Problem 80 - Calculus for Biology and Medicine

Question

Fish are indeterminate growers; that is, their length \(L(t)\) increases with age \(t\) throughout their lifetime. If we plot the growth rate \(d L / d t\) versus age \(t\) on semilog paper, a straight line with negative slope results, meaning that: $$ \frac{d L}{d t}=A e^{-k t} $$ where \(A>0\) and \(k>0\) are both coefficients that depend on the species of fish, and the habitat that it is growing in. (a) Find the solution for this differential equation (your solution will include \(A\) and \(k\) as unknown constants, as well as one additional unknown constant \(C\) from the antiderivative). (b) Find the values for the constants \(A, k, C\), that would fit the solution to the following data \(L(0)=5, L(1)=10\), and $$ \lim _{t \rightarrow \infty} L(t)=30. $$ (c) Graph the solution \(L(t)\) as a function of \(t\).

Step-by-Step Solution

Answer

The solution is \( L(t) = -25 e^{-0.223t} + 30 \), with \( A = 5.575 \), \( k = 0.223 \), and \( C = 30 \).

Step 1: Set up the Differential Equation

The differential equation given is \( \frac{dL}{dt} = A e^{-kt} \). This equation describes the growth rate of the length \( L(t) \) of a fish over time \( t \).

Step 2: Integrate to Solve for L(t)

Integrate both sides with respect to \( t \) to find \( L(t) \). \[ \int dL = \int A e^{-kt} \, dt \]This leads to\[ L(t) = -\frac{A}{k} e^{-kt} + C \]where \( C \) is the integration constant.

Step 3: Apply Initial Condition L(0)=5

Substitute \( t = 0 \) and \( L(0) = 5 \) into the equation:\[ 5 = -\frac{A}{k} e^{0} + C \]This simplifies to:\[ 5 = -\frac{A}{k} + C \]

Step 4: Apply Boundary Condition at Infinity

Given \( \lim_{t \to \infty} L(t) = 30 \):As \( t \to \infty \), \( e^{-kt} \to 0 \), so the equation becomes:\[ \lim_{t \to \infty} L(t) = C = 30 \]

Step 5: Use Condition L(1)=10

Substitute \( t = 1 \) and \( L(1) = 10 \):\[ 10 = -\frac{A}{k} e^{-k} + 30 \]Rearrange to solve for \( A \) and \( k \):\[ -\frac{A}{k} e^{-k} = -20 \]\[ \frac{A}{k} e^{-k} = 20 \]

Step 6: Solve for Constants A and k

From Steps 3 and 5:1. \( 5 = -\frac{A}{k} + 30 \rightarrow \frac{A}{k} = 25 \)2. \( \frac{A}{k} e^{-k} = 20 \)By solving \( 25 e^{-k} = 20 \), we get\[ e^{-k} = \frac{4}{5} \rightarrow k = -\ln(\frac{4}{5}) \]Thus \( k = 0.223 \) (approximately).Finally, \( A = 25k = 5.575 \).

Step 7: Write the Function L(t)

Using \( k = 0.223 \) and \( A = 5.575 \), substitute back into \( L(t) = -\frac{A}{k} e^{-kt} + C \):\[ L(t) = -25 e^{-0.223t} + 30 \].

Step 8: Graph L(t)

Plot the function \( L(t) = -25 e^{-0.223t} + 30 \) against time \( t \). The graph should show an increasing function asymptotically approaching 30, starting at \( L(0) = 5 \) and passing through \( L(1) = 10 \).

Key Concepts

Indeterminate Growth in Fish

In the world of biology, certain species exhibit what is known as indeterminate growth. This means they continue to grow in size throughout their lifetime—something particularly observed in fish. Their body length, denoted as \( L(t) \) where \( t \) is the age or time, does not stop growing even after reaching maturity. The growth rate \( \frac{dL}{dt} \) of these organisms is often a key focus as it directly influences their lifelong development.

To understand this growth mathematically, scientists use differential equations. For instance, in our problem, we encounter the equation:

• \( \frac{dL}{dt} = A e^{-kt} \)

Here, both \( A \) and \( k \) are coefficients that vary based on the species and its specific environment. Therefore, when graphing the growth rate of such fish on semilog paper, a linear trajectory with a negative slope may appear, emphasizing how logarithmic transformations provide clarity in trends over time.
This approach not only aids in understanding the continuous growth pattern but also simplifies predictions of future growth at any given age \( t \). This is crucial for ecologists, fishery managers, and conservationists because it assists in making informed decisions regarding fish populations and their habitats.

The Role of Integration in Solving Differential Equations

Integration is a mathematical process commonly used to solve differential equations. Essentially, it's the reverse process of differentiation. Given a rate of change, such as \( \frac{dL}{dt} = A e^{-kt} \), integration allows us to determine the function \( L(t) \), expressed in terms of its initial conditions and boundary constraints.

To find \( L(t) \), we perform the integration:

• \( \int dL = \int A e^{-kt} \, dt \)

This leads to the solution:

• \( L(t) = -\frac{A}{k} e^{-kt} + C \)

Here, \( C \) is an integration constant that can be determined using known initial conditions, such as \( L(0) = 5 \) or certain limits as \( t \) approaches infinity.
Integration not only provides the explicit expression for the growth of fish but also plays a critical role in ensuring that solutions are complete. By applying integration, we can tailor solutions to fit specific data points, making it an invaluable tool in practical applications such as modeling growth over time.

Understanding Exponential Decay

Exponential decay is the mathematical description of a process whereby a quantity decreases at a rate proportional to its current value. It's commonly used to model phenomena in nature, including the decrease in growth rate of indeterminate growers like fish.

In our context, \( e^{-kt} \) signifies exponential decay. As time \( t \) progresses:

• \( e^{-kt} \to 0 \) as \( t \to \infty \)

This means that while fish continue to grow, their rate of growth slows down exponentially, eventually approaching zero but never completely stopping. Such patterns are not only relevant in biological contexts but are pervasive in fields ranging from physics to finance.
Understanding exponential decay aids in interpreting fish growth dynamics and offers insights when applying models to predict long-term outcomes. It ensures that predictions are both realistic and aligned with biological realities, proving the robustness of mathematical models in representing natural processes.