Problem 6 - Calculus

Question

A woman on a dock is pulling in a rope fastened to the bow of a small boat. If the woman's hands are 10 feet higher than the point where the rope is attached to the boat and if she is retrieving the rope at a rate of 2 feet per second, how fast is the boat approaching the dock when 25 feet of rope is still out?

Step-by-Step Solution

Answer

The boat is approaching the dock at approximately 2.18 ft/s.

Step 1: Understand the Problem

We have a right triangle formed by the dock, the rope, and the distance horizontally to the boat. The woman's hands' elevation creates a vertical component of this right triangle. We need to find out how fast the horizontal distance is changing.

Step 2: Set Up Known Values and Variables

Let \( x \) be the horizontal distance between the dock and the boat, \( y \) be the height difference of 10 feet, and \( z \) be the length of the rope. We know that \( z = 25 \) feet at the moment considered and \( \frac{dz}{dt} = -2 \) ft/s (as the rope is being pulled in, the length decreases).

Step 3: Use the Pythagorean Theorem

Since we have a right triangle, we can use the Pythagorean Theorem: \[ z^2 = x^2 + y^2 \]Substitute \( y = 10 \):\[ z^2 = x^2 + 10^2 \]

Step 4: Differentiate with Respect to Time

Differentiate both sides with respect to time \( t \):\[ 2z \frac{dz}{dt} = 2x \frac{dx}{dt} \]Since \( y \) is constant, its derivative is zero.

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

Re-arrange the differentiated equation to solve for \( \frac{dx}{dt} \):\[ \frac{dx}{dt} = \frac{z \cdot \frac{dz}{dt}}{x} \]

Step 6: Substitute Known Values to Find \( x \)

First, find \( x \) when \( z = 25 \):\[ 25^2 = x^2 + 10^2 \]\[ 625 = x^2 + 100 \]\[ x^2 = 525 \]\[ x = \sqrt{525} = 5\sqrt{21} \approx 22.91 \text{ feet} \]

Step 7: Calculate \( \frac{dx}{dt} \)

Substitute \( z = 25 \), \( \frac{dz}{dt} = -2 \), and \( x \approx 22.91 \) into the equation:\[ \frac{dx}{dt} = \frac{25 \cdot (-2)}{22.91} \]Calculate this value:\[ \frac{dx}{dt} \approx \frac{-50}{22.91} \approx -2.18 \text{ ft/s} \]

Step 8: Conclusion

The negative sign indicates the boat is moving towards the dock, at a rate of approximately 2.18 feet per second.

Key Concepts

Pythagorean theorem

The Pythagorean theorem is a fundamental concept in geometry that relates the lengths of the sides in a right triangle. This theorem is expressed as \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse, and \( a \) and \( b \) are the other two sides. In our problem, this theorem helps us relate the vertical height, the horizontal distance, and the length of the rope.

In the exercise, we create a scenario with a right triangle formed by the dock's height, the boat's horizontal distance, and the rope's length. The dock's height (10 feet) and the rope (25 feet) act as two sides of a triangle.

The horizontal distance \( x \) can be found by rearranging the equation to solve for \( x \), given that the other lengths are known:

• Plug in the known lengths into the equation: \( z^2 = x^2 + 10^2 \).
• Solve for \( x \) when \( z = 25 \) feet.

This gives us the equation \( 625 = x^2 + 100 \). Solving this equation lets us find how far the boat is horizontally from the dock when 25 feet of rope is out.

differentiation

Differentiation is a crucial concept in calculus that allows us to understand how a function changes over time. When dealing with related rates exercises, like our dock and boat problem, differentiation helps determine how the rate of one changing quantity relates to another.

Let's break it down with our example:

• We know that the rope's length \( z \) is changing at a rate of \( -2 \) feet per second, indicated by \( \frac{dz}{dt} \).
• To find \( \frac{dx}{dt} \), the rate at which the horizontal distance changes, we need to differentiate the equation \( z^2 = x^2 + 10^2 \) with respect to time \( t \).

This gives us the expression \( 2z \frac{dz}{dt} = 2x \frac{dx}{dt} \). Differentiation simplifies how we handle the changing dimensions, allowing us to deduce the pace at which the boat nears the dock as rope is reeled in.

By plugging in the known values for \( z \), \( \frac{dz}{dt} \), and \( x \), we can solve for \( \frac{dx}{dt} \), providing us with the rate of approach towards the dock.

right triangles

Right triangles are a specific type of triangle that includes a 90-degree angle. They are essential in problems that involve distances and are frequently associated with the Pythagorean theorem.

In our dock and boat scenario, we can visualize a right triangle:

• One side is the vertical height difference of 10 feet.
• The other side is the horizontal distance from the dock to the boat, which we denote as \( x \).
• The hypotenuse, the rope, is the third side with a length \( z \).

Understanding the properties of right triangles helps us model situations where components like height, distance, and length are key elements.

By representing the elements as a right triangle, we use geometric principles to relate these components that allow us to calculate unknown quantities when given certain measurements.
Keeping in mind these properties and the geometric relationships, we can easily utilize additional mathematical tools such as differentiation to find valuable information, like how quickly the horizontal distance \( x \) is changing over time.