Chapter 11.2, Problem 42E

(a)

To determine

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

Answer to Problem 42E

The required position of the particle is $\langle 15.586,24.586\rangle$ .

Explanation of Solution

Given information:

The velocity vector: $\overrightarrow{v}\left(t\right)=\langle e^t-t,e^t+t\rangle$

The position of the particle = $\left(1,1\right)$

Calculation:

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

The position of the particle at $t=0$ is $\langle 1,1\rangle$ .

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:

  $\begin{array}{c}\langle {\displaystyle \underset{a}{\overset{b}{\int }}{\overrightarrow{v}}_1\left(t\right)dt},{\displaystyle \underset{a}{\overset{b}{\int }}{\overrightarrow{v}}_2\left(t\right)dt}\rangle =\langle {\displaystyle \underset{0}{\overset{3}{\int }}{e}^t-tdt},{\displaystyle \underset{0}{\overset{3}{\int }}{e}^t+tdt}\rangle \\ =\langle {\left(e^t-\frac{t^2}{2}\right)|}_0^3,{\left(e^t+\frac{t^2}{2}\right)|}_0^3\rangle \\ =\langle 14.586,23.586\rangle \end{array}$

It is seen that the particle is at $\langle 1,1\rangle$ when $t=0$ and the displacement of the particle is $\langle 14.586,23.586\rangle$ for $t=0$ to $t=3$ .

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

  $\begin{array}{c}\langle 1,1\rangle +\langle 14.586,23.586\rangle =\langle 1+14.586,1+23.586\rangle \\ =\langle 15.586,24.586\rangle \end{array}$

Hence, the required position of the particle is $\langle 15.586,24.586\rangle$ .

(b)

To determine

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

Answer to Problem 42E

The required distance is $27.791$ .

Explanation of Solution

Given information:

The velocity vector: $\overrightarrow{v}\left(t\right)=\langle e^t-t,e^t+t\rangle$

The position of the particle = $\left(1,1\right)$

Calculation:

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

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:

  $\begin{array}{c}{\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{{\left({\overrightarrow{v}}_1\left(t\right)\right)}^2+{\left({\overrightarrow{v}}_2\left(t\right)\right)}^2}}dt={\displaystyle \underset{0}{\overset{3}{\int }}\sqrt{{\left(e^t-t\right)}^2+{\left(e^t+t\right)}^2}}dt\\ \approx 27.791\end{array}$

Hence, the required distance is $27.791$ .