Problem 23 - Calculus for Biology and Medicine

Question

Denote by \(L(t)\) the length of a certain fish at time \(t\), and assume that this fish grows according to the von Bertalanffy equation $$\frac{d L}{d t}=k\left(L_{\infty}-L(t)\right) \quad \text { with } L(0)=1$$ where \(k\) and \(L_{\infty}\) are positive constants. A study showed that the asymptotic length is equal to 123 in and that it takes this fish 27 months to reach half its asymptotic length. (a) Use this information to determine the constants \(k\) and \(L_{\infty}\) in (8.45). [Hint: Solve (8.45).] (b) Determine the length of the fish after 10 months. (c) How long will it take until the fish reaches \(90 \%\) of its asymptotic length?

Step-by-Step Solution

Answer

Constants: \( k \approx 0.0257 \), \( L_\infty = 123 \). Length after 10 months: \( 28.5 \) inches. Reaches 90% length in \( about 89.7 \) months.

Step 1: Solve the Differential Equation

We have the equation \( \frac{dL}{dt} = k(L_\infty - L(t)) \) with the initial condition \( L(0) = 1 \). This is a separable differential equation. Begin by rewriting it as \( \frac{1}{L_\infty - L} \frac{dL}{dt} = k \). Integrating both sides with respect to \( t \) gives \( -\ln|L_\infty - L| = kt + C \). Solving for \( L \), we find \( L(t) = L_\infty - (L_\infty - 1)e^{-kt} \) using the initial condition \( L(0) = 1 \).

Step 2: Use Given Conditions to Find Constants

We know \( L_\infty = 123 \) and that it takes the fish 27 months to reach half its asymptotic length, i.e., \( L(27) = \frac{L_\infty}{2} = 61.5 \). Substitute into \( 123 - 122e^{-k \times 27} = 61.5 \). Solving this equation gives \( e^{-27k} \approx 0.5 \), so \( k \approx \frac{\ln 2}{27} \approx 0.0257 \).

Step 3: Calculate Length After 10 Months

Substitute \( t=10 \) into the equation \( L(t) = 123 - 122e^{-kt} \) using \( k \approx 0.0257 \): \( L(10) = 123 - 122e^{-0.0257 \times 10} \). Simplifying gives \( L(10) \approx 123 - 122 \times 0.766 \approx 28.5 \).

Step 4: Calculate Time to Reach 90% of Asymptotic Length

We want \( L(t) = 0.9 \times 123 = 110.7 \). Set up the equation: \( 110.7 = 123 - 122e^{-kt} \). Solving for \( t \), we find \( e^{-kt} = \frac{123 - 110.7}{122} \approx 0.1 \). Thus, \( -kt = \ln 0.1 \) and solving for \( t \) gives \( t \approx -\frac{\ln 0.1}{0.0257} \approx 89.7 \) months.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are powerful tools used to model real-world phenomena where change is continuous, like the growth of a population or the cooling of a hot object over time. In general, a differential equation expresses a relationship between a function and its derivative, which reflects how the function changes.
In the context of the von Bertalanffy equation, a specific type of differential equation models the growth of fish. It’s called a separable differential equation because it can be separated into two different parts, making it easier to solve. You can think of it like a puzzle where you rearrange pieces until they fit neatly into two distinct groups, then solve each group individually.

• **Separable Differential Equations:** These equations allow us to rearrange terms so we can integrate both sides separately.
• **Initial Conditions:** These are the starting points or values of the function that help us solve the equation completely.
• **Integration:** This is a mathematical process used to find the function from its derivative.

Understanding differential equations is key to solving many problems in science and engineering, including modeling how a fish grows over time, as seen in this exercise.

Fish Growth Model

The von Bertalanffy equation is a classic fish growth model used to describe how fish grow over time. This model helps researchers understand and predict growth patterns, which is crucial for fisheries management and conservation efforts. The general form of the equation is given by \( \frac{dL}{dt} = k(L_\infty - L(t)) \).Here’s a breakdown of its components:

• **\( L(t) \):** This represents the length of the fish at any given time \( t \). As time progresses, \( L(t) \) approaches the asymptotic length \( L_\infty \).
• **\( L_\infty \):** Known as the asymptotic length, this is the maximum potential length a fish can reach if it grew indefinitely. In this exercise, it is given as 123 inches.
• **\( k \):** This constant determines the rate of growth. A larger \( k \) implies faster growth towards the asymptotic length.
• **Time \( t \):** Time is usually measured in months or years, depending on the species being studied.

The von Bertalanffy equation balances growth with the natural limit set by \( L_\infty \), giving us a realistic model of how natural growth rates slow as they reach a maximum size.

Asymptotic Length

Asymptotic length, denoted by \( L_\infty \) in the von Bertalanffy equation, is a crucial parameter in growth models. It represents the hypothetical maximum size that an organism, like a fish, could potentially reach if environmental conditions were perfect and no other limiting factors were present.
In practice, no fish will typically reach its exact asymptotic length due to the influence of various factors like food availability, predation, and disease. However, \( L_\infty \) provides a useful benchmark for comparing growth across different individuals or species.

• **Biological Significance:** The asymptotic length helps researchers estimate the growth potential of fish populations and assists in managing sustainable fisheries.
• **Calculation:** It is typically estimated using empirical data or derived from long-term observations. In our exercise, \( L_\infty \) is given as 123 inches.
• **Modeling Growth Rates:** Knowing \( L_\infty \) allows for the prediction of growth rates and how long it might take a fish to reach certain sizes relative to its maximum potential size.

This concept is essential in ecology and helps biologists make informed decisions about species management and conservation strategies.