Problem 21 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Two light bulbs of different intensities are a distance, \(d,\) apart. At any point, the light intensity from one of the bulbs is proportional to the intensity of the bulb and inversely proportional to the square of the distance from the bulb. Find the point between the two bulbs at which the sum of the intensities of light from the two bulbs is minimum.

Step-by-Step Solution

Answer

The point is located at \(\frac{d}{1 + \sqrt[3]{\frac{I_2}{I_1}}}\) from bulb 1.

Step 1: Understand the problem

We have two light sources with different intensities placed at a distance of \(d\) apart. We need to find a point between them where the sum of their light intensities is minimized according to the inverse square law.

Step 2: Express intensities mathematically

Let's denote the intensities of the two bulbs as \(I_1\) and \(I_2\), and let \(x\) represent the distance from the first bulb to the point in question. Hence, \(d - x\) is the distance from the second bulb to this point.

Step 3: Write light intensity expressions

According to the inverse square law, the intensity at the point from bulb 1 is \(\frac{I_1}{x^2}\) and from bulb 2 is \(\frac{I_2}{(d-x)^2}\).

Step 4: Formulate the sum of intensities

The total intensity \(T\) at the point \(x\) is the sum of individual intensities: \[ T = \frac{I_1}{x^2} + \frac{I_2}{(d-x)^2} \]

Step 5: Find the critical points

To find where \(T\) is minimized, we need to differentiate \(T\) with respect to \(x\) and set the derivative equal to zero. Calculate \(\frac{dT}{dx}\): \[ \frac{dT}{dx} = -2\frac{I_1}{x^3} + 2\frac{I_2}{(d-x)^3} \]. Set this equation to zero to find critical points: \[ \frac{I_1}{x^3} = \frac{I_2}{(d-x)^3} \]

Step 6: Solve critical points equation

Cross multiply to solve for \(x\): \[ I_1 (d-x)^3 = I_2 x^3 \]. Assuming any positive real number solutions, solve \[ (d-x) = \sqrt[3]{\frac{I_2}{I_1}} x \].

Step 7: Evaluate and conclude

Substitute \(y = \sqrt[3]{\frac{I_2}{I_1}} x\) and express \(x\) in terms of \(d\): \[ x + y = d \Rightarrow x + \sqrt[3]{\frac{I_2}{I_1}} x = d \]. Solving for \(x\), we find: \[ x = \frac{d}{1 + \sqrt[3]{\frac{I_2}{I_1}}} \]. This gives the location of the point where the light intensities sum to their minimum.

Key Concepts

Inverse Square Law

The inverse square law is a fundamental principle in physics, particularly applicable in scenarios involving light, sound, and gravitation. It states that the intensity of a physical quantity, such as light, is inversely proportional to the square of the distance from its source. This means that as you move away from a light source, the intensity decreases rapidly.
For example, if you double the distance from a light source, the intensity becomes one-fourth of what it was initially. So, mathematically, if a light source has intensity \(I\) at a distance \(d\), moving twice the distance would make the intensity \(\frac{I}{d^2}\).
In our problem, the light intensities from two bulbs depend on their respective distances, which is crucial when calculating at which point between them the combined intensity is minimized.

Critical Points

Critical points of a function are those values of the variable where the function's derivative is zero or undefined. They are vital in calculus as they often correspond to the maximum or minimum values of a function. When you're optimizing a function—seeking to find where something reaches its extreme values—locating critical points is essential.
In our exercise, we seek to find the critical points of the total light intensity function. By differentiating the intensity function with respect to distance and setting that derivative to zero, we find where the sum of the intensities has a minimum value between the two bulbs.
Critical points provide meaningful insights into the behavior of functions, particularly useful in scenarios like ours where resource allocation, cost minimization, or efficiency maximization are required.

Light Intensity

Light intensity is a measure of the energy of light received on a surface at a certain distance from a light source. It is often measured in units like lux or lumens. In mathematical modeling, intensities are used to compare different light sources or to analyze the distribution of light.
In this problem, the intensities \(I_1\) and \(I_2\), represent the strength of two different bulbs. By expressing intensity in terms of inverse square law, we learn how the intensity changes with distance from the source.
Understanding light intensity helps in designing effective lighting systems, ensuring proper illumination, and minimizing unnecessary use of energy by pinpointing exactly where to position light sources for maximum or minimum collective intensity.

Derivative

A derivative represents the rate of change of a function with respect to one of its variables. In calculus, it is a critical tool used to find slopes of curves, velocities, accelerations, and optimize values—like in our exercise! The act of finding a derivative is known as differentiation.
In this calculus optimization problem, the derivative of the total light intensity function helps identify where the light intensity between the two bulbs is at its minimum. Specifically, \[\frac{dT}{dx} = -2\frac{I_1}{x^3} + 2\frac{I_2}{(d-x)^3}\] is derived to track how changes in position affect intensity. Setting it to zero uncovers the critical points.
Leveraging derivatives is not just for theoretical bouncing; it translates into practical applications like maximizing power efficiency and ensuring proper lighting conditions.