Problem 37 - Essential Calculus Early Transcendentals

Question

The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one three times as strong as the other, are placed 10 ft apart, where should an object be placed on the line between the sources so as to receive the least illumination?

Step-by-Step Solution

Answer

Place the object 2.5 ft from the weaker light source.

Step 1: Understand the Problem

The problem states that the illumination (I) of an object is directly proportional to the strength (S) of the light source and inversely proportional to the square of the distance (d) from the source. For two light sources, one stronger than the other, we need to find where to place the object to get the least illumination.

Step 2: Set Up the Proportionality Equation

According to the direct and inverse proportionality, the illumination from a single light source can be expressed as \( I = \frac{S}{d^2} \), where \(S\) is the strength of the light source and \(d\) is the distance from the source.

Step 3: Define Variables and Distances

Let \(S_1\) be the strength of the weaker light source and \(S_2 = 3S_1\) be the strength of the stronger source. Place the weaker source at point 0 and the stronger source at point 10 on a line. Let \(x\) be the distance from the weaker source to the object, so the distance from the stronger source to the object is \(10 - x\).

Step 4: Create the Illumination Equation

The total illumination at the object is the sum of the illuminations from both sources: \[ I_{total} = \frac{S_1}{x^2} + \frac{3S_1}{(10-x)^2} \].

Step 5: Simplify the Illumination Equation

Factor out \(S_1\) from the equation because it is common to both terms: \( I_{total} = S_1 \left( \frac{1}{x^2} + \frac{3}{(10-x)^2} \right) \). This allows us to focus on minimizing \( I = \frac{1}{x^2} + \frac{3}{(10-x)^2} \).

Step 6: Differentiate the Illumination Function

Find the function's critical points by differentiating \( I \) with respect to \(x\): \[ \frac{dI}{dx} = -\frac{2}{x^3} + \frac{6}{(10-x)^3} \].

Step 7: Find the Critical Points

Set the derivative equal to zero to find critical points: \[ -\frac{2}{x^3} + \frac{6}{(10-x)^3} = 0 \]. Solving this gives \(-2(10-x)^3 = 6x^3\) which simplifies to \(x^3 = 3(10-x)^3\).

Step 8: Solve the Critical Points Equation

Solve \(x^3 = 3(10-x)^3\). First, express as \((10-x)^3 = \frac{x^3}{3}\). Solving gives \(x = 2.5\), so the object should be 2.5 ft from the weaker source.

Key Concepts

Illumination and Inverse Square Law

When discussing illumination in physics, it is important to understand how light intensity behaves over distances. The illumination of an object by a light source is directly proportional to the strength of the source but is inversely proportional to the square of the distance from the source. This principle is known as the Inverse Square Law.

• "Directly proportional" means that as the strength of the light source increases, the illumination increases linearly.
• "Inversely proportional" means that as the distance from the light source increases, the illumination decreases rapidly—in fact, with the square of the distance.

Imagine shining a flashlight toward a wall. As you move further away, the light becomes dimmer because it spreads out. This spread causes the same amount of light to cover a larger area, reducing the intensity.

In mathematical terms, if \( I \) denotes the illumination, \( S \) the strength of the source, and \( d \) the distance from the source, the equation is \( I = \frac{S}{d^2} \). This expression shows how even a small increase in distance leads to a significant reduction in illumination. In optimization problems, understanding this relationship helps determine the optimal position for achieving desired lighting outcomes.

Critical Points and Differentiation

In optimization, finding the minimum or maximum of a function is crucial, and this is where critical points and differentiation come into play.

• A critical point of a function occurs where the derivative is zero or undefined, showing potential local maxima or minima.
• Using differentiation, you find how a function changes at any point, which helps identify critical points.

For the exercise given, we needed to find the position that would minimize the illumination from two light sources. We expressed the total illumination as a function of distance, \( I(x) = \frac{1}{x^2} + \frac{3}{(10-x)^2} \), where \( x \) is the distance from the weaker light source.

We then differentiated this function: \( \frac{dI}{dx} = -\frac{2}{x^3} + \frac{6}{(10-x)^3} \). Setting this derivative equal to zero allows us to isolate the critical points. Solving \( -\frac{2}{x^3} + \frac{6}{(10-x)^3} = 0 \) gives us \( x = 2.5 \). This means that placing the object here will give the least illumination exposure due to the balance of intensities from each source.

Proportional Relationships

The concept of proportional relationships is fundamental to understanding how variables change relative to each other. In this exercise, illumination was governed by two types of proportional relationships:

• Direct proportionality between illumination and the strength of the source.
• Inverse proportionality between illumination and the square of the distance.

Direct proportionality means if you increase the light source's strength, the illumination increases at the same rate—double the strength means double the illumination.

Inverse square proportionality means that doubling the distance causes the illumination to drop to a quarter of its original value. This makes understanding and calculating illumination crucial when multiple light sources are involved.

By modeling these relationships with equations, you can explore how changes in one variable affect another. This is particularly useful in real-world applications like setting illumination levels in a room or adjusting the focus in photography. Recognizing and applying these concepts allows for precise control over the desired outcomes.