관련 변화율

Let $x$ represent the distance from the runner to third base along the path from second to third, let $z$ be the distance from the umpire at home plate to the runner, and call $\theta$ the angle between the third baseline and the line of sight from the umpire to the runner. See the figure below.

Each of $x$, $z$, and $\theta$ is implicitly a function of $t$. We are given that the runner is advancing at a rate of $24$ feet per second, so $\frac{dx}{dt} = -24$. Since we want to know $\left. \frac{d\theta}{dt} \right|_{x = 30}$, we want to relate $x$ and $\theta$.

We first observe that by definition,

$$\tan(\theta) = \frac{x}{90} = \frac{1}{90}x\,.$$ Having related $x$ and $\theta$, we can differentiate both sides of the preceding equation with respect to $t$. By the chain rule,

$$\sec^2 (\theta) \frac{d\theta}{dt} = \frac{1}{90} \frac{dx}{dt}\,,$$ and thus

$$\frac{d\theta}{dt} = \frac{1}{90 \sec^2 (\theta)} \frac{dx}{dt}$$

At the instant $x = 30$, since $x^2 + 90^2 = z^2$, $z = \sqrt{90^2 + 30^2} = \sqrt{9000} = 30\sqrt{10}$. Hence, at this same instance, $\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{30\sqrt{10}}{90} = \frac{\sqrt{10}}{3}$. Recall we also know that $\frac{dx}{dt} = -24$. Applying this information to the equation that relates the rates of change of $x$ and $\theta$,

$$\left. \frac{d\theta}{dt} \right|_{x = 30} = \frac{1}{90 \left( \frac{\sqrt{10}}{3} \right)^2} (-24) = -\frac{1}{100}(24) = -0.24$$

radians per second.