Problem 10 - Calculus for Biology and Medicine

Question

Fish Growth The rate of growth of a fish is sometimes modeled by the equation $$ d L / d t=L_{0} e^{-k t} $$ where \(L\) is the length of the fish, and \(k\) and \(L_{0}\) are positive constants. (a) Interpret in words the quantity \(\int_{0}^{3} \frac{d L}{d f} d t\). (b) Calculate the integral from part (a); your answer will include the constants \(k\) and \(L_{0}\).

Step-by-Step Solution

Answer

Total change in fish length from 0 to 3 is \( \frac{L_{0}}{k} (1 - e^{-3k}) \).

Step 1: Interpret the Integral

The integral \( \int_{0}^{3} \frac{dL}{df} dt \) represents the total change in length of the fish from time \( t = 0 \) to \( t = 3 \). This integral accumulates the incremental changes in the fish's length over the specified time period, based on the growth rate equation provided.

Step 2: Set Up the Integral

We need to calculate \( \int_{0}^{3} L_{0} e^{-kt} dt \). This integral represents the accumulation of the instantaneous growth rate of the fish over the interval from 0 to 3.

Step 3: Evaluate the Integral

First, integrate \( L_{0} e^{-kt} \) with respect to \( t \). The antiderivative of \( e^{-kt} \) is \( \frac{-1}{k} e^{-kt} \). Thus, the integral to evaluate is:\[\int_{0}^{3} L_{0} e^{-kt} dt = L_{0} \left[ \frac{-1}{k} e^{-kt} \right]_{0}^{3}.\]

Step 4: Calculate Definite Integral

Substitute the limits of integration. - For \( t = 3 \): \[ L_{0} \left( \frac{-1}{k} e^{-3k} \right) \]- For \( t = 0 \): \[ L_{0} \left( \frac{-1}{k} e^{0} \right) = L_{0} \left( \frac{-1}{k} \right). \]Combine these to get:\[L_{0} \left( \frac{-1}{k} e^{-3k} - \frac{-1}{k} \right) = L_{0} \left( \frac{-1}{k} e^{-3k} + \frac{1}{k} \right).\]

Step 5: Simplify the Expression

Simplify the expression from the previous step:\[L_{0} \left( \frac{-1}{k} e^{-3k} + \frac{1}{k} \right) = L_{0} \left( \frac{1}{k} \right) \left( 1 - e^{-3k} \right).\] So, the solution to the integral is:\[ \frac{L_{0}}{k} \left( 1 - e^{-3k} \right). \]

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives, which represent rates of change. In the context of the fish growth model, the differential equation \( \frac{dL}{dt} = L_0 e^{-kt} \) describes how the length \( L \) of a fish changes over time based on certain conditions. Here, \( L_0 \) is the initial length rate constant and \( k \) is a decay constant, both positive. This equation provides us with a dynamic model of growth, indicating that as time increases, the rate of growth influenced by \( e^{-kt} \) diminishes exponentially due to the presence of the negative exponent. Such equations are fundamental in modeling real-world processes as they capture the continuous change across conditions.
This specific type of differential equation, being first-order, implies that the solution will typically involve an understanding of exponential behaviors. When analyzing fish growth, the differential aspect allows us to focus not on the static value of \( L \), but on its development across time \( t \). This results in a powerful tool to predict future values or to backtrack the growth process from past observations.

Integral Calculus

Integral calculus plays a crucial role when we want to find the total accumulation of changes described by our differential equation. In the exercise, we find ourselves calculating the integral \( \int_{0}^{3} L_0 e^{-kt} \, dt \); this represents the total change in the fish's length over the given period from \( t = 0 \) to \( t = 3 \).
By integrating the function, we accumulate all the tiny changes in the fish's length over the specified time interval. This is important in applications where understanding total growth or accumulation is necessary, such as determining how much a fish has grown over a certain period. Calculating the integral involves finding an antiderivative, which in practical terms translates to reversing the differentiation process.

• The antiderivative of \( e^{-kt} \) is \( \frac{-1}{k} e^{-kt} \), accounting for the constant of decay \( k \).
• The evaluation at the upper and lower limits \( t=3 \) and \( t=0 \), respectively, allows us to find the net growth over the specific period.
• The result, \( \frac{L_0}{k} \left( 1 - e^{-3k} \right) \), delivers the solution to the integral, showing how much length the fish gains from time \( 0 \) to time \( 3 \).

Integral calculus is the mathematical tool that provides us with the ability to understand these total changes in a continuous function like this one.

Exponential Functions

Exponential functions are characterized by a constant raised to the power of a variable. In the fish growth model, \( e^{-kt} \) is an exponential function reflecting decay, which moderates how the fish grows over time. This exponential decay is particularly interesting in biological models where growth slows down over time due to various limiting factors.
The general form \( e^{-kt} \) causes the growth rate to decrease swiftly as \( t \) increases, due to the nature of the negative exponent \(-kt\). Here are some properties and implications:

• The negative sign in the exponent signifies a decay process, opposite to exponential growth which would have a positive exponent.
• As time \( t \) grows larger, the term \( e^{-kt} \) gets smaller, meaning the contribution of each additional time unit to growth becomes less significant.
• This aligns with many natural growth processes where initial rapid growth eventually levels off, leading to a more sustainable long-term growth rate.

Understanding the behavior of exponential functions in differential equations like this one is key to capturing realistic growth phenomena and predicting future trends based on initial conditions.