Question

The illumination at a point is inversely proportional to the square of the distance of the point from the light source and directly proportional to the intensity of the light source. If two light sources are \(s\) feet apart and their intensities are \(I_{1}\) and \(I_{2}\), respectively, at what point between them will the sum of their illuminations be a minimum?

Short answer

The illumination sum is minimized at \( x = \frac{s}{1 + \left(\frac{I_2}{I_1}\right)^{1/3}} \).

Step-by-step solution

Express Illumination of Each Light Source

If the illumination due to a light source is inversely proportional to the square of the distance, then for light source 1 whose intensity is \( I_1 \), the illumination at a point \( x \) feet from it is given by \( \frac{I_1}{x^2} \). Similarly, for light source 2 at a distance \( s-x \) from the point, the illumination is \( \frac{I_2}{(s-x)^2} \).

Set Up the Total Illumination Function

The total illumination at the point is the sum of the two illuminations, which is given by the function: \[ f(x) = \frac{I_1}{x^2} + \frac{I_2}{(s-x)^2} \] Our goal is to find the value of \( x \) that minimizes this function.

Differentiate the Illumination Function

To find the minimum point, differentiate \( f(x) \) with respect to \( x \): \[ f'(x) = -2\frac{I_1}{x^3} + 2\frac{I_2}{(s-x)^3} \]

Find Critical Points

Set the derivative equal to zero to find critical points: \[ -2\frac{I_1}{x^3} + 2\frac{I_2}{(s-x)^3} = 0 \] Simplify this to get \[ \frac{I_1}{x^3} = \frac{I_2}{(s-x)^3} \] Taking cube roots, we solve: \[ \frac{x}{s-x} = \left(\frac{I_1}{I_2}\right)^{1/3} \] so: \[ x = \frac{s}{1 + \left(\frac{I_2}{I_1}\right)^{1/3}} \]

Verify the Minimum

We need to ensure that the critical point found is indeed a minimum. Check the second derivative \( f''(x) \): \[ f''(x) = 3\frac{I_1}{x^4} + 3\frac{I_2}{(s-x)^4} \] Since \( f''(x) \) is positive at \( x = \frac{s}{1 + \left(\frac{I_2}{I_1}\right)^{1/3}} \), it confirms a local minimum.

Key concepts

Inverse Square Law

The inverse square law is a fundamental principle often encountered in physics, particularly with light and sound. Essentially, this law states that a physical quantity or force is inversely proportional to the square of the distance from the source of that physical quantity.

For illumination, this means that as one moves away from a light source, the illumination decreases rapidly. It is given by the formula:
\[ I = \frac{P}{d^2} \]where \( I \) is the illumination, \( P \) is the power or intensity of the light source, and \( d \) is the distance from the source. This formula highlights why light intensity decreases so dramatically with distance, which is crucial for solving optimization problems in calculus involving multiple light sources.

• The inverse relationship makes analyzing light distribution a lot more precise in practical applications, such as lighting design and optical engineering.
• This principle also sets the foundation for calculating total illumination when multiple light sources are present, as seen in the given problem.

Illumination Model

The illumination model in the given exercise involves calculating how light distributes from two sources. By following the inverse square law, each light’s illumination at a point is computed based on its distance from that point.

For light source 1 with intensity \( I_1 \), the intensity at distance \( x \) is \( \frac{I_1}{x^2} \). Similarly, light source 2 at a distance \( s-x \) from the point has an illumination of \( \frac{I_2}{(s-x)^2} \). The total illumination at any point is the sum of these two values, forming an illumination model:
\[ f(x) = \frac{I_1}{x^2} + \frac{I_2}{(s-x)^2} \]

• This model helps calculate the cumulative light intensity at a specific point. Understanding this model is key to solving calculus problems where minimizing or maximizing illumination is required.
• In real-world scenarios, this model is useful for optimizing lighting conditions in environments to achieve the desired ambiance or energy efficiency.

Critical Points

Critical points play a vital role in calculus, especially in optimization problems. These are the points where the derivative of a function is zero or undefined. In the context of the exercise, critical points indicate potential locations where the total illumination might be minimized or maximized.

To find the critical points for our illumination function \( f(x) = \frac{I_1}{x^2} + \frac{I_2}{(s-x)^2} \), we calculate the derivative \( f'(x) \), and set it to zero:
\[ -2\frac{I_1}{x^3} + 2\frac{I_2}{(s-x)^3} = 0 \]
This results in simplified equation:
\[ \frac{I_1}{x^3} = \frac{I_2}{(s-x)^3} \]

• By solving this equation, we find the value of \( x \) that could minimize or maximize illumination, revealing potential optimal positions between the light sources.
• Identifying critical points is crucial as they guide us towards understanding the function’s behavior and potential solutions.

Derivative Test

The derivative test is a method used to determine whether the critical points found are indeed points of minimum or maximum value. In this exercise, after determining the critical point, it's necessary to ensure it's a minimum by using the second derivative test.

The second derivative is calculated as follows:
\[ f''(x) = 3\frac{I_1}{x^4} + 3\frac{I_2}{(s-x)^4} \]
A positive second derivative at the critical point indicates a local minimum. If \( f''(x) \) is positive, it verifies that the illumination is minimized at the point calculated.

• This test reassures us that the solution is correct by providing a mathematical confirmation through concavity analysis.
• Understanding how to apply the derivative test ensures efficient solving of calculus optimization problems in various fields, including physics and engineering.