Chapter 6.3, Problem 70E

Solution

To find: the minimum distance between the Chang and Kuan.

The minimum distance between the Chang and Kuan is approximately 10.48 ft.

Given information:

The center of the first wheel is $\left(0,20\right)$ and the second wheel is $\left(15,15\right)$ .

For one revolution ( $2\pi$ radians) First wheel takes 12 seconds and second wheel takes 8 seconds.

At $t=0$ Chang’s position is at $\left(20,20\right)$ , and Kuan’s positon is at $\left(15,0\right)$ , since Kuan position at 6 O’clock.

Formula used:

• The parametric equations for a point on the circle with centered $\left(h,k\right)$ having radius $r$ and making angle $\theta$ with the horizontal is $x=h+r\cos \theta$ , $y=k+r\sin \theta$ .
• If $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are two points. Then the distance between these points is $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Calculation:

The path of Chang’s position on the first wheel:

The angle covered by the first wheel per 1 sec is $\frac{2\pi }{12}=\frac{\pi }{6}$ radians. The path of the Chang position can be computed by substituting the values $\left(h,k\right)=\left(0,20\right)$ , $r=20$ and $\theta =\frac{\pi }{6}t$ in the parametric equations: $x=h+r\cos \theta$ , $y=k+r\sin \theta$

  $\begin{array}{c}\left(x_1,y_1\right)=\left(h+r\cos \theta ,k+r\sin \theta \right)\\ =\left(20\cos \left(\frac{\pi }{6}t\right),20+20\sin \left(\frac{\pi }{6}t\right)\right)\end{array}$

Thus, the path of the Chang’s position on the first wheel is given in parametric equations is

  $x_1=20\cos \left(\frac{\pi }{6}t\right)$ , $y_1=20+20\sin \left(\frac{\pi }{6}t\right)$ , $t\ge 0$ .

The path of Kuan’s position on the second wheel:

The angle covered by the second wheel per 1 sec is $\frac{2\pi }{8}=\frac{\pi }{4}$ radians. The path of Kuan position can be computed by substituting the values $\left(h,k\right)=\left(15,15\right)$ , $r=15$ and $\theta =\frac{\pi }{4}t$ in the parametric equations: $x=h+r\cos \theta$ , $y=k+r\sin \theta$ . But the initial Kuan’s position is at $\left(15,0\right)$ , which implies that $\theta =\frac{\pi }{4}t-\frac{\pi }{2}$

  $\begin{array}{c}\left(x_2,y_2\right)=\left(h+r\cos \theta ,k+r\sin \theta \right)\\ =\left(15+15\cos \left(\frac{\pi }{4}t-\frac{\pi }{2}\right),15+15\sin \left(\frac{\pi }{4}t-\frac{\pi }{2}\right)\right)\\ =\left(15+15\sin \left(\frac{\pi }{4}t\right),15-15\cos \left(\frac{\pi }{4}t\right)\right)\end{array}$

Thus, the path of the Kuan’s position on the second wheel is given in parametric equations is

  $x_2=15+15\sin \left(\frac{\pi }{4}t\right)$ , $y_2=15-15\cos \left(\frac{\pi }{4}t\right)$ , $t\ge 0$ .

Distance between Chang and Kuan at any time:

Substitute the values $x_1=20\cos \left(\frac{\pi }{6}t\right)$ , $y_1=20+20\sin \left(\frac{\pi }{6}t\right)$ , $x_2=15+15\sin \left(\frac{\pi }{4}t\right)$ , and $y_2=15-15\cos \left(\frac{\pi }{4}t\right)$ in $d\left(t\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

  $\begin{array}{c}d\left(t\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(15+15\sin \left(\frac{\pi }{4}t\right)-20\cos \left(\frac{\pi }{6}t\right)\right)}^2+{\left(-5-20\sin \left(\frac{\pi }{6}t\right)-15\cos \left(\frac{\pi }{4}t\right)\right)}^2}\\ =\sqrt{{\left(15+15\sin \left(\frac{\pi }{4}t\right)-20\cos \left(\frac{\pi }{6}t\right)\right)}^2+{\left(5+20\sin \left(\frac{\pi }{6}t\right)+15\cos \left(\frac{\pi }{4}t\right)\right)}^2}\end{array}$

Minimum distance between Chang and Kaun:

By observing the graph (see figure (1)) of the parametric equations $x_3=t$ , $y_3=d\left(t\right)$ , conclude that at $t=12.3$ the value of $d\left(t\right)$ is minimum which is approximately 10.48 ft.

  

Figure (1)