Problem 5 - 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 \(\mathrm{ft}\) 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

You can work without artificial light up to about 60 feet deep.

Step 1: Understand the Differential Equation

The given differential equation \( \frac{dL}{dx} = -kL \) describes exponential decay in light intensity with depth. The constant \( k \) determines the rate of decrease.

Step 2: Solve the Differential Equation

The differential equation \( \frac{dL}{dx} = -kL \) is a separable first-order linear differential equation. Its general solution is \( L(x) = L_0 e^{-kx} \), where \( L_0 \) is the intensity at the surface (\( x = 0 \)).

Step 3: Apply the Given Condition

We know that at \( x = 18 \) ft, the intensity is half of the surface intensity. So, \( L(18) = \frac{1}{2} L_0 \). Substituting into the equation gives us \( \frac{1}{2} L_0 = L_0 e^{-18k} \). Simplifying, \( \frac{1}{2} = e^{-18k} \).

Step 4: Solve for k

Taking natural logarithms of both sides, \( \ln(\frac{1}{2}) = -18k \). So, \( k = -\frac{\ln(\frac{1}{2})}{18} = \frac{\ln(2)}{18} \).

Step 5: Set Intensity Condition for Working Without Artificial Light

You can work without artificial light as long as the intensity \( L(x) \) is not less than \( \frac{1}{10} L_0 \) (one-tenth of the surface intensity). Thus, \( L(x) = \frac{1}{10} L_0 = L_0 e^{-kx} \).

Step 6: Calculate Maximum Depth x for Natural Light Using Solved k

Using the derived \( k = \frac{\ln(2)}{18} \), substitute into the equation \( \frac{1}{10} = e^{-kx} \). Take natural logarithms: \( \ln(\frac{1}{10}) = -x \frac{\ln(2)}{18} \). Solve for \( x \): \( x = -\frac{18 \ln(\frac{1}{10})}{\ln(2)} \).

Step 7: Evaluate

Calculate \( x \) using the values obtained: \( x \approx \frac{18 \times \ln(10)}{\ln(2)} \approx 59.67 \) ft. Therefore, you can work up to approximately 60 feet without artificial light.

Key Concepts

Exponential Decay

Exponential decay occurs when a quantity decreases at a rate proportional to its current value. In our context, light intensity underwater decreases as the depth increases. This is described by the differential equation \( \frac{dL}{dx} = -kL \). The negative sign indicates a decrease.
Light follows this exponential pattern as it gets absorbed and scattered by water molecules. Think of it like a glass of colored water filtering more light as more dye is added, with the color intensity representing light intensity in our scenario.
Key points about exponential decay:

• Faster decay results from a larger decay constant \(k\).
• The initial condition sets the starting value, such as the light intensity at the surface.
• The decay is asymptotic, meaning it approaches zero but never quite reaches it.

Separable Equations

Differential equations that can be written in the form \( \frac{dy}{dx} = g(y) \, h(x) \) are known as separable equations. The equation we are dealing with, \( \frac{dL}{dx} = -kL \), fits the mold perfectly because it allows us to separate variables, placing all \(L\) terms on one side and \(x\) terms on the other.
The steps in solving this are:

• Rearrange to \( \frac{1}{L} \, dL = -k \, dx \).
• Integrate both sides: \( \int \frac{1}{L} \, dL = \int -k \, dx \).
• The solutions are found by solving these integrals, leading us to an expression that uses the natural exponential \( e \) function.

By separating and solving, we obtain the general solution \( L(x) = L_0 \, e^{-kx} \), where \( L_0 \) is the initial intensity. This separation and solving process is a powerful method in calculus for addressing various real-world phenomena.

Natural Logarithms

Natural logarithms are a type of logarithm with the base \( e \), where \( e \approx 2.71828 \). They're particularly useful in calculus because of their properties with exponential functions, especially those involving \( e \).
In our problem, after finding \( e^{-18k} = \frac{1}{2} \), we take the natural logarithm of both sides for simplicity:

• \( \ln(\frac{1}{2}) = -18k \) translates the exponent problem into a multiplication.
• Solve \( k = \frac{-\ln(\frac{1}{2})}{18} = \frac{\ln(2)}{18} \), as logarithmic properties allow us to manipulate the equation conveniently.
• These logs condense complex calculations into manageable arithmetic operations.

The natural logarithm simplifies solving the exponential decay model, relating depth directly to light intensity, making it a crucial tool in the equation-solving arsenal.