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

Question

Trout, Moose, and Bear lakes are connected into a chain by a stream that runs into Trout Lake, out of Trout Lake into Moose Lake, out of Moose Lake and into Bear Lake and out of Bear Lake. The volumes of all of the three lakes are the same, and stream flow is constant into and out of all lakes. A load of waste is dumped into Trout Lake. With \(t\) measured in days and concentration measured in \(\mathrm{mg} / \mathrm{l}\), the concentration of wastes in the three lakes is projected to be $$ \begin{array}{ll} \text { Trout Lake: } & C_{\mathrm{T}}(t)=0.01 e^{-0.05 t} \\ \text { Moose Lake: } & C_{\mathrm{M}}(t)=0.005 t e^{-0.05 t} \\ \text { Bear Lake: } & C_{\mathrm{B}}(t)=0.000025 \frac{t^{2}}{2} e^{-0.05 t} \end{array} $$ For each lake, find the time interval, it any, on which the concentration in the lake is increasing.

Step-by-Step Solution

Answer

Moose Lake: increasing on \(0 < t < 20\); Bear Lake: increasing on \(0 < t < 40\); Trout Lake: always decreasing.

Step 1: Understand the Problem

We need to find the time intervals where the concentration of waste in each lake is increasing. This requires us to find the derivative of each concentration function and determine where these derivatives are positive.

Step 2: Differentiation of Trout Lake Concentration

The function for Trout Lake is given as \( C_{\mathrm{T}}(t) = 0.01 e^{-0.05t} \). The derivative, \( C_{\mathrm{T}}'(t) \), is found using the chain rule: \( C_{\mathrm{T}}'(t) = 0.01 \times (-0.05) e^{-0.05t} = -0.0005 e^{-0.05t} \). This derivative is always negative, so the concentration in Trout Lake is always decreasing.

Step 3: Differentiation of Moose Lake Concentration

For Moose Lake, the function is \( C_{\mathrm{M}}(t) = 0.005 t e^{-0.05t} \). Applying the product rule, we have: \[ C_{\mathrm{M}}'(t) = 0.005 e^{-0.05t} + 0.005 t (-0.05) e^{-0.05t} = 0.005 e^{-0.05t} (1 - 0.05t). \] The concentration increases when \( C_{\mathrm{M}}'(t) > 0 \), i.e., when \( 1 - 0.05t > 0 \). Solving for \( t \), we get \( t < 20 \). Hence, the concentration is increasing for \( 0 < t < 20 \).

Step 4: Differentiation of Bear Lake Concentration

The function for Bear Lake is \( C_{\mathrm{B}}(t) = 0.000025 \frac{t^2}{2} e^{-0.05t} \). Applying the product rule: \[ C_{\mathrm{B}}'(t) = 0.000025 \frac{t^2}{2}(-0.05) e^{-0.05t} + 0.000025 t e^{-0.05t}. \] Simplifying gives \[ C_{\mathrm{B}}'(t) = 0.000025 e^{-0.05t} \left(t - 0.025t^2\right). \] Setting \( C_{\mathrm{B}}'(t) > 0 \) means \( t - 0.025t^2 > 0 \). Solving gives \( t (1 - 0.025t) > 0 \), leading to \( 0 < t < 40 \). Therefore, the concentration increases for \( 0 < t < 40 \).

Key Concepts

Derivatives

Derivatives are a fundamental concept in differential calculus. They represent the rate at which a function is changing at any given point. In the context of lake concentrations, the derivative tells us how the concentration of waste in each lake is changing with respect to time. For example, if the derivative is positive, the concentration is increasing, while a negative derivative indicates a decreasing concentration.

• To find the derivative of a function, we need to understand different rules and techniques, such as the Chain Rule and Product Rule, which are essential for tackling complex functions.
• In the given problem, we are interested in identifying when the lake's concentration increases, determined by finding where the derivative is positive.

Understanding derivatives allow us to predict changes and make informed decisions based on mathematical models like the waste concentration in lakes.

Chain Rule

The Chain Rule is a method in calculus used when differentiating composite functions, where one function is inside of another. It allows us to see how a change in one part of the function affects the entire function.

• In the formula form, if we have a composite function like \( g(h(x)) \), the derivative is \( g'(h(x)) \cdot h'(x) \).
• In our problem with Trout Lake, the concentration function is \( C_{\mathrm{T}}(t) = 0.01 e^{-0.05t} \), which is a simple exponential function with a constant.
• Taking its derivative involves using the Chain Rule, where we first differentiate the outer function, the exponential, and then multiply it by the derivative of the inner function, which is \(-0.05t\).

This technique simplifies the process of differentiating more complex expressions by breaking them down into more manageable parts.

Product Rule

The Product Rule is essential when differentiating functions that are products of two or more functions. It states that if you have a function \( u(t) \cdot v(t) \), the derivative is \( u'(t) \cdot v(t) + u(t) \cdot v'(t) \).

• For Moose Lake, the concentration function \( C_{\mathrm{M}}(t) = 0.005 t e^{-0.05t} \) is a product of \( 0.005t \) and \( e^{-0.05t} \).
• Using the Product Rule helps us differentiate by taking the derivative of one function while keeping the other constant, then swapping, and adding the results.
• This ensures that we properly account for all parts of the function, leading to accurate results and allowing us to determine when the concentration in the lake is increasing.

Utilizing the Product Rule simplifies working with complex multipliers, making differentiation quick and simple.

Exponential Functions

Exponential functions are mathematical functions of the form \( a \cdot e^{kt} \), where \( e \) is the base of natural logarithms, approximately equal to 2.718. They are crucial in modeling growth and decay processes.

• In the context of lake concentrations, each one of our lake functions incorporated an exponential decay \( e^{-0.05t} \), showing how concentration decreases over time.
• They are particularly important in this scenario as they help model how different factors like the initial waste dump and time affect pollution levels in the lakes.
• The behavior of exponential functions, specifically their decay or growth rates, can be determined through the sign and value of the exponent, making it easier to predict long-term changes in the system.

Understanding exponential functions provides insights into how systems naturally evolve, exhibiting fast changes initially and stabilizing over time.