Problem 9 - Advanced Engineering Mathematics

Question

When a vertical beam of light passes through a transparent medium, therate at whichits intensity I decreasesis propartional to \(I(t)\), where trepresents the thickness of the medium (in feet). In clear seawater, the intensity 3 feet below the surface is \(25 \%\) of the initial intensity \(I_{0}\) of the incident beam. What is the intensity of the beam 15 feet below the surface?

Step-by-Step Solution

Answer

Intensity at 15 feet below is \( \frac{I_0}{1024} \).

Step 1: Define the Differential Equation

The problem states that the rate at which the intensity \( I \) decreases is proportional to the intensity \( I(t) \). This can be modeled by the differential equation \( \frac{dI}{dt} = -kI \), where \( k \) is a constant of proportionality.

Step 2: Solve the Differential Equation

The differential equation \( \frac{dI}{dt} = -kI \) is separable. Integrating both sides, we have \( \int \frac{1}{I} \, dI = -\int k \, dt \), which gives \( \ln I = -kt + C \). To solve for \( I \), exponentiate both sides to get \( I = e^{-kt+C} = Ce^{-kt} \), where \( C \) is an integration constant.

Step 3: Determine the Constant Using Initial Conditions

Given that \( I(0) = I_0 \), plug \( t = 0 \) into \( I = Ce^{-kt} \), giving \( I_0 = Ce^{0} \). This implies \( C = I_0 \), so \( I(t) = I_0 e^{-kt} \).

Step 4: Use Known Depth to Find Constant k

We know that when \( t = 3 \) feet, the intensity is \( 25\% \) of the initial intensity: \( I(3) = 0.25 I_0 \). Substitute into the equation: \( 0.25 I_0 = I_0 e^{-3k} \). Solving for \( k \), we get \( e^{-3k} = 0.25 \), leading to \( -3k = \ln(0.25) \), and further \( k = -\frac{\ln(0.25)}{3} \).

Step 5: Calculate the Intensity at 15 Feet

Substitute \( k \) back into the equation \( I(t) = I_0 e^{-kt} \) and find \( I(15) \): \( I(15) = I_0 e^{-\frac{\ln(0.25)}{3} \times 15} \). Simplifying gives \( I(15) = I_0 e^{-5\ln(0.25)} = I_0 (0.25)^5 = I_0 \times \frac{1}{1024} \).

Step 6: Final Calculation and Interpretation

The intensity of the beam 15 feet below the surface is \( \frac{I_0}{1024} \). This is a very small fraction of the initial intensity, indicating significant attenuation in seawater at this depth.

Key Concepts

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. This is often seen in nature and occurs in various contexts such as cooling, radioactive decay, and as in our example, light intensity reduction through a medium.

• The key feature of exponential decay is that it demonstrates a rapid drop in value initially, which slows down over time.
• This is defined by the equation \( I(t) = I_0 e^{-kt} \), where \( I_0 \) is the initial quantity, and \( k \) is the decay constant that determines the rate of decay.

In the light intensity problem, the beam loses intensity as it travels through seawater. Exponential decay helps to predict how much light remains after traveling a certain distance.
This concept is essential not only for solving numerical problems but also for understanding the natural phenomena where gradual reduction occurs.

Rate of Change

The rate of change in a quantity measures how quickly it increases or decreases over time. This is often described with a differential equation, which shows the relationship between the rate of change and the initial quantity.

• For light intensity, the differential equation is \( \frac{dI}{dt} = -kI \), meaning the rate of change of the light intensity is proportional to its current intensity.
• The negative sign indicates the quantity is decreasing over time.

Understanding the rate of change is crucial when dealing with processes that evolve over time, like in our case with light traveling through water. It provides insight into how quickly the initial value diminishes and assists in forming solutions for such phenomena.

Separable Equations

Separable equations are a type of differential equations that can be divided into two parts; one only involving the dependent variable, and the other the independent variable.

• Such equations can be rearranged to allow each variable to be integrated separately. This simplifies solving the equation.
• In our example, we have the equation \( \frac{dI}{dt} = -kI \), which can be separated and rewritten as \( \int \frac{1}{I} \, dI = -\int k \, dt \).

This method allows us to find the solution in terms of a function that describes how the intensity changes over time. The separation of variables is a powerful tool for solving many real-world problems, aiding in understanding complex processes through simpler means.