Chapter 7.1, Problem 7E

a.

To determine

To find: When the particle is moving to the right, when moving to the left and when it is stopped.

Answer to Problem 7E

The particle is moving to the right for $0\le t<\frac{\pi }{2}$ and $\frac{3\pi }{2}<t\le 2\pi$ , moving to the left for $\frac{\pi }{2}<t<\frac{3\pi }{2}$ , and stopped at $t=\frac{\pi }{2},\frac{3\pi }{2}$ .

Explanation of Solution

Given information:

The function $v(t)=e^{\sin t}\cos t$ and the interval $0\le t\le 2\pi$ .

Calculation:

Graph the given function over the given interval.

  

The graph of $v\left(t\right)$ intersects the $t$ -axis at $t=\frac{\pi }{2},\frac{3\pi }{2}$ . So, the particle stops at $t=\frac{\pi }{2},\frac{3\pi }{2}$ .

The velocity is positive for $0\le t<\frac{\pi }{2}$ and $\frac{3\pi }{2}<t\le 2\pi$ . So, the particle is moving to the right for $0\le t<\frac{\pi }{2}$ and $\frac{3\pi }{2}<t\le 2\pi$ .

The velocity is negative in the interval $\frac{\pi }{2}<t<\frac{3\pi }{2}$ . So, the particle moves to the left for $\frac{\pi }{2}<t<\frac{3\pi }{2}$ .

Conclusion:

The particle is moving to the right for $0\le t<\frac{\pi }{2}$ and $\frac{3\pi }{2}<t\le 2\pi$ , moving to the left for $\frac{\pi }{2}<t<\frac{3\pi }{2}$ , and stopped at $t=\frac{\pi }{2},\frac{3\pi }{2}$ .

b.

To determine

To find: The particle’s displacement over the given interval, and the final position of the particle if $s\left(0\right)=3$ .

Answer to Problem 7E

The displacement of the particle is 0. If $s\left(0\right)=3$ , the final position of the particle is 3.

Explanation of Solution

Given information:

The function $v(t)=e^{\sin t}\cos t$ and the interval $0\le t\le 2\pi$ .

Calculation:

Integrate the function $v(t)=e^{\sin t}\cos t$ over the interval $0\le t\le 2\pi$ to find the displacement of the particle.

  $\begin{array}{c}\text{Displacement}={\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(e^{\sin t}\cos t\right)dt}\\ ={\left[e^{\sin t}\right]}_0^{2\pi }\\ =\left[e^{\sin \left(2\pi \right)}-e^{\sin \left(0\right)}\right]\\ =e^0-e^0\\ =0\end{array}$

The displacement of the particle is 0.

If $s\left(0\right)=3$ , the final position of the particle will be:

  $\begin{array}{c}\text{Final position}=\text{initial position}+\text{ displacement}\\ =3+0\\ =3\end{array}$

Conclusion:

The displacement of the particle is 0. If $s\left(0\right)=3$ , the final position of the particle is 3.

c.

To determine

To find: The total distance travelled by the particle.

Answer to Problem 7E

The total distance traveled by the particle is $2\left(e-e^{-1}-1\right)$ .

Explanation of Solution

Given information:

The function $v(t)=e^{\sin t}\cos t$ and the interval $0\le t\le 2\pi$ .

Calculation:

Find the total distance traveled by the particle over the given interval as follows:

  $\begin{array}{c}\text{Distance traveled}={\displaystyle \underset{0}{\overset{2\pi }{\int }}\left|e^{\sin t}\cos t\right|dt}\\ ={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\left(e^{\sin t}\cos t\right)dt}-{\displaystyle \underset{\pi /2}{\overset{3\pi /2}{\int }}\left(e^{\sin t}\cos t\right)dt}+{\displaystyle \underset{3\pi /2}{\overset{2\pi }{\int }}\left(e^{\sin t}\cos t\right)dt}\\ ={\left[e^{\sin t}\right]}_0^{\pi /2}-{\left[e^{\sin t}\right]}_{\pi /2}^{3\pi /2}+{\left[e^{\sin t}\right]}_{3\pi /2}^{2\pi }\\ =\left(e^{\sin \left(\frac{\pi }{2}\right)}-e^{\sin \left(0\right)}\right)-\left(e^{\sin \left(\frac{3\pi }{2}\right)}-e^{\sin \left(\frac{\pi }{2}\right)}\right)+\left(e^{\sin \left(2\pi \right)}-e^{\sin \left(\frac{3\pi }{2}\right)}\right)\\ =\left(e^1-e^0\right)-\left(e^{-1}-e^1\right)+\left(e^0-e^{-1}\right)\\ =\left(e-1\right)-\left(\frac{1}{e}-e\right)+\left(1-\frac{1}{e}\right)\\ =2e-\frac{2}{e}-2\\ =2\left(e-e^{-1}-1\right)\end{array}$

Conclusion:

The total distance traveled by the particle is $2\left(e-e^{-1}-1\right)$ .