Chapter 6.3, Problem 67E

Solution

a)

To prove: the parametric equations for Jane’s path are $x_1=20\cos \left(\frac{\pi }{6}t\right)$ , $y_1=20+20\sin \left(\frac{\pi }{6}t\right)$ , $t\ge 0$ .

It is proved that the parametric equations for Jane’s path are $x_1=20\cos \left(\frac{\pi }{6}t\right)$ , $y_1=20+20\sin \left(\frac{\pi }{6}t\right)$ , $t\ge 0$ .

Given information:

The center of the wheel is $\left(0,20\right)$ and the radius of the wheel is 20 ft.

Speed of the wheel is 12 sec/revolution.

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 $x=h+r\cos \theta$ , $y=k+r\sin \theta$ .

Calculation:

Since the speed of the wheel is 12 sec per one revolution, which means 12 seconds per $2\pi$ radians. That is per second the angle making with horizontal is $\frac{2\pi }{12}=\frac{\pi }{6}$ .

Assume that $t\ge 0$ seconds will take Jane to receive the ball which is released by Ferris. Thus, in $t$ seconds, Jane making angle horizontal is $\frac{\pi }{6}t$ .

Substitute 0 for $h$ , 20 for $k$ , 20 for $r$ , and $\frac{\pi }{6}t$ for $\theta$ in the equations:

  $x=h+r\cos \theta$ , $y=k+r\sin \theta$

  $\begin{array}{c}\left(x,y\right)=\left(h+r\cos \theta ,k+r\sin \theta \right)\\ =\left(0+20\cos \left(\frac{\pi }{6}t\right),20+20\sin \left(\frac{\pi }{6}t\right)\right)\\ =\left(20\cos \left(\frac{\pi }{6}t\right),20+20\sin \left(\frac{\pi }{6}t\right)\right)\end{array}$

Thus, parametric equations for Jane’s path are $x_1=20\cos \left(\frac{\pi }{6}t\right)$ , $y_1=20+20\sin \left(\frac{\pi }{6}t\right)$ , $t\ge 0$ .

b)

To find:

The parametric equations for the path of the ball are $x_2=-30t+75$ , $y_2=-16t^2+30\sqrt{3}t$ , $t\ge 0$ .

Given information:

Eric releases the ball at the point $\left(75,0\right)$ with the speed 60 ft/sec and an angle of $120°$ with the horizontal.

Formula used:

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 $x=\left(v_0\cos \theta \right)t+x_0,$

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

Calculation:

Substitute 60 for $v_0$ , $120°$ for $\theta$ , 75 for $x_0$ , and 0 for $y_0$ in the parametric equations: $x_2=\left(v_0\cos \theta \right)t+x_0,$

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

  $\begin{array}{l}\left(x_2,y_2\right)=\left( \left(v_0\cos \theta \right)t+x_0,-16t^2+\left(v_0\sin \theta \right)t+y_0\right)\\ =\left(60\cos \left(120°\right)t+75,-16t^2+60\sin \left(120°\right)t+0\right)\\ =\left(60\cos \left(90°+30°\right)t+75,-16t^2+60\sin \left(90°+30°\right)t\right)\\ =\left(-60\sin \left(30°\right)t+75,-16t^2+60\cos \left(30°\right)t\right)\\ =\left(-30t+75,-16t^2+60\left(\frac{\sqrt{3}}{2}\right)t\right)\\ =\left(-30t+75,-16t^2+30\sqrt{3}t\right)\end{array}$

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

  $x_2=-30t+75$ , $y_2=-16t^2+30\sqrt{3}t$

c)

To find: determine whether Jane and the ball arrive at the point of intersection of the two paths at the same time by graphing simultaneously for two paths.

Jane and the ball will be close to each other, but not at the exact same point, at $t=2.2$ sec. The graphs shown in the figures (1), (2) and (3).

Given information:

From part (a), the path of the Jane is given as $x_1=20\cos \left(\frac{\pi }{6}t\right)$ , $y_1=20+20\sin \left(\frac{\pi }{6}t\right)$ , $t\ge 0$ .

From part (b), the path of the ball is given as $x_2=-30t+75$ , $y_2=-16t^2+30\sqrt{3}t$ , $t\ge 0$ .

Calculation:

For $t=2$ , 2.1 and 2.2, compute the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ .

$t$	2	2.1	2.2
$\left(x_1,y_1\right)=\left(20\cos \left(\frac{\pi }{6}t\right),20+20\sin \left(\frac{\pi }{6}t\right)\right)$	$\left(10,37.3205\right)$	$\left(9.0798,37.8202\right)$	$\left(8.1347,38.2709\right)$
$\left(x_2,y_2\right)=\left(-30t+75,-16t^2+30\sqrt{3}t\right)$	$\left(15,39.92\right)$	$\left(12,38.5592\right)$	$\left(9,36.8753\right)$

Observe that Jane and the ball will be close to each other, but not at the exact same point. By the time 2.2 seconds they cross each other. This can be seen in below figures (1), (2) and (3).

  

At $t=2$

Figure (1)

  

At $t=2.1$

Figure (2)

  

At $t=2.2$

Figure (3)

d)

To find: the distance $d\left(t\right)$ between Jane and ball at any time $t$ .

The distance $d\left(t\right)$ between Jane and ball at any time $t$ is

  $d\left(t\right)=\sqrt{{\left(20\cos \left(\frac{\pi }{6}t\right)+30t-75\right)}^2+{\left(20+20\sin \left(\frac{\pi }{6}t\right)+16t^2-30\sqrt{3}t\right)}^2}$

Given information:

From part (a), the path of the Jane is given as $x_1=20\cos \left(\frac{\pi }{6}t\right)$ , $y_1=20+20\sin \left(\frac{\pi }{6}t\right)$ , $t\ge 0$ .

From part (b), the path of the ball is given as $x_2=-30t+75$ , $y_2=-16t^2+30\sqrt{3}t$ , $t\ge 0$ .

Formula used:

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:

Substitute the values $x_1=20\cos \left(\frac{\pi }{6}t\right)$ , $y_1=20+20\sin \left(\frac{\pi }{6}t\right)$ , $x_2=-30t+75$ , and $y_2=-16t^2+30\sqrt{3}t$ in $d\left(t\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

  $\begin{array}{c}d\left(t\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(20\cos \left(\frac{\pi }{6}t\right)+30t-75\right)}^2+{\left(20+20\sin \left(\frac{\pi }{6}t\right)+16t^2-30\sqrt{3}t\right)}^2}\end{array}$

e)

To find: estimate the minimum distance between Jane and the ball by graphing $\left(t,d\left(t\right)\right)$

The minimum distance between Jane and ball occur at 2.2 seconds and the distance between them is 1.64 ft.

Given information:

From part (d), the distance between the Jane and the ball is $d\left(t\right)=\sqrt{{\left(\Delta x\right)}^2+{\left(\Delta y\right)}^2}$ , where

  $\Delta x=20\cos \left(\frac{\pi }{6}t\right)+30t-75$ and $\Delta y=20\sin \left(\frac{\pi }{6}t\right)+16t^2-30\sqrt{3}t+20$ .

Calculation:

By observing the graph (see figure (4)) 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 1.64 ft. Jane will have a good chance of catching the ball.

  

Figure (4)