Chapter 11.2, Problem 39E

(a)

To determine

To calculate: The position of the particle at time $t=3$ .

Answer to Problem 39E

The required position of the particle is $\langle 20,9\rangle$ .

Explanation of Solution

Given information:

The velocity vector: $\overrightarrow{v}\left(t\right)=\langle 3t^2-2t,1+\cos \pi t\rangle$

The position of the particle = $\left(2,3\right)$

Calculation:

The given velocity vector is $\overrightarrow{v}\left(t\right)=\langle 3t^2-2t,1+\cos \pi t\rangle$ .

Substitute 0 for t in the above expression.

  $\begin{array}{c}\overrightarrow{v}\left(0\right)=\langle 3{\left(0\right)}^2-2\left(0\right),1+\cos \pi \left(0\right)\rangle \\ =\langle 2,6\rangle \end{array}$

The position of the particle at $t=0$ is $\langle 2,6\rangle$ .

It is seen that the velocity of a particle moving along a path at any time t is $\overrightarrow{v}\left(t\right)=\langle {\overrightarrow{v}}_1\left(t\right),{\overrightarrow{v}}_2\left(t\right)\rangle$ .

Also, the displacement from $t=a$ and $t=b$ is $\langle {\displaystyle \underset{a}{\overset{b}{\int }}{\overrightarrow{v}}_1\left(t\right)dt},{\displaystyle \underset{a}{\overset{b}{\int }}{\overrightarrow{v}}_2\left(t\right)dt}\rangle$ .

Now, the displacement of the particle for the time $t=0$ to $t=3$ is:

  $\begin{array}{c}\langle {\displaystyle \underset{a}{\overset{b}{\int }}{\overrightarrow{v}}_1\left(t\right)dt},{\displaystyle \underset{a}{\overset{b}{\int }}{\overrightarrow{v}}_2\left(t\right)dt}\rangle =\langle {\displaystyle \underset{0}{\overset{3}{\int }}\left(3t^2-2t\right)dt},{\displaystyle \underset{0}{\overset{3}{\int }}\left(1+\cos \pi t\right)dt}\rangle \\ =\langle {\left(t^3-t^2\right)|}_0^3,{\left(t+\frac{1}{\pi }\sin \pi t\right)|}_0^3\rangle \\ =\langle 18,3\rangle \end{array}$

It is seen that the particle is at $\langle 2,6\rangle$ when $t=0$ and the displacement of the particle is $\langle 18,3\rangle$ for $t=0$ to $t=3$ .

Write the position of the particle at $t=3$ as:

  $\begin{array}{c}\langle 2,6\rangle +\langle 18,3\rangle =\langle 2+18,6+3\rangle \\ =\langle 20,9\rangle \end{array}$

Hence, the required position of the particle is $\langle 20,9\rangle$ .

(b)

To determine

To calculate: The distance that the particle travels from $t=0$ to $t=3$ .

Answer to Problem 39E

The required distance is 19.343.

Explanation of Solution

Given information:

The velocity vector: $\overrightarrow{v}\left(t\right)=\langle 3t^2-2t,1+\cos \pi t\rangle$

The position of the particle = $\left(2,3\right)$

Calculation:

The given velocity vector is $\overrightarrow{v}\left(t\right)=\langle 3t^2-2t,1+\cos \pi t\rangle$ .

It is seen that the distance that a particle travels from $t=a$ and $t=b$ is ${\displaystyle \underset{a}{\overset{b}{\int }}\left|\overrightarrow{v}\left(t\right)\right|dt}={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{{\left({\overrightarrow{v}}_1\left(t\right)\right)}^2+{\left({\overrightarrow{v}}_2\left(t\right)\right)}^2}}dt$ .

Now, distance traveled by the particle from time $t=0$ to $t=3$ is:

  $\begin{array}{c}{\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{{\left({\overrightarrow{v}}_1\left(t\right)\right)}^2+{\left({\overrightarrow{v}}_2\left(t\right)\right)}^2}}dt={\displaystyle \underset{0}{\overset{3}{\int }}\sqrt{{\left(3t^2-2t\right)}^2+{\left(1+\cos \pi t\right)}^2}}dt\\ \approx 19.343\end{array}$

Hence, the required distance is 19.343.