Figure 5

A person standing at the end of a pier is docking a boat by pulling a rope at the rate of $2 \,{\rm m}/{\rm s}$. The end of the rope is $3 \,{\rm m}$ above water level. See Figure 5(a). How fast is the boat approaching the base of the pier when $5 \,{\rm m}$ of rope are left to pull in? (Assume the rope never sags, and that the rope is attached to the boat at water level.)

Solution

Step 1 Draw illustrations, like Figure 5, representing the problem. Label the sides of the triangle $3$ and $x$, and label the hypotenuse of the triangle $L.$

Step 2 $x$ is the distance (in meters) from the boat to the base of the pier, $L$ is the length of the rope (in meters), and the distance between the water level and the person's hand is $3 \, {\rm m}$. Both $x$ and $L$ are changing with respect to time, so $\dfrac{\textit{dx}}{\textit{dt}}$ is the rate at which the boat approaches the pier, and $\dfrac{\textit{dL}}{ \textit{dt}}$ is the rate at which the rope is pulled in.

We are given $\dfrac{\textit{dL}}{ \textit{dt}}=2 \,{\rm m}/{\rm s}$, and we seek $\dfrac{\textit{dx}}{\textit{dt}}$ when $L=5 \,{\rm m}$.

Step 3 The lengths $3$, $x$, and $L$ are the sides of a right triangle and, by the Pythagorean Theorem, are related by the equation $$\begin{equation*} x^{2}+3^{2}=L^{2}\tag{2} \end{equation*}$$

Step 4 We differentiate both sides of equation (2) with respect to $t.$ $$\begin{eqnarray*} 2x\frac{\textit{dx}}{\textit{dt}} &=&2L\frac{\textit{dL}}{\textit{dt}} \\[3pt] \frac{\textit{dx}}{\textit{dt}} &=&\dfrac{L}{x}\frac{\textit{dL}}{\textit{dt}}\qquad \color{#0066A7}{{\hbox{ Solve for }\tfrac{\textit{dx}}{\textit{dt}}.}} \end{eqnarray*}$$

Step 5 When $L=5$, we use equation (2) to find $x=4$. Since $\dfrac{\textit{dL}}{\textit{dt}}=2$, $$\begin{eqnarray*} \frac{\textit{dx}}{\textit{dt}} &=&\dfrac{5}{4}(2) \qquad \color{#0066A7}{{\hbox{\({L=5},{x=4}, \tfrac{\textit{dL}}{\textit{dt}} =2\)}}} \\[3pt] \frac{\textit{dx}}{\textit{dt}} &=&2.5 \end{eqnarray*}$$

When $5 \,{\rm m}$ of rope are left to be pulled in, the boat is approaching the pier at the rate of $2.5 \,{\rm m}/{\rm s}$.