Chapter 5, Problem 21CT

To determine

Tofind:The time when two people are at the same height from the ground on a Ferries wheel.

Explanation of Solution

Given information:

The equations of the heights of two people on Ferris wheel are $h_1=28\cos 10t+38$ , $h_2=28\cos 10\left(t-\frac{\pi }{6}\right)+38$ , $0\le t\le 2$ , where h in feet, and t in minutes.

Formulaused:

Equate the given heights. When $\cos \alpha =\cos \theta$ then $\alpha =2k\pi \pm \theta$ , $k=0,1,2\dots$ .

Calculations:

Two people will be on same height from the ground when $h_2=h_1$

  $\Rightarrow 28\cos 10t+38=28\cos 10\left(t-\frac{\pi }{6}\right)+38$

  $\Rightarrow 28\cos 10t=28\cos 10\left(t-\frac{\pi }{6}\right)+38-38$

  $\Rightarrow 28\cos 10t=28\cos 10\left(t-\frac{\pi }{6}\right)$

  $\Rightarrow \cos 10t=\cos 10\left(t-\frac{\pi }{6}\right)$

… (1)

The solutions of equation (1) are

  $10t=2k\pi \pm 10\left(t-\frac{\pi }{6}\right)$

For $k=0$ :

  $10t=2k\pi \pm 10\left(t-\frac{\pi }{6}\right)\Rightarrow 10t=\pm 10\left(t-\frac{\pi }{6}\right)$

  $\Rightarrow t=+\left(t-\frac{\pi }{6}\right)$ , that does not give any solution.

  $\Rightarrow t=-\left(t-\frac{\pi }{6}\right)$

  $\Rightarrow t=\frac{\pi }{12}=0.262\min .$

For $k=1$ :

  $10t=2k\pi \pm 10\left(t-\frac{\pi }{6}\right)\Rightarrow 10t=2\pi \pm 10\left(t-\frac{\pi }{6}\right)$

  $\Rightarrow 10t=2\pi -10\left(t-\frac{\pi }{6}\right)$ ,

other equation does not give any solution.

  $\Rightarrow 20t=2\pi +\frac{10\pi }{6}=\frac{22\pi }{6}$

  $\Rightarrow t=\frac{22\pi }{120}=0.576\min .$

For $k=2$ :

  $10t=2k\pi \pm 10\left(t-\frac{\pi }{6}\right)\Rightarrow 10t=4\pi \pm 10\left(t-\frac{\pi }{6}\right)$

  $\Rightarrow 10t=4\pi -10\left(t-\frac{\pi }{6}\right)$ ,

other equation does not give any solution.

  $\Rightarrow 20t=4\pi +\frac{10\pi }{6}=\frac{34\pi }{6}$

  $\Rightarrow t=\frac{34\pi }{120}=0.890\min .$

For $k=3$ :

  $10t=2k\pi \pm 10\left(t-\frac{\pi }{6}\right)\Rightarrow 10t=6\pi \pm 10\left(t-\frac{\pi }{6}\right)$

  $\Rightarrow 10t=6\pi -10\left(t-\frac{\pi }{6}\right)$ ,

other equation does not give any solution.

  $\Rightarrow 20t=6\pi +\frac{10\pi }{6}=\frac{46\pi }{6}$

  $\Rightarrow t=\frac{46\pi }{120}=1.20\min .$

For $k=4$ :

  $10t=2k\pi \pm 10\left(t-\frac{\pi }{6}\right)\Rightarrow 10t=8\pi \pm 10\left(t-\frac{\pi }{6}\right)$

  $\Rightarrow 10t=8\pi -10\left(t-\frac{\pi }{6}\right)$ ,

other equation does not give any solution.

  $\Rightarrow 20t=8\pi +\frac{10\pi }{6}=\frac{58\pi }{6}$

  $\Rightarrow t=\frac{58\pi }{120}=1.52\min .$

For $k=5$ :

  $10t=2k\pi \pm 10\left(t-\frac{\pi }{6}\right)\Rightarrow 10t=10\pi \pm 10\left(t-\frac{\pi }{6}\right)$

  $\Rightarrow 10t=10\pi -10\left(t-\frac{\pi }{6}\right)$ ,

other equation does not give any solution.

  $\Rightarrow 20t=10\pi +\frac{10\pi }{6}=\frac{70\pi }{6}$

  $\Rightarrow t=\frac{70\pi }{120}=1.83\min .$

For $k=6$ :

  $10t=2k\pi \pm 10\left(t-\frac{\pi }{6}\right)\Rightarrow 10t=12\pi \pm 10\left(t-\frac{\pi }{6}\right)$

  $\Rightarrow 10t=12\pi -10\left(t-\frac{\pi }{6}\right)$ ,

other equation does not give any solution.

  $\Rightarrow 20t=12\pi +\frac{10\pi }{6}=\frac{82\pi }{6}$

  $\Rightarrow t=\frac{82\pi }{120}=2.14\min .$

Since, $0\le t\le 2$ , therefore at times ( in minutes) $t=0.262,0.576,0.890,1.20,1.52$ , and 1.83 two people will be at same height.

Conclusion:

Therefore, at times (in minutes) $t=0.262,0.576,0.890,1.20,1.52$ , and 1.83 two people will be at same height.