Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Assume that \(1000 \mathrm{w} / \mathrm{m}^{2}\) of light is striking the surface of a lake and that \(40 \%\) of that light is reflected back into the atmosphere. Solve the initial value problem $$\begin{array}{r}I^{\prime}(x)=-K I(x) \\\I(0)=600\end{array} $$Suppose the light intensity at 10 meters is \(500 \mathrm{w} / \mathrm{m}^{2}\). Find the value of \(K\).

Step-by-Step Solution

Answer

The value of \( K \) is \( -\frac{1}{10} \ln\left(\frac{5}{6}\right) \).

Step 1: Understand the Problem

You are given a differential equation for the intensity of light, \( I(x) \), as it penetrates the water surface. Additionally, you know that 40% of the light is reflected, meaning 60% enters the water. Thus, the initial condition is \( I(0) = 600 \text{ w/m}^2 \). You need to find the constant \( K \) given that the intensity at 10 meters is 500 \text{ w/m}^2.

Step 2: Set Up the Differential Equation

The differential equation given is \( I'(x) = -K I(x) \). This is a first-order linear differential equation that can be solved by separation of variables. It describes exponential decay of light intensity in the water.

Step 3: Solve the Differential Equation

Separate the variables: \( \frac{dI}{I} = -K dx \). Integrating both sides gives \( \ln|I| = -Kx + C \), where \( C \) is the integration constant.

Step 4: Exponentiate to Solve for I(x)

Exponentiating both sides, we get \( I(x) = e^{C} e^{-Kx} \). Let \( e^{C} = I_0 \), so \( I(x) = I_0 e^{-Kx} \).

Step 5: Use Initial Condition to Find \( I_0 \)

At \( x = 0 \), \( I(0) = I_0 = 600 \), so \( I(x) = 600 e^{-Kx} \).

Step 6: Apply Boundary Condition at x = 10

You know \( I(10) = 500 \). Substitute into the intensity equation: \( 500 = 600 e^{-10K} \).

Step 7: Solve for K

Divide both sides by 600 to get \( \frac{500}{600} = e^{-10K} \). Simplify to \( \frac{5}{6} = e^{-10K} \). Take the natural logarithm of both sides: \(-10K = \ln\left(\frac{5}{6}\right) \). Thus, \( K = -\frac{1}{10} \ln\left(\frac{5}{6}\right) \).

Key Concepts

Differential Equation

A differential equation is essentially a mathematical equation that involves functions and their derivatives. In essence, it connects a function with its rates of change. They are incredibly useful in describing natural phenomena and processes, such as the motion of waves or the decay of light intensity as it travels through a medium like water. The equation can specify how one variable changes with respect to another variable and can model complex systems.
The given differential equation in the exercise is a first-order linear differential equation: \[ I'(x) = -K I(x) \] This equation tells us how the intensity of light \( I(x) \) changes with depth \( x \) in water. The negative sign indicates that the intensity decreases as depth increases, which aligns with our physical understanding of light attenuation in a medium.

Exponential Decay

Exponential decay describes processes where a quantity decreases at a rate proportional to its current value. In other words, the larger the quantity, the faster it decreases. This is commonly modeled by differential equations that result in exponential functions.
In the problem, the light intensity \( I(x) \) decreases exponentially as it goes deeper into the water. The solution \( I(x) = I_0 e^{-Kx} \) represents an exponential decay model, where \( I_0 \) is the initial intensity and \( K \) is a constant that determines how quickly the intensity decreases. This reflects a natural process where light energy is absorbed or scattered by water molecules, reducing its intensity.
Understanding exponential decay is crucial in fields such as physics, biology, and finance, as it helps to predict how a certain variable will change over time or space.

Separation of Variables

Separation of variables is a mathematical method used to solve differential equations. The method involves rearranging the equation to separate the different variables onto either side of the equation, allowing the equation to be solved through integration.
In the given exercise, we separate the intensity \( I \) and the spatial variable \( x \) from the equation: \[ \frac{dI}{I} = -K dx \] By integrating both sides independently, we can find an expression for \( I(x) \). This method is particularly useful for exponential-type problems like the one provided, as it simplifies the process of solving complicated equations by breaking them down into manageable parts.
It is an essential technique in calculus, allowing students and professionals alike to approach a wide range of differential equations effectively.

Reflection of Light

The reflection of light is a phenomenon that occurs when light bounces off a surface. In this problem, it is given that 40% of the light striking the lake's surface is reflected back. This leaves 60% to enter the water and continue on its path, which is crucial for setting the initial value of the intensity equation.
Understanding reflection is key when dealing with light-related problems as it affects the energy distribution. Reflections are responsible for many everyday observations, such as seeing yourself in a mirror or light bouncing off a shiny surface.
In terms of mathematical modeling, taking reflection into account allows for accurate representations of how much light propagates into different mediums, which again impacts how we solve related differential equations. By recognizing how much of the initial energy continues through the medium, we can set up initial conditions accurately in equations modeling light intensity or other wave phenomena.