Problem 6 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Solve for \(p\) in $$\frac{p}{M-p}=\frac{p_{0}}{M-p_{0}} e^{r t}$$ to obtain Equation 17.31 , $$p(t)=\frac{M p_{0}}{p_{0}+\left(M-p_{0}\right) e^{-r t}} "$$ It will be useful to first solve for \(p\) in $$\frac{p}{M-P}=K \quad\left(K \quad \text { replaces } \quad \frac{p_{0}}{M-p_{0}} e^{r t}\right)$$ You should get $$p=M \times \frac{K}{1+K}$$ Then substitute $$K=\frac{p_{0}}{M-p_{0}} e^{r t} $$and simplify. As a final step, divide numerator and denominator by \(e^{r t} .\)

Step-by-Step Solution

Answer

Solve for \( p \) using \( K \), substitute \( K \) back, and simplify.

Step 1: Rewrite the original equation

Start with the equation \( \frac{p}{M-p} = K \), where \( K = \frac{p_{0}}{M-p_{0}} e^{r t} \). Here, we replace the given expression with \( K \) to simplify solving for \( p \).

Step 2: Cross-multiply

To eliminate the fraction, cross-multiply to obtain \( p = K(M-p) \). This helps us solve for \( p \) in terms of \( K \).

Step 3: Expand and rearrange

Expand the right side of the equation: \( p = KM - Kp \). Rearrange this to get all terms involving \( p \) on the same side: \( p + Kp = KM \).

Step 4: Factor out \( p \)

Factor \( p \) out of the left side: \( p(1 + K) = KM \). This allows us to isolate \( p \).

Step 5: Solve for \( p \)

Divide both sides of the equation by \( 1 + K \) to solve for \( p \), yielding \( p = \frac{KM}{1 + K} \). This is the expression we sought for \( p \).

Step 6: Substitute back for \( K \)

Substitute \( K = \frac{p_{0}}{M-p_{0}} e^{r t} \) back into the equation: \( p = \frac{M \left( \frac{p_{0}}{M-p_{0}} e^{r t} \right)}{1 + \frac{p_{0}}{M-p_{0}} e^{r t}} \).

Step 7: Simplify the expression

Simplify the expression by multiplying the numerator and denominator by \( M-p_{0} \): \( p = \frac{M p_{0} e^{r t}}{M-p_{0} + p_{0} e^{r t}} \).

Step 8: Final simplification

Finally, divide both the numerator and the denominator by \( e^{r t} \) to get the final expression: \( p(t) = \frac{M p_{0}}{p_{0} + (M-p_{0}) e^{-r t}} \). This matches the given Equation 17.31.

Key Concepts

Logistic Growth Model

The logistic growth model is a classic model used to describe how populations grow in an environment with limited resources. Unlike exponential growth, which assumes unlimited resources, the logistic growth model factors in limits on growth, resulting in an S-shaped curve known as a sigmoid. This type of growth is more realistic for biological populations.
The model is based on the differential equation:\[ \frac{dp}{dt} = rp \left( 1 - \frac{p}{M} \right) \]where:

• \( p \) is the population size,
• \( r \) is the intrinsic growth rate, and
• \( M \) is the carrying capacity.

As the population size approaches the carrying capacity \( M \), the growth rate slows down, eventually reaching zero when the population size equals \( M \). The logistic model provides a more comprehensive understanding of how populations stabilize over time in a resource-limited habitat.

Calculus for Life Sciences

Calculus plays a vital role in the life sciences, particularly when modeling biological processes, such as population dynamics. In the context of differential equations like the logistic growth model, calculus helps us understand how populations change over time.
Differential equations are used to describe rates of change, which are crucial for biological systems that are continuous and dynamic. In this case, they allow us to calculate population sizes at any given time based on specific growth parameters.
In life sciences, calculus is not only used for population models but also for understanding processes such as:

• Drug concentration levels in pharmacokinetics,
• Rates of enzyme reactions in biochemistry, and
• Growth patterns in ecology.

By applying calculus, scientists can make predictions about biological systems, optimize processes, and even design better strategies for managing biological resources.

Exponential Decay

Exponential decay is a process where a quantity decreases at a rate proportional to its current value. This concept is often visualized as a rapidly decreasing curve that never quite reaches zero.
The mathematical representation of exponential decay is:\[ p(t) = p_0 e^{-rt} \]where:

• \( p(t) \) is the quantity of interest at time \( t \),
• \( p_0 \) is the initial quantity, and
• \( r \) is the decay rate.

Exponential decay is prevalent in processes such as radioactive decay, cooling of objects, and certain chemical reactions. In the logistic growth model context, this concept helps in adjusting the interpreted growth when the population size is near the carrying capacity, as the excess growth diminishes increasingly fast.
Understanding exponential decay is crucial for fields like ecology, pharmacology, and environmental science, where predicting how fast substances break down or how populations decline can have significant implications. By mastering this concept, students can better analyze natural phenomena that involve gradual reduction over time.