Problem 30 - Thomas Calculus

Question

A moving shadow \(A\) man 6 \(\mathrm{ft}\) tall walks at the rate of 5 \(\mathrm{ft} / \mathrm{sec}\) toward a streetlight that is 16 \(\mathrm{ft}\) above the ground. At what rate is the tip of his shadow moving? At what rate is the length of his shadow changing when he is 10 \(\mathrm{ft}\) from the base of the light?

Step-by-Step Solution

Answer

The tip of the shadow moves at -8 ft/sec, and the shadow's length changes at -3 ft/sec.

Step 1: Identifying Relations

Let the distance of the man from the streetlight be \( x \) feet, the length of his shadow be \( s \) feet, and the distance from the light pole to the tip of his shadow be \( y = x + s \). Then we have two similar triangles: one formed between the streetlight and the ground, and the other formed between the man and the ground.

Step 2: Setting Up Similar Triangles

Using the concept of similar triangles, we have the ratios \( \frac{16}{y} = \frac{6}{s} \). Hence, we can express \( s \) as \( s = \frac{6y}{16} = \frac{3y}{8} \).

Step 3: Differentiate with respect to Time

Differentiate the equation \( y = x + s \) with respect to time \( t \). So, \( \frac{dy}{dt} = \frac{dx}{dt} + \frac{ds}{dt} \). Given \( \frac{dx}{dt} = -5 \) ft/sec (since the man is approaching the light), we have our first relationship.

Step 4: Find the Derivative of Triangle Expression

Differentiating \( s = \frac{3y}{8} \) with respect to time gives \( \frac{ds}{dt} = \frac{3}{8} \frac{dy}{dt} \). Substitute into the relationship from the previous step: \( \frac{dy}{dt} = \frac{dx}{dt} + \frac{3}{8} \frac{dy}{dt} \).

Step 5: Solve for \( \frac{dy}{dt} \), the Rate of Change of Tip of Shadow

Rearrange the equation to solve for \( \frac{dy}{dt} \), leading to \( \frac{dy}{dt} - \frac{3}{8} \frac{dy}{dt} = \frac{dx}{dt} \). Therefore, \( \frac{5}{8} \frac{dy}{dt} = -5 \), making \( \frac{dy}{dt} = -8 \) ft/sec.

Step 6: Calculate \( \frac{ds}{dt} \), the Rate of Shadow's Length Change

Now compute \( \frac{ds}{dt} \): since \( \frac{ds}{dt} = \frac{3}{8} \frac{dy}{dt} \), substitute \( \frac{dy}{dt} = -8 \) into this, resulting in \( \frac{ds}{dt} = \frac{3}{8} \times (-8) = -3 \) ft/sec.

Key Concepts

Similar Triangles

The concept of similar triangles is crucial in solving problems involving proportions, especially in scenarios with geometric figures. Similar triangles have the same shape but differ in size. This means their corresponding angles are equal, and the ratios of their corresponding sides are constant.
In the context of our shadow problem, similar triangles are formed by:

• The streetlight, the ground, and the shadow's tip.
• The man, the ground, and the point where the man’s shadow meets the ground.

These geometric relationships allow us to write a proportion between the sides of these triangles. By setting up the equation \( \frac{16}{y} = \frac{6}{s} \), we compare the height of the streetlight to the total length from the light to the tip of the shadow, with the man’s height to his shadow’s length.
These relationships can help to express unknown variables like the length of the shadow \( s \) using known distances, vital for finding rates of change in related rates problems.

Differentiation

Differentiation is a fundamental tool in calculus, used to determine how a function changes as its inputs change. When applied to related rates problems, it helps us find out how one quantity changes with respect to another over time.
In this exercise, differentiation is used on the equation \( y = x + s \), which links the man's distance from the light \( x \) and the length of his shadow \( s \) to the distance from the light to the tip of the shadow \( y \). By differentiating with respect to time, we can represent the rates of change for each quantity.
The derivative \( \frac{dy}{dt} = \frac{dx}{dt} + \frac{ds}{dt} \) illustrates the relationship between their rates of change. Since we know the man's walking speed, \( \frac{dx}{dt} = -5 \text{ ft/sec} \), this becomes part of our calculation to find \( \frac{dy}{dt} \), the rate at which the tip of his shadow moves.

Rate of Change

The rate of change tells us how quickly a particular quantity changes over time. This concept is often used in related rates problems to determine how one quantity affects another.
In our problem, there are two primary rates of change of interest:

• \( \frac{dy}{dt} \): The rate at which the tip of the shadow moves.
• \( \frac{ds}{dt} \): The rate at which the length of the man's shadow changes.

We solve for \( \frac{dy}{dt} \) using the differentiated equation to understand how the entire system's geometry changes as the man walks towards the light. The given value \( \frac{dy}{dt} = -8 \text{ ft/sec} \) indicates that the shadow's tip moves faster than the man's actual walking rate due to the geometric relationship.
Similarly, \( \frac{ds}{dt} \) is calculated to be \(-3 \text{ ft/sec}\), revealing that the shadow length decreases at this rate as the man approaches the light.

Calculus Problem Solving

Calculus problem solving involves breaking complex real-world problems into manageable mathematical relationships and solving them systematically.
In related rates problems, like this one, the process requires:

• Identifying all relevant variables and their relationships, often using geometric or algebraic equations.
• Differentiating equations with respect to time to find how each variable's rate of change relates to others.
• Solving for the desired rate of change, using given conditions such as speeds or distances.

This exercise showcases how understanding the situation, forming a mathematical model using tools like similar triangles, and applying calculus skills like differentiation enables us to solve real-life problems by finding how fast different parts of a system change over time.
Mastering these steps enhances our ability to tackle diverse and dynamic problems that can arise in various fields like physics, engineering, and beyond.