Find the position vector $\vec{r}(t)$ and the velocity vector $\vec{v}(t)$ of a particle that moves with acceleration vector
function $\vec{a}(t)=\langle 0,0,-10\rangle m/sec^{2}$, initial velocity $\vec{v}(0)=\langle 0,1,2\rangle$ m/sec and initial position
$\vec{r}(0)=\langle 0,0,3\rangle$ m.
Solution:
If the position vector (or trajectory) of the particle is $\vec{r}(t)$ then velocity vector (by definition) is $\vec{v}(t)=
\vec{r}'(t)$ and acceleration vector (by definition) is $\vec{a}(t)=\vec{r}''(t)$
By condition we have $\vec{r}''(t)=\langle 0,0,-10\rangle.$ Therefore, after first integration we will get
$\vec{r}'(t)=\langle 0,0,-10\rangle t + \vec{c}$
(where $\vec{c}$ is some constant vector). After one more times integration we will obtain
$$
\vec{r}(t) = \langle 0,0,-10\rangle \frac{t^{2}}{2}+\vec{c}t+\vec{d}
$$
where $\vec{d}$ again is some constant vector. Since $\vec{r}'(0)=\langle 0,1,2\rangle$ therefore $\vec{c}=\langle 0,1,2\rangle$.
Since $\vec{r}(0)=\langle 0,0,3\rangle$ therefore $\vec{d} = \langle 0,0,3\rangle$. So finally we have
Answer:
\begin{align*}
\vec{r}(t) &= \langle 0,0,-10\rangle \frac{t^{2}}{2}+\langle 0,1,2\rangle t+\langle 0,0,3\rangle\;\; \text{ - position vector}\\
\vec{r}'(t) &= \langle 0,0,-10\rangle t+\langle 0,1,2\rangle \;\; \text{ - velocity}
\end{align*}
Find the length of the indicated portion of the curve:
\begin{align*}
\vec{r}(t) =\vec{j} \cos^{3}t + \vec{k} \sin^{3}t , \; \; 0\leq t\leq \frac{\pi}{2}
\end{align*}
Solution:
\begin{align*}
\text{Length} = \int_{0}^{\pi/2}|\vec{r}'(t)|dt = \int_{0}^{\pi/2}\sqrt{(-3\cos^{2}t\sin t)^{2}+(3\sin^{2} t \cos t)^{2} }dt=\\
\int_{0}^{\pi/2}\sqrt{9\cos^{4}t\sin^{2} t+9\sin^{4} t \cos^{2} t }dt=\int_{0}^{\pi/2}3|\cos t \sin t |\sqrt{\cos^{2}t+\sin^{2} t }dt=\\
\int_{0}^{\pi/2}3\cos t \sin t dt = 3\int_{0}^{\pi/2}\sin t d\sin t = "(\text{Let } u=\sin t)" =
3 \int_{\sin(0)}^{\sin(\pi/2)}u du = 3 \int_{0}^{1}u du = \frac{3}{2}
\end{align*}