Answer
The rate at which the distance between the two players is changing is $ - 26.8$ ft/s.
Work Step by Step
From Figure 7, we see that the distance between the two players $d$ is given by
$d = \sqrt {{x^2} + {y^2}} $
The partial derivatives are
$\frac{{\partial d}}{{\partial x}} = \frac{x}{{\sqrt {{x^2} + {y^2}} }}$, ${\ \ \ }$ $\frac{{\partial d}}{{\partial y}} = \frac{y}{{\sqrt {{x^2} + {y^2}} }}$
Thus, the rate at which the distance between the two players is changing:
$\frac{{\partial d}}{{\partial t}} = \frac{{\partial d}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial d}}{{\partial y}}\frac{{dy}}{{dt}}$
$\frac{{\partial d}}{{\partial t}} = \frac{x}{{\sqrt {{x^2} + {y^2}} }}\frac{{dx}}{{dt}} + \frac{y}{{\sqrt {{x^2} + {y^2}} }}\frac{{dy}}{{dt}}$
At a moment when the hitter is 8 ft from first base and the first baseman is 6 ft from first base, we are given the speeds:
$\frac{{dx}}{{dt}}{|_{\left( {x,y} \right) = \left( {6,8} \right)}} = 18$, ${\ \ }$ $\frac{{dy}}{{dt}}{|_{\left( {x,y} \right) = \left( {6,8} \right)}} = 20$
Substituting these in $\frac{{\partial d}}{{\partial t}}$ gives
$\frac{{\partial d}}{{\partial t}}{|_{\left( {x,y} \right) = \left( {6,8} \right)}} = \frac{6}{{\sqrt {{6^2} + {8^2}} }}\cdot18 + \frac{8}{{\sqrt {{6^2} + {8^2}} }}\cdot20 \simeq 26.8$
Since both players are running toward the same first base $P$, we need to assign a negative sign to the rate. Thus, the rate at which the distance between the two players is changing is $ - 26.8$ ft/s.
