Problem 22 - Calculus for Biology and Medicine

Question

Denote by \(L(t)\) the length of a fish at time \(t\), and assume that the fish grows according to the von Bertalanffy equation $$\frac{d L}{d t}=k(34-L(t)) \quad \text { with } L(0)=2$$ (a) Solve (8.44). (b) Use your solution in (a) to determine \(k\) under the assumption that \(L(4)=10\). Sketch the graph of \(L(t)\) for this value of \(k\). (c) Find the length of the fish when \(t=10\). (d) Find the asymptotic length of the fish; that is, find \(\lim _{t \rightarrow \infty} L(t)\)

Step-by-Step Solution

Answer

(a) \( L(t) = 34 - 32e^{-kt} \). (b) \( k = -\frac{1}{4} \ln(\frac{3}{4}) \). (c) \( L(10) \approx 20.5 \). (d) Asymptotic length is 34.

Step 1: Solve the Differential Equation

The given differential equation is \( \frac{dL}{dt} = k(34 - L(t)) \) with the initial condition \( L(0) = 2 \). This is a first-order linear differential equation and can be solved using separation of variables.Rearrange the terms: \( \frac{dL}{34-L} = k \, dt \).Integrate both sides: \( \int \frac{1}{34-L} \, dL = \int k \, dt \), yielding \( -\ln|34-L| = kt + C \), where \( C \) is the integration constant.Solving for \( L(t) \), we get \(|34 - L| = e^{-kt+C} \). Thus, \( L(t) = 34 - e^{C}e^{-kt} \).Using the initial condition \( L(0) = 2 \), we find that \( 34 - 2 = e^C \), so \( e^C = 32 \). Therefore, the solution is \( L(t) = 34 - 32e^{-kt} \).

Step 2: Determine the Constant k

Using the information \( L(4) = 10 \), substitute into the solution \( L(t) = 34 - 32e^{-kt} \).So, \( 10 = 34 - 32e^{-4k} \).This implies \( 32e^{-4k} = 24 \), and therefore \( e^{-4k} = \frac{3}{4} \).Taking the natural logarithm, \( -4k = \ln(\frac{3}{4}) \), so \( k = -\frac{1}{4} \ln(\frac{3}{4}) \).

Step 3: Sketch the Graph of L(t)

Using the expression for \( L(t) = 34 - 32e^{-kt} \) with the calculated \( k \), plot the function over a suitable range of \( t \). It will show a growth starting at \( L(0) = 2 \) and approaches 34 over time, reflecting the fish growing towards an asymptotic maximum length (the intercept at \( L = 10 \) when \( t = 4 \) should also be shown).

Step 4: Find the Length at t=10

Now, substitute \( t = 10 \) into \( L(t) = 34 - 32e^{-kt} \) using the value of \( k \) obtained earlier.\( L(10) = 34 - 32e^{-10k} \).Calculate this using \( k = -\frac{1}{4} \ln(\frac{3}{4}) \). The numerical solution yields \( L(10) \approx 20.5 \).

Step 5: Determine Asymptotic Length as t → ∞

The asymptotic length, \( \lim_{t \to \infty} L(t) \), is the length approached by the fish as time goes to infinity.From \( L(t) = 34 - 32e^{-kt} \), as \( t \rightarrow \infty \) , \( e^{-kt} \rightarrow 0 \), thus \( L(t) \rightarrow 34 \).So, the asymptotic length of the fish is 34 units.

Key Concepts

von Bertalanffy growth model

The von Bertalanffy growth model is a popular way to describe how certain biological creatures, like fish, grow over time. This model is an example of a first-order differential equation, which helps us predict changes in the length of an organism as time progresses. The equation we're dealing with here is:

• \[ \frac{d L}{d t} = k(34 - L(t)) \]
• This equation means the rate at which the fish's length \( L(t) \) changes with time \( t \) is proportional to how far it is from a maximum theoretical length, in this case, 34.
• The constant \( k \) affects how quickly the fish approaches this maximum length.

By solving this differential equation, we can develop an expression that predicts the fish's length at any given time, providing a clear understanding of its growth pattern.

asymptotic analysis

When we talk about asymptotic analysis in the context of the von Bertalanffy model, we are referring to the behavior of the fish's growth as time gets very large. In simpler terms, it's like asking "how long will the fish get as time goes on forever?" In mathematics, we achieve this by examining the limit of the length function \( L(t) \) as \( t \to \infty \).

• For the equation \( L(t) = 34 - 32e^{-kt} \), as \( t \to \infty \), the term \( e^{-kt} \) becomes very small because the exponential function decreases rapidly for negative values.
• This means that \( 32e^{-kt} \) approaches zero, and so \( L(t) \) approaches 34.
• The number 34 represents the asymptotic maximum size of the fish — it can never exceed this length.

This kind of analysis is crucial for understanding the long-term behavior and constraints in natural growth processes.

initial value problem

An initial value problem is a differential equation paired with a specific value at the start of the observation, called the initial condition. This allows us to find a unique solution to the differential equation tailored to a particular situation. In our fish growth model, this is represented as:

• The differential equation: \( \frac{dL}{dt} = k(34 - L(t)) \)
• Initial condition: \( L(0) = 2 \)

The initial value \( L(0) = 2 \) specifies that the length of the fish at time \( t = 0 \) is 2 units. Using this condition, we can solve the equation to find the exact growth pattern of the fish — a process known as "solving the initial value problem." This specificity is vital for making accurate predictions about the fish's size at any given time.

separation of variables

Separation of variables is a mathematical technique used to solve differential equations like our von Bertalanffy growth equation. This method works by rearranging the equation to isolate terms involving the dependent variable (here, \( L \)) from those involving the independent variable (here, \( t \)). This allows us to integrate each side separately. Here's how it's done:

• Start with the differential equation: \( \frac{dL}{dt} = k(34 - L) \).
• Rearrange to separate variables: \( \frac{dL}{34-L} = k \, dt \).
• Integrate both sides: \( \int \frac{1}{34-L} \, dL = \int k \, dt \).

This process produces a solution that can then be solved further using initial conditions. It’s a powerful tool because it simplifies complex equations into simpler forms we can solve with basic calculus methods. This way, we achieve a clear and usable solution to predict the fish's growth accurately.