Problem 4 - Active Calculus

Question

A sailboat is sitting at rest near its dock. A rope attached to the bow of the boat is drawn in over a pulley that stands on a post on the end of the dock that is 5 feet higher than the bow. If the rope is being pulled in at a rate of 2 feet per second, how fast is the boat approaching the dock when the length of rope from bow to pulley is 13 feet?

Step-by-Step Solution

Answer

The boat is approaching the dock at a rate of approximately -2.17 feet per second.

Step 1 - Understand the Problem

We need to find out how fast the sailboat is approaching the dock given the rate at which the rope is being pulled in and the fixed height difference between the dock and the bow.

Step 2 - Set Up the Variables and Diagram

Let the height of the post be 5 feet. Let the length of the rope from the bow to the pulley be represented by the variable \(l\). Let the horizontal distance from the boat to the dock be \(x\). We know that \( \frac{dl}{dt} = -2\) feet per second because the rope is being pulled in.

Step 3 - Apply the Pythagorean Theorem

Since the height of the post and the horizontal distance form a right triangle with the rope, we use the Pythagorean theorem: \[ x^2 + 5^2 = l^2 \] which simplifies to: \[ x^2 + 25 = l^2 \]

Step 4 - Differentiate with Respect to Time

Differentiate both sides of the equation \( x^2 + 25 = l^2 \) with respect to time (t): \[ 2x \frac{dx}{dt} = 2l \frac{dl}{dt} \] which simplifies to: \[ x \frac{dx}{dt} = l \frac{dl}{dt} \]

Step 5 - Solve for \( \frac{dx}{dt} \)

We're given \( l = 13 \) feet and \( \frac{dl}{dt} = -2 \) ft/s. We need to find \( x \) when \( l = 13 \). Using the Pythagorean theorem: \[ x^2 + 25 = 13^2 \] which simplifies to: \[ x^2 + 25 = 169 \] \[ x^2 = 144 \] \[ x = 12 \] feet.Substitute back into the differentiated equation: \[ 12 \frac{dx}{dt} = 13 \times (-2) \] \[ 12 \frac{dx}{dt} = -26 \] \[ \frac{dx}{dt} = -\frac{26}{12} \approx -2.17 \] feet per second.

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is fundamental in solving this related rates problem. It describes how the lengths of the sides of a right triangle are connected. The theorem states: \[ a^2 + b^2 = c^2 \] where \(a\) and \(b\) are the legs of the triangle, and \(c\) is the hypotenuse. In our exercise, the height of the post (\(5\) feet) and the horizontal distance from the boat to the dock (\(x\)) form the two legs, while the length of the rope (\(l\)) is the hypotenuse. To find the horizontal distance \(x\), we set up the equation: \[ x^2 + 5^2 = l^2 \]
This simplifies to:\[ x^2 + 25 = l^2 \] Using the known length of the rope (\(l = 13\) feet), we solve for \(x\): \[ x^2 + 25 = 13^2 \] \[ x^2 + 25 = 169 \] \[ x^2 = 144 \] \[ x = 12 \] feet. Understanding this step is crucial before moving on to the rate of change calculations.

Differentiation with Respect to Time

In our problem, as the sailboat moves, the lengths and distances change over time. To find the rate at which the boat approaches the dock, we need to differentiate the equation from the Pythagorean Theorem \( x^2 + 25 = l^2 \) with respect to time \( t \). This involves knowing how to use implicit differentiation. Let's differentiate both sides of the equation: \[ 2x \frac{dx}{dt} = 2l \frac{dl}{dt} \] Here,

• \(x\) is the horizontal distance from the boat to the dock,
• \(\frac{dx}{dt}\) is the rate at which this distance changes,
• \(l\) is the length of the rope,
• \(\frac{dl}{dt}\) is the rate at which the rope is being pulled in.

Since the rope is being pulled in at a rate of 2 feet per second, we have \( \frac{dl}{dt} = -2 \) feet per second (negative because the rope shortens). Incorporating these into our differentiated equation allows us to link the rates of change for different parts of the problem.

Solving for Rate of Change

Now that we have differentiated the equation with respect to time, our next step is to use the values provided to solve for \( \frac{dx}{dt} \) – the rate at which the boat is approaching the dock. From the earlier steps, we know:

• \( l = 13 \) feet,
• \( x = 12 \) feet,
• \( \frac{dl}{dt} = -2 \) feet per second.

Using the differentiated equation: \[ x \frac{dx}{dt} = l \frac{dl}{dt} \] we substitute the given values: \[ 12 \frac{dx}{dt} = 13 \times (-2) \] \[ 12 \frac{dx}{dt} = -26 \] Solving for \( \frac{dx}{dt} \) gives us: \[ \frac{dx}{dt} = -\frac{26}{12} \] This simplifies to: \[ \frac{dx}{dt} \thickapprox -2.17 \] feet per second. This means the boat is approaching the dock at an approximate rate of \( -2.17 \) feet per second. Note the negative sign indicates the distance is decreasing. Understanding these calculations is essential for solving similar related rates problems.