Question 2.8.6: Modeling the Rate of Change of a Ballplayer’s Gaze One of th...
Modeling the Rate of Change of a Ballplayer’s Gaze
One of the guiding principles of most sports is to “keep your eye on the ball.” In baseball, a batter stands 2 feet from home plate as a pitch is thrown with a velocity of 130 ft/s (about 90 mph). Assuming that the ball only moves horizontally, at what rate does the batter’s angle of gaze need to change to follow the ball as it crosses home plate?
First, look at the triangle shown in Figure 2.44 (on the following page). We denote the distance from the ball to home plate by d and the angle of gaze by θ. Since the distance is changing with time, we write d = d(t). The velocity of 130 ft/s means that d′(t) = −130. [Why would d′(t) be negative?] From Figure 2.44, notice that
\theta(t)=\tan ^{-1}\left[\frac{d(t)}{2}\right].
The rate of change of the angle is then
\theta^{\prime}(t)=\frac{1}{1+\left[\frac{d(t)}{2}\right]^2} \frac{d^{\prime}(t)}{2}=\frac{2 d^{\prime}(t)}{4+[d(t)]^2} radians/second.
When d(t) = 0 (i.e., when the ball is crossing home plate), the rate of change is then
\theta^{\prime}(t)=\frac{2(-130)}{4}=-65 \text { radians/second }One problem with this is that most humans can accurately track objects only at the rate of about 3 radians/second. Keeping your eye on the ball in this case is thus physically impossible. (See Watts and Bahill, Keep Your Eye on the Ball.)
