미분법 전체 문제

Using the given information, we construct the figure shown below.

As pictured, we know that $5$ is the vertical height from the pulley to the level of the bow of the boat, and we let $z$ be the length of the rope from the pulley to the bow of the boat, and $x$ the horizontal distance from the dock to the bow of the boat.

We are given that the rope is being pulled in at $2$ feet per second, and thus $\frac{dz}{dt} = -2$ feet per second. Because we want to know how fast the boat is approaching the dock when the length of rope from bow to pulley is 13 feet, we want to know $\left. \frac{dx}{dt} \right|_{z=13}$. Thus, we need to relate the changing quantities $z$ and $x$.

Because the rope, the post (vertically extended), and the horizontal distance from the bow of the boat to the post on the dock form a right triangle at all times, it follows that

$$x^2 + 5^2 = z^2\,.$$ Having now related $z$ and $x$, we differentiate this equation with respect to $t$. By the chain rule, we now see that

$$2x \frac{dx}{dt} = 2z \frac{dz}{dt}\,.$$ At the instant $z = 13$, $x^2 + 5^2 = 13^2$, and thus $x = 12$. Using all of the given information at the instant $z = 13$ (including that $\frac{dz}{dt} = -2$),

$$2(12) \left. \frac{dx}{dt} \right|_{z = 13} = 2(13)(-2)\,.$$ Solving for $\left. \frac{dx}{dt} \right|_{z = 13}$,

$$\left. \frac{dx}{dt} \right|_{z = 13} = -\frac{13}{6}$$

feet per second. Thus the boat is approaching the dock at a rate of $\frac{13}{6} \approx 2.167$ feet per second.

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.