Answer to Question #274209 in Calculus for chavindu

Question

A boat is pulled into a dock by means of a rope attached to a pulley on the dock, Figure 2.1. The rope is attached to the bow of the boat at a point 1 m below the pulley. If the rope is pulled through the pulley at a rate of 1 m/sec, at what rate will the boat be approaching the dock when 10 m of rope is out

Answer 1

We have a right triangle. The hypotenuse is the rope. One side drops vertically from the pulley to the water, and the other side is the horizontal path along which the bow of the boat travels.

Let $y$ be the length of the rope from the dock to the bow of the boat. Let $x$ be the horizontal distance from the dock to the boat.

Using the diagram, we find the relation between $x$ and $y$

$$y^2=x^2+1^2$$

Differentiate both sides with respect to $t$

$$\dfrac{d}{dt}(y^2)=\dfrac{d}{dt}(x^2+1)$$

$$2y\dfrac{dy}{dt}=2x\dfrac{dx}{dt}$$

Solve for

$$\dfrac{dx}{dt}=(\dfrac{y}{x})\dfrac{dy}{dt}$$

$$\dfrac{dx}{dt}=(\dfrac{y}{\sqrt{y^2-1}})\dfrac{dy}{dt}$$

Given $\dfrac{dy}{dt}=-1\ m/s$

When $y=10\ m$

$$\dfrac{dx}{dt}=(\dfrac{10}{\sqrt{10^2-1}})(-1\ m/s)=-\dfrac{10\sqrt{11}}{33}\ m/s$$

The boat will be approaching the dock at rate $\dfrac{10\sqrt{11}}{33}\ m/s$ when 10 m of rope is out.