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Chapter 3: Problem 14

A man on a dock is pulling in a boat at the rate of \(50 \mathrm{ft} / \mathrm{min}\) by means of a rope attached to the boat at water level. If the man's hands are \(16 \mathrm{ft}\) above the water level, how fast is the boat approaching the dock when the amount of rope out is \(20 \mathrm{ft}\) ?

Short Answer

Expert verified
-83.33 ft/min

Step by step solution

01

Understand the problem and set up the relationship

We are given that a man pulls the boat with a rope at a rate of 50 ft/min. The rope forms a right triangle where the man's hands are 16 ft above the water. Let L be the length of the rope, y be the horizontal distance between the boat and the dock, and h = 16 ft be the vertical height. We want to find dy/dt when L = 20 ft.
02

Write the Pythagorean theorem for this scenario

The rope, the height, and the distance form a right triangle. From the Pythagorean theorem: \[ L^2 = h^2 + y^2 \] Given h = 16 ft, we have: \[ L^2 = 16^2 + y^2 \] \[ L^2 = 256 + y^2 \]
03

Differentiate with respect to time

Differentiate both sides of the Pythagorean equation with respect to time t: \[ 2L \frac{dL}{dt} = 2y \frac{dy}{dt} \] Since we are given \( \frac{dL}{dt} = -50 \text{ ft/min} \) (negative because the length of the rope is decreasing): \[ L \frac{dL}{dt} = y \frac{dy}{dt} \] Substitute L = 20 ft and solve for y from \[ L^2 = 256 + y^2 \].
04

Solve for y

We know that \( L = 20 \text{ ft} \) and \( L^2 = 256 + y^2 \): \[ 20^2 = 256 + y^2 \] \[ 400 = 256 + y^2 \] \[ y^2 = 144 \] y = 12 ft.
05

Substitute the values to find dy/dt

Use the relationship from differentiation: \[ L \frac{dL}{dt} = y \frac{dy}{dt} \] Substitute L = 20 ft, y = 12 ft, and \( \frac{dL}{dt} = -50 \) ft/min: \[ 20 \times (-50) = 12 \frac{dy}{dt} \] \[ -1000 = 12 \frac{dy}{dt} \] Solving for dy/dt: \[ \frac{dy}{dt} = \frac{-1000}{12} = -\frac{250}{3} \] \( \frac{dy}{dt} \approx -83.33 \text{ ft/min} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pythagorean theorem
The Pythagorean theorem is a fundamental principle in geometry. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Mathematically, it is represented as: \[ a^2 + b^2 = c^2 \]
In our problem, the scenario involves a right-angled triangle formed by the length of the rope (hypotenuse), the horizontal distance between the boat and the dock, and the vertical height of the man's hands above the water. Given that the height is constant at 16 ft, we can plug this into the Pythagorean theorem equation as follows: \[ L^2 = h^2 + y^2 \]
Here, \( L \) represents the length of the rope, and \( y \) represents the horizontal distance. By knowing one side of the triangle and the hypotenuse, we can solve for the unknown side.
differentiation with respect to time
Differentiation with respect to time is a powerful tool in calculus. When we differentiate an equation involving multiple variables with respect to time, we get the rate of change of those variables. In physics and engineering problems, this is crucial for understanding how different quantities change over time.
In our problem, we need to find out how fast the boat is approaching the dock as the man pulls the rope. We start with the Pythagorean theorem and differentiate it with respect to time \( t \): \[ 2L \frac{dL}{dt} = 2y \frac{dy}{dt} \]
Here, \( \frac{dL}{dt} \) is the rate at which the length of the rope changes (which is given), and \( \frac{dy}{dt} \) is what we need to find. This type of differentiation helps relate the rates of change of all variables involved.
calculus problem-solving
Solving related rates problems in calculus often involves several steps. It is essential to first understand the problem clearly, identify the relationship between the variables, and then use calculus techniques like differentiation to find the needed rates of change. Let's break down our problem step by step:
  • **Step 1**: Understand the problem and identify the variables. We have a rope, a boat, and a vertical height forming a right triangle.
  • **Step 2**: Apply the Pythagorean theorem to express the relationship between these variables: \( L^2 = h^2 + y^2 \).
  • **Step 3**: Differentiate with respect to time to establish a relationship between their rates of change: \( 2L \frac{dL}{dt} = 2y \frac{dy}{dt} \).
  • **Step 4**: Plug in the known values and solve the resulting equation for the desired rate \( \frac{dy}{dt} \).
By following these logical steps and carefully applying calculus principles, we can efficiently solve related rates problems and understand the dynamics of the situation.

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