Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 18

a.) By using Pythagorean Theorem, we have...
$x^2 + 90^2 = z^2$; when $x = 45$ft; $z = \sqrt{45^2 + 90^2} = 45 \sqrt{5}$ft


Taking the derivative with respect to time,
$\displaystyle 2x \frac{dx}{dt} + 0 = 2z \frac{dz}{dt}$

$
\begin{equation}
\begin{aligned}
x \frac{dx}{dt} &= z \frac{dz}{dt}\\
\\
\frac{dz}{dt} &= \frac{x}{z} \frac{dx}{dt}
\end{aligned}
\end{equation}
$

Plugging in all the values we have,

$
\begin{equation}
\begin{aligned}
\frac{dz}{dt} &= \frac{\cancel{45}}{\cancel{45}\sqrt{5}} (24)\\
\\
\frac{dz}{dt} &= \frac{24}{\sqrt{5}} \text{ or } \frac{24\sqrt{5}}{5} \frac{\text{ft}}{s}
\end{aligned}
\end{equation}
$


The distance of the battler from the second base is decreasing at a rate of $\displaystyle\frac{24\sqrt{5}}{5}\frac{\text{ft}}{s}$
b.)



Again, by using Pythagorean Theorem,
$x^2 + 90^2 = z^2$; when $x = 45$ft; $z = \sqrt{45^2 + 90^2} = 45 \sqrt{5}$ft
Taking the derivative with respect to time,

$
\begin{equation}
\begin{aligned}
0 + 2x \frac{dx}{dt} &= 2z \frac{dz}{dt}\\
\\
x \frac{dx}{dt} & = z \frac{dz}{dt}\\
\\
\frac{dz}{dt} & = \frac{x}{z} \frac{dx}{dt}
\end{aligned}
\end{equation}
$


Plugging all the values we obtain,


$
\begin{equation}
\begin{aligned}
\frac{dz}{dt} &= \frac{45}{45\sqrt{5}} (24)\\
\\
\frac{dz}{dt} &= \frac{24}{\sqrt{5}} \text{ or } \frac{24\sqrt{5}}{5} \frac{\text{ft}}{s}
\end{aligned}
\end{equation}
$

Thus shows that the distance of the batter from the third base is increasing at a rate equal to the decreasing rate of the batter's distance from the second base.