Problem 41 - Thomas Calculus in SI Units

Question

The intensity of illumination at any point from a light source is proportional to the square of the reciprocal of the distance between the point and the light source. Two lights, one having an intensity eight times that of the other, are \(6 \mathrm{m}\) apart. How far from the stronger light is the total illumination least?

Step-by-Step Solution

Answer

The total illumination is least 4 meters from the stronger light.

Step 1: Identify the Variables

Let's denote the distance from the stronger light to the point in question as \( x \). Thus, the distance from the weaker light will be \( 6 - x \) because the lights are 6 meters apart. Let the intensity of the stronger light source be \( I_s \) and the intensity of the weaker light source be \( I_w = \frac{I_s}{8} \).

Step 2: Express Illumination from Each Light

The intensity of illumination from the stronger light at the point is proportional to \( \frac{I_s}{x^2} \). Similarly, the intensity of illumination from the weaker light is proportional to \( \frac{I_w}{(6 - x)^2} \).

Step 3: Write the Total Illumination Function

The total illumination at the point can be expressed as \( I(x) = \frac{I_s}{x^2} + \frac{I_s/8}{(6-x)^2} \).

Step 4: Simplify the Illumination Function

Combine terms in the function to get \( I(x) = \frac{I_s}{x^2} + \frac{I_s}{8(6-x)^2} \). This function needs to be minimized to find the position where the total illumination is least.

Step 5: Take the Derivative

To minimize the function \( I(x) \), take the derivative \( I'(x) \): \( I'(x) = -\frac{2I_s}{x^3} - \frac{I_s}{4(6-x)^3} \).

Step 6: Solve for Critical Points

Set the derivative equal to zero to find critical points: \( -\frac{2I_s}{x^3} = -\frac{I_s}{4(6-x)^3} \). Simplify and solve for \( x \).

Step 7: Solve the Equation

Simplifying the derivative equation gives \( 8(6-x)^3 = x^3 \), leading to \( (6-x)^3 = \frac{1}{8}x^3 \). Solve for \( x \).

Step 8: Find the Critical Point

Solve \( (6-x)^3 = \frac{1}{8}x^3 \). By solving \( (6-x) = \frac{x}{2} \), we get \( x = 4 \).

Step 9: Verify the Minimum

To ensure \( x = 4 \) is a minimum, use the second derivative test. Calculate \( I''(x) \) and check its sign at \( x = 4 \). If positive, it indicates a minimum.

Key Concepts

Illumination Intensity

Illumination intensity is a measure of how much light is received by a surface at a given point. Imagine a light bulb shining light. The farther you move away from it, the dimmer the light seems. This behavior is a result of how light spreads out over an area.

In our problem, the intensity of illumination from a light source lessens as you move away. More formally, it is proportional to the square of the inverse of the distance from the light source. This characteristic is crucial in determining how effectively a light covers an area.

Understanding how illumination changes with distance helps in everyday scenarios, like placing lights in a room for optimal brightness. It also serves as a foundation in calculus optimization problems, where understanding these relationships can determine the best location for the desired lighting outcome.

Distance Formula

The distance formula is a mathematical method used to calculate the distance between two points. In the context of the exercise, it helps define distances from a hypothetical point to our two light sources.

Here, the problem involves calculating how far a point is from each of these light sources. We are given two lights apart by 6 meters. If we let the distance from the stronger light to some point be represented by \( x \), it follows that the distance from the weaker light is \( 6 - x \).

• If you know one distance, you can easily find the other using the formula above.
• This relationship is pivotal in setting up the equation to find the critical points where total illumination is minimal.

As you can see, understanding the distance between objects is key in efficiently solving many physical and mathematical problems. In our scenario, it contributes significantly to setting up the function we need to minimize.

Inverse Square Law

The inverse square law describes a principle where a physical quantity is inversely proportional to the square of the distance from the source.

In lighting, it explains why light becomes dimmer as one moves away from the source. If you double the distance, the illumination is spread over four times the area, thus reducing its intensity by a factor of four.

• This law applies to various forces and effects in physics, including gravity and electromagnetic fields.
• In our illumination problem, this concept guides how each light's intensity diminishes away from its source, becoming a fundamental part of modeling illumination mathematically.

The inverse square law is vital because it describes not just light, but any propagation from a point source in three-dimensional space. Understanding it provides insight into numerous real-world applications and it is essential for optimizing light placement for desired illumination levels.

Critical Points

Critical points in calculus are where a function's derivative is zero or undefined. In optimization problems, like finding the minimum illumination, these points help us identify potential solutions.

For example, in the illumination problem, we calculated the derivative of the illumination function to find values of \( x \) where the rate of change is zero. Solving \( I'(x) = 0 \) helped find such a critical point, leading to the conclusion that\( x = 4 \), which indicates where total illumination is least.

• Critical points allow us to test different values within a function.
• They help determine whether a function is reaching a minimum, maximum, or a point of inflection.

To confirm whether these critical points are indeed minima, maxima, or saddle points, further tests, like the second derivative test, can be employed. This process is an essential step in achieving precise and optimized solutions in problems encountered in calculus and various real-life scenarios.