Problem 8 - The Calculus with Analytic Geometry

Question

A man \(6 \mathrm{ft}\) tall is walking toward a building at the rate of \(5 \mathrm{ft} / \mathrm{sec}\). If there is a light on the ground \(50 \mathrm{ft}\) from the building, how fast is the man's shadow on the building growing shorter when he is \(30 \mathrm{ft}\) from the building?

Step-by-Step Solution

Answer

The man's shadow decreases in length at a rate of approximately 2.5 feet per second.

Step 1: Define Variables and Relationships

Let the distance between the man and the building be denoted as \(x\) feet. The height of the man is \(6\) feet, and he is walking towards the building at a rate of \(-5\) feet per second (negative because the distance \(x\) is decreasing).

Step 2: Set Up the Relationship Using Similar Triangles

The light on the ground \(50\) feet from the building, the height of the building's shadow, and the height of the man form similar triangles. Let \(y\) be the length of the man’s shadow on the building at a given time. By the similar triangles relationship, we have: \[ \frac{6}{x + y} = \frac{y}{50} \]

Step 3: Solve for \( y \)

Rearrange the equation to express \( y \) in terms of \( x \): \[ 6 \times 50 = y \times (x + y) \] \[ 300 = yx + y^2 \]

Step 4: Implicit Differentiation with Respect to Time

Differentiate both sides of the equation \( 300 = yx + y^2 \) with respect to time \( t \): \[ 0 = y \frac{dx}{dt} + x \frac{dy}{dt} + 2y \frac{dy}{dt} \]

Step 5: Substitute Known Values

Plug in the known values: \( x = 30 \), \( \frac{dx}{dt} = -5 \), and solve for \( \frac{dy}{dt} \): \[ 0 = y (-5) + 30 \frac{dy}{dt} + 2y \frac{dy}{dt} \]Simplify: \[ 0 = -5y + (30 + 2y) \frac{dy}{dt} \]

Step 6: Solve the Simplified Equation

Solve the above equation for \( \frac{dy}{dt} \): \[ 5y = (30 + 2y) \frac{dy}{dt} \] \[ \frac{dy}{dt} = \frac{5y}{30 + 2y} \]

Step 7: Find \( y \) When \( x = 30 \)

Use the similar triangles relationship again: \[ 50 = \frac{6(x + y)}{y} \] \[ 50 = 6 \left( \frac{x}{y} + 1 \right) \]When \( x = 30 \): \[ y = \sqrt{270} \approx 16.43 \]

Step 8: Substitute \( y \) and Solve

Finally we substitute \( y = 16.43 \) into the previous equation for \( \frac{dy}{dt} \): \[ \frac{dy}{dt} = \frac{5(16.43)}{30 + 2(16.43)} \approx 2.5\]

Key Concepts

related rates

Related rates problems involve finding the rate at which one quantity changes by relating it to other quantities whose rates of change are known. These problems often arise in real-world scenarios where different quantities are linked by some relationship, like geometric relationships. In our exercise, we have the height of a man, the speed at which he's walking, and the length of his shadow on the building. Our goal is to determine how fast the shadow is changing. To tackle related rates problems, follow these steps:

• Identify all the variables involved and know which are constants
• Write an equation that relates these variables
• Differentiate the equation with respect to time (using implicit differentiation)
• Substitute the known values and solve for the unknown rate

implicit differentiation

Implicit differentiation is a technique used to differentiate equations where the variables are intermingled and not easily separated. Instead of explicitly solving for one variable in terms of another, we differentiate both sides of the equation with respect to time (or another variable). This technique is particularly useful in related rates problems. In our example, we started with the equation from similar triangles and implicitly differentiated it with respect to time:
\[ 0 = y \frac{dx}{dt} + x \frac{dy}{dt} + 2y \frac{dy}{dt} \]
Here, we used implicit differentiation because both 'x' (the distance of the man from the building) and 'y' (the length of the shadow) are functions of time. After differentiation, we substitute the known values to solve for the unknown rate of change.

similar triangles

The concept of similar triangles is crucial in many related rates problems. Similar triangles have the same shape but different sizes, and their corresponding sides are proportional. This property allows us to set up equations to relate different parts of the geometric figures. In our problem, similar triangles are formed by the man, the light source on the ground, and his shadow on the building. By setting up a proportion between the heights and corresponding horizontal distances, we developed the equation:
\[ \frac{6}{x + y} = \frac{y}{50} \]
This equation connected the distance 'x' and the shadow length 'y' through the properties of similar triangles, helping us to find the changing rates.