Chapter 5.4, Problem 46E

a.

To determine

To find the times in the interval do the masses pass each other.

Answer to Problem 46E

Both mases will cross each other when $t$ is an integral multiple of $\pi$ is $t=k\pi .$

Explanation of Solution

Given position of two masses $s_1=\sin t,s_2=\sin \left(t+\frac{\pi }{3}\right).$

Calculation: It is given in the question that position of two masses is given by

  $s_1=\sin t,s_2=\sin \left(t+\frac{\pi }{3}\right),0\le t\le 2\pi$

Both masses will cross each other when

  $\begin{array}{l}{s}_1=s_2\\ \sin t=\sin \left(t+\frac{\pi }{3}\right)\\ \sin t=\sin t\cos \frac{\pi }{3}+\cos t\sin \frac{\pi }{3}\\ \sin t=\frac{1}{2}\sin t+\cos t\frac{\sqrt{3}}{2}\\ \frac{1}{2}\sin t=\cos t\frac{\sqrt{3}}{2}\\ \sin t=\sqrt{3}\cos t\\ \tan t=\sqrt{3}\\ t=\frac{\pi }{6}\end{array}$

It occurs for $t=\frac{\pi }{6}$ .

Thus, both masses will cross each other when $t$ is an integral multiple of $\pi$ is $t=\frac{\pi }{6}.$

b.

To determine

What is the farthest apart that the particles ever get.

Answer to Problem 46E

The distance is greatest at $t=\frac{4\pi }{3}$ sec and the distance is $\frac{3\sqrt{3}}{2}.$

Explanation of Solution

Given position of two masses $s_1=\sin t,s_2=\sin \left(t+\frac{\pi }{3}\right).$

Calculation:

Vertical distance between two bodies are,

  $\begin{array}{l}f\left(t\right)=s_1-s_2\\ =\sin \left(t+\frac{\pi }{3}\right)-\sin t\\ =\sin t\cos \frac{\pi }{3}+\cos t\sin \frac{\pi }{3}-\sin t\\ =\frac{1}{2}\sin t+\cos t\frac{\sqrt{3}}{2}-\sin t\\ f\left(t\right)=\cos t\frac{\sqrt{3}}{2}-\frac{1}{2}\sin t\\ f^{\prime }\left(t\right)=-\frac{\sqrt{3}}{2}\left(\sin t\right)-\frac{1}{2}\cos t\end{array}$

Critical points occur at

  $\begin{array}{l}{f}^{\prime }\left(t\right)=0\\ -\frac{\sqrt{3}}{2}\left(\sin t\right)-\frac{1}{2}\cos t=0\\ \tan t=\frac{1}{\sqrt{3}}\\ t=\frac{5\pi }{6},\frac{11\pi }{6}\end{array}$

It occurs at $t=\frac{5\pi }{6}$

  $\begin{array}{l}f\left(t\right)=\cos t\frac{\sqrt{3}}{2}-\frac{1}{2}\sin t\\ =-1\end{array}$

At $t=\frac{11\pi }{6}$

  $\begin{array}{l}f\left(t\right)=\cos t\frac{\sqrt{3}}{2}-\frac{1}{2}\sin t\\ =1\end{array}$

Here critical point and endpoint lies in $\left(0,\frac{\sqrt{3}}{2}\right),\left(2\pi ,\frac{\sqrt{3}}{2}\right),\left(\frac{5\pi }{6},-1\right),\left(\frac{4\pi }{3},1\right)$

Thus, the distance is greatest at $t=\frac{5\pi }{6},\frac{11\pi }{6}$ sec and the distance is $1m$ .

c.

To determine

Find the interval.

Answer to Problem 46E

So the distance is changed the fastest any particular time but for $t=\frac{\pi }{3},\frac{4\pi }{3}$ the distance is change faster than any other instant in time interval.

Explanation of Solution

Given position of two masses $s_1=\sin t,s_2=\sin \left(t+\frac{\pi }{3}\right).$

Calculation:

  $f^{\prime }\left(t\right)$ will maximum when ${f^{\prime }}^{\prime }\left(t\right)=0.$

  $\begin{array}{l}{f}^{\prime }\left(t\right)=-\frac{\sqrt{3}}{2}\left(\sin t\right)-\frac{1}{2}\cos t\\ {f^{\prime }}^{\prime }\left(t\right)=-\frac{\sqrt{3}}{2}\left(\cos t\right)+\frac{1}{2}\sin t\\ \sin t=\sqrt{3}\cos t\\ \tan t=\sqrt{3}\\ t=\frac{\pi }{3},\frac{4\pi }{3}\end{array}$

For $y=f^{\prime }\left(t\right)$ the critical point and endpoint lies in $\left(0,\frac{1}{2}\right),\left(2\pi ,-\frac{1}{2}\right),\left(\frac{\pi }{3},-1\right),\left(\frac{4\pi }{3},1\right)$

at the time $t=\frac{\pi }{3},\frac{4\pi }{3}$ two body pass each other and the graph of $y=\left|f\left(t\right)\right|$ has comer at these points, while $y=\frac{d\left|f\left(t\right)\right|}{dt}$ is not define at these time.

Thus, the distance is changed the fastest any particular time but for $t=\frac{\pi }{3},\frac{4\pi }{3}$ the distance is change faster than any other instant in time interval.