Chapter 5, Problem 56RE

a.

To determine

To find the farthest apart the particles.

Answer to Problem 56RE

The farthest apart the particles ever get $D=0.765$ .

Explanation of Solution

Given equation $D=\cos t-\cos \left(t+\frac{\pi }{4}\right).$

Calculation:

Let $D=s_1-s_2$ be the distance between the two particles.

From the given, $D=\cos t-\cos \left(t+\frac{\pi }{4}\right)$

Derivative with respect to $t$ on both sids.

  $\begin{array}{l}{D}^{\prime }=-\sin t+\sin \left(t+\frac{\pi }{4}\right)\\ D^{\prime }=\frac{1}{\sqrt{2}}\left(-\sqrt{2}\sin t+\cos t+\sin t\right)\\ D^{\prime }=0\\ \frac{1}{\sqrt{2}}\left(-\sqrt{2}\sin t+\cos t+\sin t\right)=0\\ -\sqrt{2}\sin t+\cos t+\sin t=0\\ Tant=\frac{1}{\sqrt{2}-1}\\ t=\frac{3\pi }{8}\end{array}$

Plug $t=\frac{3\pi }{8}$ value in $D=\cos t-\cos \left(t+\frac{\pi }{4}\right)$

  $\begin{array}{l}D=\cos \frac{3\pi }{8}-\cos \left(\frac{3\pi }{8}+\frac{\pi }{4}\right)\\ D=0.765\end{array}$

Thus the farthest apart the particles ever get $D=0.765$ .

b.

To determine

To find the particles collide.

Answer to Problem 56RE

The particle collide is $t=-\frac{\pi }{8}+n\pi ,\text{Where is }n\text{ any integer}.$

Explanation of Solution

Given equation $D=\cos t-\cos \left(t+\frac{\pi }{4}\right).$

Calculation:

From the given, collide when $s_1=s_2$

  $\begin{array}{l}\cos t=\cos \left(t+\frac{\pi }{4}\right)\\ \cos t=\frac{1}{\sqrt{2}}\left(\cos t-\sin t\right)\\ \sqrt{2}\cos t-\cos t=-\sin t\\ 1-\sqrt{2}=\tan t\\ t=-\frac{\pi }{8}+n\pi ,\text{Where is }n\text{ any integer}.\end{array}$

Thus the particle collide is $t=-\frac{\pi }{8}+n\pi ,\text{Where is }n\text{ any integer}.$