A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat

Question

A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast (in m/s) is the boat approaching the dock when it is 7 m from the dock? (Round your answer to two decimal places.)

Answer 1

Let x denote the horizontal distance (measured in meters) the boat is away from the dock (treat this a level surface)

Let y denote the vertical distance (measured in meters) above the surface of the water level.

Let S denote the length (measured in meters) of rope (from the pulley to the bow of the boat).

Let t denote the time (measured in seconds)

Assumptions: the dock is vertical and the water level is horizontal (or it is at 90 degrees with the dock) and the bow of the boat is at water level

Changing quantities with time t are the variables x, S, but y remains constant as the bow of the boat approaches the dock.

Rates of Change quantities dS/dt, dx/dt.

Given that dS/dt = -1 m/s (this is negative since S is decreasing as the boat approaches the dock)

Goal: Compute dx/dt when x = 7 m and y = 1 m and S = 5√2

Related Rates

Pythagoras: S² = x² + y² When y =1 m and x = 7 m S = 5√2

Differentiate with respect to t, but keep in mind that S and x are functions of time t, so remember the Chain Rule.

Hence,

2S dS/dt = 2x dx/dt or S dS/dt = x dx/dt

Now we solve for dx/dt

This gives dx/dt = (S dS/dt) / x

Finally, dx/dt = ( 5√2 (-1) ) / 7 ≈ -1.01 m/s (Remember this negative because x is decreasing as the bow of the boat approaches the dock!!!! This makes sense,)

However, we want the rate of this decrease which is ≈ 1.01 m/s (to two decimal places)