Chapter 11.2, Problem 41E

(a)

To determine

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

Answer to Problem 41E

The required position of the particle is $\langle 3+\ln 4,-\frac{17}{10}\rangle$ .

Explanation of Solution

Given information:

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

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

Calculation:

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

The position of the particle at $t=0$ is $\langle 3,-2\rangle$ .

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:

  $\begin{array}{c}\langle {\displaystyle \underset{a}{\overset{b}{\int }}{\overrightarrow{v}}_1\left(t\right)dt},{\displaystyle \underset{a}{\overset{b}{\int }}{\overrightarrow{v}}_2\left(t\right)dt}\rangle =\langle {\displaystyle \underset{0}{\overset{3}{\int }}{\left(t+1\right)}^{-1}dt},{\displaystyle \underset{0}{\overset{3}{\int }}{\left(t+2\right)}^{-2}dt}\rangle \\ =\langle {\left(\ln \left(t+1\right)\right)|}_0^3,{\left(-{\left(t+2\right)}^{-1}\right)|}_0^3\rangle \\ =\langle \ln 4,\frac{3}{10}\rangle \end{array}$

It is seen that the particle is at $\langle 3,-2\rangle$ when $t=0$ and the displacement of the particle is $\langle \ln 4,\frac{3}{10}\rangle$ for $t=0$ to $t=3$ .

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

  $\begin{array}{c}\langle 3,-2\rangle +\langle \ln 4,\frac{3}{10}\rangle =\langle 3+\ln 4,-2+\frac{3}{10}\rangle \\ =\langle 3+\ln 4,-\frac{17}{10}\rangle \end{array}$

Hence, the required position of the particle is $\langle 3+\ln 4,-\frac{17}{10}\rangle$ .

(b)

To determine

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

Answer to Problem 41E

The required distance is $1.419$ .

Explanation of Solution

Given information:

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

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

Calculation:

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

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:

  $\begin{array}{c}{\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{{\left({\overrightarrow{v}}_1\left(t\right)\right)}^2+{\left({\overrightarrow{v}}_2\left(t\right)\right)}^2}}dt={\displaystyle \underset{0}{\overset{3}{\int }}\sqrt{{\left({\left(t+1\right)}^{-1}\right)}^2+{\left({\left(t+2\right)}^{-2}\right)}^2}}dt\\ \approx 1.419\end{array}$

Hence, the required distance is $1.419$ .