$
C) $\pi \sqrt { 3 }$
D) $10 \pi$

Question

Quizplus · Accepted Answer

The distance traveled by a particle moving along a curve given by a vector function $\vec{r}(t)$ from $t = a$ to $t = b$ is found by integrating the magnitude of the velocity vector $\vec{v}(t) = \vec{r}'(t)$ over the interval $[a, b]$. The velocity vector is the derivative of the position vector, so first, we find $\vec{r}'(t)$:$$\vec{r}'(t) = \langle 3\cos(3t), -3\sin(3t), 1 \rangle$$The magnitude of $\vec{r}'(t)$ is:$$|\vec{r}'(t)| = \sqrt{(3\cos(3t))^2 + (-3\sin(3t))^2 + 1^2} = \sqrt{9\cos^2(3t) + 9\sin^2(3t) + 1}$$Using the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$, we simplify this to:$$|\vec{r}'(t)| = \sqrt{9(1) + 1} = \sqrt{10}$$The distance traveled from $t = 0$ to $t = \pi$ is:$$\int_{0}^{\pi} |\vec{r}'(t)| dt = \int_{0}^{\pi} \sqrt{10} dt = \sqrt{10} \cdot \pi = \pi\sqrt{10}$$

Find the Distance Traveled by a Particle Moving on the Curve