Problem 45 - Thomas Calculus

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 when the point is 2 meters away from the stronger light.

Step 1: Define Variables

Let the distance from the stronger light to the point of interest be \( x \) meters. Since the lights are 6 meters apart, the distance from the weaker light to the same point is \( 6 - x \) meters.

Step 2: Establish Relationship of Intensity

For the stronger light, let its initial intensity be \( 8I \), where \( I \) is the intensity of the weaker light. The intensity from each light at a point is given by the formula \( \text{Intensity} = \frac{k}{d^2} \), where \( k \) is a proportionality constant and \( d \) is the distance.

Step 3: Write Intensity Expressions

The intensity from the stronger light at the point is \( \frac{8I}{x^2} \). The intensity from the weaker light at the point is \( \frac{I}{(6-x)^2} \).

Step 4: Formulate Total Illumination Equation

The total illumination \( I_{total} \) at the point is the sum of intensities from both lights: \[ I_{total} = \frac{8I}{x^2} + \frac{I}{(6-x)^2} \].

Step 5: Differentiate Total Illumination

Differentiate the total illumination expression with respect to \( x \) to find the point where illumination is minimum. This gives us: \[ \frac{dI_{total}}{dx} = -\frac{16I}{x^3} + \frac{2I}{(6-x)^3} \].

Step 6: Solve for Minimum Illumination

Set the derivative equal to zero to find the critical points: \[ -\frac{16}{x^3} + \frac{2}{(6-x)^3} = 0 \]. Solving this equation, cross-multiply and simplify to find \( 8x^3 = (6-x)^3 \).

Step 7: Solve the Equation

Solve \( 8x^3 = (6-x)^3 \) by finding cube roots and performing algebraic manipulations. This simplifies to \( 2x = 6-x \), leading to \( x = 2 \) meters.

Step 8: Verify Minimum Condition

Check the second derivative to ensure it is positive at \( x = 2 \) to confirm it provides a minimum illumination, not a maximum.

Step 9: Conclusion

Thus, the point of least total illumination is 2 meters from the stronger light.

Key Concepts

Illumination Intensity

Illumination intensity refers to the amount of light that falls on a given surface area. In practical terms, it is how brightly an area is lit by a light source. It's an important concept in many fields such as photography, interior design, and physics.

For our problem, the illumination intensity at a point is influenced by two main factors:

• The strength of the light source - A stronger light will provide more intense illumination.
• The distance from the light source - The farther away from the light source, the less intense the illumination.

In mathematical terms, if a light source has an intensity of \( I \) and the distance from the light source is \( d \), the intensity of illumination \( E \) at that distance can be expressed as:
\[ E = \frac{k}{d^2} \]
Where \( k \) is a constant that depends on the medium and other conditions. In essence, the intensity of illumination is an inverse function of the square of the distance, as described by the Inverse Square Law.

Inverse Square Law

The Inverse Square Law is a principle in physics stating that the intensity of an effect, such as illumination, sound, or force, diminishes with the square of the distance from the source. This is a key reason why, as you move away from a light source, the brightness decreases rapidly.

In the context of our exercise, the law can be described with the formula:
\[ E = \frac{k}{d^2} \]
For example, if a light source is twice as far away, the illumination intensity will be spread over an area four times as large, hence the intensity is only a quarter as strong. The formula captures this concept, making it a vital tool in calculating how a change in distance affects light intensity.

The Inverse Square Law not only applies to light but also various fields such as gravitational force and electromagnetic fields, showcasing fundamental aspects of physics.

Differentiation

Differentiation is a mathematical process used to determine how a function changes as its input changes. It is a fundamental tool in calculus and plays a critical role in optimization problems, such as finding minimum or maximum values.

In optimization problems, like the one presented, differentiation helps us find points known as critical points where the slope of a function is zero. When we differentiate a function that models total illumination, we aim to find the rate at which illumination changes with varying distance \( x \) from the light source.

• We can use the first derivative to locate critical points where the function might have a minimum or maximum value.
• The second derivative is often used to assess whether these critical points correspond to a minimum (concave up) or maximum (concave down).

For example, in our problem, after differentiating the total illumination expression, setting it to zero gives a solution where the illumination is at its minimum. Checking the second derivative ensures that this value is indeed a minimum. This analytical approach underscores the power of differentiation in solving real-world problems efficiently.