Chapter 4.4, Problem 46E

a.

To determine

At what time in the interval $0\le t\le 2\pi$ do the particles meet?

Answer to Problem 46E

The particles meet when $t=\frac{\pi }{3},\frac{4\pi }{3}.$

Explanation of Solution

Given:

Position of one particle:- $s_1=\sin \left(t\right)$

Position of second particle:- $s_2=\sin \left(t+\frac{\pi }{3}\right)$

Here, $s_1,s_2$ are in meters and time $t$ is in seconds.

Calculation:

If both the particles meet then $s_1=s_2$ ,

  $\begin{array}{l}{s}_1=s_2\\ \sin \left(t\right)=\sin \left(t+\frac{\pi }{3}\right)\\ 0=\sin \left(t+\frac{\pi }{3}\right)-\sin \left(t\right)\\ \sin \left(t+\frac{\pi }{3}\right)-\sin \left(t\right)=0\\ \sin \left(t\right)\cos \left(\frac{\pi }{3}\right)+\cos \left(t\right)\sin \left(\frac{\pi }{3}\right)-\sin \left(t\right)=0\\ \sin \left(t\right)\cos \left(\frac{\pi }{3}\right)-\sin \left(t\right)+\cos \left(t\right)\sin \left(\frac{\pi }{3}\right)=0\\ \sin \left(t\right)\left(\cos \left(\frac{\pi }{3}\right)-1\right)+\cos \left(t\right)\sin \left(\frac{\pi }{3}\right)=0\\ \sin \left(t\right)\left(\frac{1}{2}-1\right)+\cos \left(t\right)\sin \left(\frac{\pi }{3}\right)=0\left(\because \cos \left(\frac{\pi }{3}\right)=\frac{1}{2}\right)\\ \sin \left(t\right)\left(\frac{-1}{2}\right)+\cos \left(t\right)\sin \left(\frac{\pi }{3}\right)=0\\ -\sin \left(t\right)\left(\frac{1}{2}\right)+\cos \left(t\right)\sin \left(\frac{\pi }{3}\right)=0\\ -\sin \left(t\right)\cos \left(\frac{\pi }{3}\right)+\cos \left(t\right)\sin \left(\frac{\pi }{3}\right)=0\left(\because \cos \left(\frac{\pi }{3}\right)=\frac{1}{2}\right)\\ \cos \left(t\right)\sin \left(\frac{\pi }{3}\right)-\sin \left(t\right)\cos \left(\frac{\pi }{3}\right)=0\\ \sin \left(t\right)\cos \left(\frac{\pi }{3}\right)-\cos \left(t\right)\sin \left(\frac{\pi }{3}\right)=0\\ \sin \left(t-\frac{\pi }{3}\right)=0\\ \left(t-\frac{\pi }{3}\right)=n\pi ;n\in \mathbb{Z}\\ t=n\pi +\frac{\pi }{3};n\in \mathbb{Z}\end{array}$

But, given that $0\le t\le 2\pi$ then

Case 1:- when $n=0$ ,

  $\begin{array}{l}t=n\pi +\frac{\pi }{3}\\ t=0+\frac{\pi }{3}=\frac{\pi }{3}\\ t=\frac{\pi }{3}.\end{array}$

Case 2:- when $n=1$ ,

  $\begin{array}{l}t=n\pi +\frac{\pi }{3}\\ t=\pi +\frac{\pi }{3}=\frac{4\pi }{3}<2\pi \\ t=\frac{4\pi }{3}.\end{array}$

Case 3:- when $n=2$ ,

  $\begin{array}{l}t=n\pi +\frac{\pi }{3}\\ t=2\pi +\frac{\pi }{3}=\frac{7\pi }{3}>2\pi \\ t=\frac{7\pi }{3}\notin \left[0,2\pi \right].\end{array}$

Conclusion:

The particles meet when $t=\frac{\pi }{3},\frac{4\pi }{3}.$

b.

To determine

What is the farthest apart that the particles ever get?

Answer to Problem 46E

Time $t=\frac{5\pi }{6},\frac{11\pi }{6}.$

Explanation of Solution

Given:

Position of one particle:- $s_1=\sin \left(t\right)$

Position of second particle:- $s_2=\sin \left(t+\frac{\pi }{3}\right)$

Calculation:

Distance between the particles:-

  $\begin{array}{l}d=s_1-s_2\\ =\sin \left(t\right)-\sin \left(t+\frac{\pi }{3}\right)\\ =\sin \left(t\right)-\left(\sin \left(t\right)\cos \left(\frac{\pi }{3}\right)+\cos \left(t\right)\sin \left(\frac{\pi }{3}\right)\right)\\ =\sin \left(t\right)\left(1-\cos \left(\frac{\pi }{3}\right)\right)-\cos \left(t\right)\sin \left(\frac{\pi }{3}\right)\\ =\sin \left(t\right)\left(1-\left(\frac{1}{2}\right)\right)-\cos \left(t\right)\sin \left(\frac{\pi }{3}\right)\left(\because \cos \left(\frac{\pi }{3}\right)=\frac{1}{2}\right)\\ =\sin \left(t\right)\left(\frac{1}{2}\right)-\cos \left(t\right)\sin \left(\frac{\pi }{3}\right)\\ =\sin \left(t\right)\cos \left(\frac{\pi }{3}\right)-\cos \left(t\right)\sin \left(\frac{\pi }{3}\right)\left(\because \cos \left(\frac{\pi }{3}\right)=\frac{1}{2}\right)\\ =\sin \left(t-\frac{\pi }{3}\right)\\ \text{So, }d=\sin \left(t-\frac{\pi }{3}\right)\end{array}$

Differentiate the $d$ with respect to $t$,

  $\begin{array}{l}d=\sin \left(t-\frac{\pi }{3}\right)\\ d\text{'}=\cos \left(t-\frac{\pi }{3}\right)\end{array}$

For maxima minima put $d\text{'}=0$ ,

  $\begin{array}{l}d\text{'}=0\\ \cos \left(t-\frac{\pi }{3}\right)=0\\ \left(t-\frac{\pi }{3}\right)=\left(2n+1\right)\frac{\pi }{2}\\ t=\left(2n+1\right)\frac{\pi }{2}+\frac{\pi }{3}\end{array}$

Case 1:- when $n=0$ ,

  $\begin{array}{l}t=\left(2n+1\right)\frac{\pi }{2}+\frac{\pi }{3}\\ t=\frac{\pi }{2}+\frac{\pi }{3}=\frac{5\pi }{6}\\ t=\frac{5\pi }{6}.\end{array}$

Case 2:- when $n=1$ ,

  $\begin{array}{l}t=\left(2n+1\right)\frac{\pi }{2}+\frac{\pi }{3}\\ t=\frac{3\pi }{2}+\frac{\pi }{3}=\frac{11\pi }{6}<2\pi \\ t=\frac{11\pi }{6}.\end{array}$

Conclusion:

Time $t=\frac{5\pi }{6},\frac{11\pi }{6}.$

c.

To determine

When in the interval $0\le t\le 2\pi$ is the distance between the particles changing the fastest?

Answer to Problem 46E

Time $t=\frac{\pi }{3},\frac{4\pi }{3}.$

Explanation of Solution

Given:

Position of one particle:- $s_1=\sin \left(t\right)$

Position of second particle:- $s_2=\sin \left(t+\frac{\pi }{3}\right)$

From part (b):- $d=\sin \left(t-\frac{\pi }{3}\right)$ and $d\text{'}=\cos \left(t-\frac{\pi }{3}\right)$ .

Calculation:

Differentiate the $d\text{'}$ with respect to $t$,

  $\begin{array}{l}d=\sin \left(t-\frac{\pi }{3}\right)\\ d\text{'}=\cos \left(t-\frac{\pi }{3}\right)\\ d\text{'}\text{'}=-\sin \left(t-\frac{\pi }{3}\right)\end{array}$

To find time when the distance between the particles changes fastest put $d\text{'}\text{'}=0$ ,

  $\begin{array}{l}d\text{'}\text{'}=0\\ -\sin \left(t-\frac{\pi }{3}\right)=0\\ \sin \left(\pi +t-\frac{\pi }{3}\right)=0\\ \sin \left(t+\frac{2\pi }{3}\right)=0\\ \left(t+\frac{2\pi }{3}\right)=n\pi ;n\in \mathbb{Z}\\ t=n\pi -\frac{2\pi }{3};n\in \mathbb{Z}\end{array}$

Case 1:- when $n=0$ ,

  $\begin{array}{l}t=n\pi -\frac{2\pi }{3}\\ t=0-\frac{2\pi }{3}=-\frac{2\pi }{3}\\ t=\frac{-2\pi }{3}\notin \left[0,2\pi \right].\end{array}$

Case 2:- when $n=1$ ,

  $\begin{array}{l}t=n\pi -\frac{2\pi }{3}\\ t=\pi -\frac{2\pi }{3}=\frac{\pi }{3}\in \left[0,2\pi \right]\\ t=\frac{\pi }{3}.\end{array}$

Case 3:- when $n=2$ ,

  $\begin{array}{l}t=n\pi -\frac{2\pi }{3}\\ t=2\pi -\frac{2\pi }{3}=\frac{4\pi }{3}\in \left[0,2\pi \right]\\ t=\frac{4\pi }{3}.\end{array}$

Conclusion:

Time $t=\frac{\pi }{3},\frac{4\pi }{3}.$