Chapter 6.3, Problem 68E

Solution

To find: the minimum distance between the ball and Matthew

The minimum distance between the ball and Matthew is approximately 3.47 ft.

Given information:

The center of the wheel is $\left(0,71\right)$

Wheel turns one revolution ( $2\pi$ radians) every 20 seconds.

Tony stands at the point $\left(90,0\right)$ .

Tony throws a ball towards wheel with an initial velocity 88 ft/sec at an angle with the horizontal of $100°$ .

Mathew position at the time $t=0$ is $\left(71,71\right)$ .

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 $x=h+r\cos \theta$ , $y=k+r\sin \theta$ .
• Suppose that an object is thrown from a point $\left(x_0,y_0\right)$ with an initial speed of $v_0$ ft/sec at an angle $\theta$ with the horizontal. The path of the object is modeled by parametric equations $x=\left(v_0\cos \theta \right)t+x_0,$ $y=-16t^2+\left(v_0\sin \theta \right)t+y_0$ .
• 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 Mathew’s position on the wheel:

The angle covered by the wheel per 1 sec is $\frac{2\pi }{20}=\frac{\pi }{10}$ radians. The path of the Mathew position can be computed by substituting the values $\left(h,k\right)=\left(0,71\right)$ , $r=71$ and $\theta =\frac{\pi }{10}t$ in the parametric equations: $x=h+r\cos \theta$ , $y=k+r\sin \theta$

  $\begin{array}{c}\left(x_1,y_1\right)=\left(h+r\cos \theta ,k+r\sin \theta \right)\\ =\left(71\cos \left(\frac{\pi }{10}t\right),71+71\sin \left(\frac{\pi }{10}t\right)\right)\end{array}$

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

  $x_1=71\cos \left(\frac{\pi }{10}t\right)$ , $y_1=71+71\sin \left(\frac{\pi }{10}t\right)$ , $t\ge 0$ .

The path of the ball:

Substitute 88 for $v_0$ , $100°$ for $\theta$ , 90 for $x_0$ , and 0 for $y_0$ in the parametric equations: $x_2=\left(v_0\cos \theta \right)t+x_0,$

  $y_2=-16t^2+\left(v_0\sin \theta \right)t+y_0$

  $\begin{array}{l}\left(x_2,y_2\right)=\left( \left(v_0\cos \theta \right)t+x_0,-16t^2+\left(v_0\sin \theta \right)t+y_0\right)\\ =\left(88\cos \left(100°\right)t+90,-16t^2+88\sin \left(100°\right)t+0\right)\\ =\left(88\left(-0.174\right)t+90,-16t^2+88\left(0.985\right)t\right)\\ =\left(-15.312t+90,-16t^2+86.68t\right)\end{array}$

Thus, the parametric equations for the path of the ball are

  $x_2=-15.31t+90$ , $y_2=-16t^2+86.68t$

Distance between Mathew and ball at any time:

Substitute the values $x_1=71\cos \left(\frac{\pi }{10}t\right)$ , $y_1=71+71\sin \left(\frac{\pi }{10}t\right)$ , $x_2=-15.31t+90$ , and $y_2=-16t^2+86.68t$ in $d\left(t\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

  $\begin{array}{c}d\left(t\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(71\cos \left(\frac{\pi }{10}t\right)+15.31t-90\right)}^2+{\left(71\sin \left(\frac{\pi }{6}t\right)+16t^2-86.68t+71\right)}^2}\end{array}$

Minimum distance between the Mathew and the ball:

By observing the graph (see figure (1)) of the parametric equations $x_3=t$ , $y_3=d\left(t\right)$ , conclude that at $t=2.2$ the value of $d\left(t\right)$ is minimum which is approximately 3.47 ft.

  

Figure (1)