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

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

Plugging in all the values we have,

\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}

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{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}

Plugging all the values we obtain,

\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}

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.