Problem 27 - Thomas Calculus

Question

Working underwater The intensity \(L(x)\) of light \(x\) feet beneath the surface of the ocean satisfies the differential equation $$\frac{d L}{d x}=-k L$$ As a diver, you know from experience that diving to 18 ft in the Caribbean Sea cuts the intensity in half. You cannot work without artificial light when the intensity falls below one-tenth of the surface value. About how deep can you expect to work without artificial light?

Step-by-Step Solution

Answer

The diver can work up to about 60 feet without artificial light.

Step 1: Understand the Differential Equation

The given differential equation \( \frac{d L}{d x} = -k L \) indicates an exponential decay of intensity \( L(x) \) as a function of depth \( x \). This is a separable differential equation.

Step 2: Solve the Differential Equation

To solve \( \frac{d L}{d x} = -k L \), rearrange to obtain \( \frac{dL}{L} = -k \, dx \). Integrate both sides to get \( \ln L = -kx + C \), where \( C \) is the integration constant. Express \( L \) in terms of \( x \): \( L(x) = e^{-kx + C} = Ce^{-kx} \).

Step 3: Apply Initial Conditions

According to the problem, diving to 18 ft reduces the intensity to half of its surface value. Assume the surface intensity is \( L_0 \). At \( x = 18 \), \( L(18) = \frac{1}{2} L_0 \). Substitute these into the intensity equation: \( \frac{1}{2} L_0 = C e^{-18k} \). Simplify to find \( C = L_0 \), so \( L(x) = L_0 e^{-kx} \).

Step 4: Solve for Constant k

From \( L(18) = \frac{1}{2} L_0 = L_0 e^{-18k} \), we can solve for \( k \). Divide both sides by \( L_0 \): \( \frac{1}{2} = e^{-18k} \). Taking natural logarithms, \( \ln(\frac{1}{2}) = -18k \). Thus \( k = -\frac{\ln(\frac{1}{2})}{18} = \frac{\ln 2}{18} \).

Step 5: Determine Maximum Work Depth

You need the intensity \( L(x) \) to be at least one-tenth of \( L_0 \). Set \( L(x) = \frac{1}{10} L_0 \). Substitute into the equation: \( \frac{1}{10} L_0 = L_0 e^{-kx} \). Simplify: \( \frac{1}{10} = e^{-kx} \). Take natural logs: \( \ln(\frac{1}{10}) = -kx \). Substitute \( k = \frac{\ln 2}{18} \) and solve for \( x \): \( x = -\frac{\ln(\frac{1}{10})}{k} = 18 \frac{\ln 10}{\ln 2} \).

Step 6: Calculate Final Depth

Calculate the expression \( 18 \frac{\ln 10}{\ln 2} \) using logarithm values: \( \ln 10 \approx 2.302 \) and \( \ln 2 \approx 0.693 \). Then \( x = 18 \times \frac{2.302}{0.693} \approx 60 \) ft.

Key Concepts

Exponential Decay

Exponential decay describes a process where some quantity, like light intensity underwater, decreases at a rate proportional to its current value. This concept is central in many natural and scientific phenomena, including radioactive decay and cooling of objects. An exponential decay is generally represented by the formula \[ N(t) = N_0 e^{-kt} \]where:

• \( N(t) \) is the quantity at time \( t \).
• \( N_0 \) is the initial quantity (here, initial light intensity on the surface \( L_0 \)).
• \( k \) is the decay constant indicating how fast the decay occurs.

The decay constant \( k \) affects how rapidly the light diminishes as one goes deeper underwater. A larger \( k \) implies a faster reduction in intensity. In this exercise, the decay constant \( k \) was calculated using the fact that the light intensity halves after 18 feet.

Light Intensity Underwater

As you dive deeper beneath the ocean surface, the light intensity decreases due to absorption and scattering. This decrease in intensity follows an exponential decay model, which means light intensity drops quickly at first, and then more slowly as you go deeper.
The materials in the water, such as salt and small particles, absorb and scatter light, making it less intense. As a diver, understanding how light intensity changes based on depth is crucial for ensuring safety and effectiveness.
In this problem, you start with a known surface light intensity (\( L_0 \)). By using exponential decay, you can calculate how deep you can dive (approximately 60 feet, in this case) before the intensity falls below a usable level, such as one-tenth of the surface light.

Solving Separable Differential Equations

The differential equation given is separable, which means it can be divided into two separate integrals. It's written as: \[ \frac{dL}{dx} = -kL \]
Separable differential equations can be solved by separating the variables, resulting in:\[ \frac{dL}{L} = -k dx \]Now, you integrate both sides:

• Right side: Integral of \( \frac{dL}{L} \) is \( \ln L \).
• Left side: Integral of \( -k dx \) is \( -kx \).

This equation leads to the solution for light intensity as:\[ L(x) = Ce^{-kx} \]The integration constant \( C \) is found using initial conditions, where you substitute known values to solve for \( C \). This makes the solution reflect actual conditions, like initial light intensity at the water's surface.

Initial Conditions in Differential Equations

Initial conditions are vital for determining the specific solution of a differential equation. They specify the value of the function and sometimes its derivatives at certain points. Without these, we'd only have a general solution frilled with unknown constants.
In this exercise, the initial condition used is the knowledge that at 18 feet depth, half of the surface light intensity \( L_0 \) remains. This helps solve for the integration constant \( C \) and further refines the solution to match real-life conditions.
Using initial conditions enables you to tailor a differential equation to a particular scenario. It allows solving for unknown parameters like \( k \), which represents the rate of change. This process helps produce results directly applicable to practical situations such as determining safe dive depths.