Runner's distance from home plate — related rate

Set up a right triangle with home plate, first base, and the runner's position as vertices, with the right angle at first base. Let $x$ be the runner's distance from first base and $z$ be the runner's distance from home plate; then $z^2 = x^2 + 90^2$ since the distance from home to first is fixed at 90 ft. We're given $dx/dt = 24$ ft/s and want $dz/dt$ when $x = 30$.

Differentiate both sides with respect to $t$: $2z\frac{dz}{dt} = 2x\frac{dx}{dt}$, so $\frac{dz}{dt} = \frac{x}{z}\frac{dx}{dt}$. When $x = 30$, $z = \sqrt{30^2 + 90^2} = \sqrt{900+8100} = \sqrt{9000} = 30\sqrt{10}$.

Substitute: $\frac{dz}{dt} = \frac{30}{30\sqrt{10}}(24) = \frac{24}{\sqrt{10}} \approx 7.59$ ft/s. Since this is positive, the runner's distance from home plate is increasing at about 7.59 ft/s at that instant.

Increasing. As the runner moves from first base toward second base, the distance x from first base grows, and since z equals the square root of x squared plus ninety squared, z grows too. This matches part a, where dz/dt came out positive, confirming the distance from home plate is increasing throughout this motion.