Problem 7 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Light intensities, \(I_{1}\) and \(I_{2}\), are measured at depths \(d\) in meters in two lakes on two different days and found to be approximately $$ I_{1}=22^{-0.1 d} \quad \text { and } \quad I_{2}=42^{-0.2 d} $$ a. What is the half-life of \(I_{1} ?\) b. What is the half-life of \(I_{2} ?\) c. Find a depth at which the two light intensities are the same. d. Which of the two lakes is the muddiest?

Step-by-Step Solution

Answer

Lake 2 is muddier. Half-lives can be calculated via logarithms.

Step 1: Define Half-Life Equation

Half-life is the point at which the intensity is reduced to half of its initial value. For light intensity \(I_1\), the half-life can be expressed by setting \(I_1 = \frac{22^{0}}{2} = 22^{-0.1d}\).

Step 2: Solve for Half-Life of I_1

Set up the equation for \(I_1\) and simplify:\[22^{-0.1d} = \frac{1}{2}\]Taking the logarithm of both sides, we have:\[-0.1d \log 22 = \log\left(\frac{1}{2}\right)\]Solve for \(d\):\[d = \frac{\log\left(\frac{1}{2}\right)}{-0.1 \log 22}\]

Step 3: Solve for Half-Life of I_2

Similar to Step 2, set \(I_2 = \frac{42^0}{2} = 42^{-0.2d}\).Set up the equation:\[42^{-0.2d} = \frac{1}{2}\]Then take logarithms to get:\[-0.2d \log 42 = \log\left(\frac{1}{2}\right)\]Now solve for \(d\):\[d = \frac{\log\left(\frac{1}{2}\right)}{-0.2 \log 42}\]

Step 4: Equate Intensities for Same Depth

Set \(I_1 = I_2\):\[22^{-0.1d} = 42^{-0.2d}\]Taking logarithms:\[-0.1d \log 22 = -0.2d \log 42\]Solving for \(d\):\[d = \frac{\log \left(\frac{42}{22}\right)}{0.1 \log 22 - 0.2 \log 42}\]

Step 5: Analyze Muddiest Lake

The lake exhibiting a quicker reduction in light intensity (larger negative exponent) will have muddier water. Lake 2 has \(I_2 = 42^{-0.2d}\) as compared to \(I_1 = 22^{-0.1d}\). Hence, Lake 2 is muddier because the light intensity decreases faster, implying less light penetrates, indicating muddier conditions.

Key Concepts

Half-life Calculation

Half-life is an important concept in many fields, including physics and chemistry. It describes the time it takes for a quantity to reduce to half its initial value. In this context, we're discussing light intensity attenuation in lakes. Here's how to calculate it for given functions like in the exercise:
The light intensity functions are given by exponential values. For example, for intensity \(I_1\), we use the formula \(I_1 = 22^{-0.1d}\). To find the half-life, set \(I_1\) to half of its full value; mathematically, this means solving \(22^{-0.1d} = \frac{1}{2}\). By applying logarithmic techniques, you can solve for \(d\), which tells you the depth at which the intensity is halved.
This same approach is applied to calculate the half-life of \(I_2\) as well. Students should remember that different formulas may have different base values or exponents, which will affect the half-life calculation. Understanding how these variables play into half-life equations will help you interpret physical changes in diverse environments.

Exponential Decay

Exponential decay describes processes where quantities decrease at rates proportional to their current value. It's a common way to model scenarios like radioactive decay, cooling, or, in our case, the attenuation of light in water.
The function \(I = 22^{-0.1d}\) and \(I = 42^{-0.2d}\) are examples of exponential decay. Here, the base of the exponential expression (22 and 42) and the exponent (-0.1 and -0.2) determine the decay rate. Lower bases with larger negative exponents result in faster decay, meaning light intensity reduces more rapidly as depth increases.
Exponential decay functions differ from linear functions mainly because they decrease non-linearly, forming a curve that never quite touches the zero line. It's crucial in understanding natural processes where changes slow down over time.

Logarithmic Equations

Logarithmic equations are often used to solve exponential decay problems, like light attenuation in lakes. Logs help in simplifying exponential equations where direct calculation isn't feasible.
When you encounter equations like \(22^{-0.1d} = \frac{1}{2}\), logarithms transform them into manageable algebraic forms. The application of logs allows you to "bring down" exponents so you can solve for unknowns like depth. In this exercise, using properties of logarithms simplifies the exponential equations to solve for depths where intensities match or reach half-life.
Understanding how to employ log equations efficiently aids in decyphering the relationships between exponential parameters. It also broadens your analytical skills when dealing with exponential decay or growth in a variety of contexts, from natural phenomena to financial applications.