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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 \(e^{r t} .\)

Short Answer

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

Step by step solution

01

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 \( p \).
02

Cross-multiply

To eliminate the fraction, cross-multiply to obtain \( p = K(M-p) \). This helps us solve for \( p \) in terms of \( K \).
03

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: \( p + Kp = KM \).
04

Factor out \( p \)

Factor \( p \) out of the left side: \( p(1 + K) = KM \). This allows us to isolate \( p \).
05

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 \( p \).
06

Substitute back for \( K \)

Substitute \( K = \frac{p_{0}}{M-p_{0}} e^{r t} \) back into the equation: \( p = \frac{M \left( \frac{p_{0}}{M-p_{0}} e^{r t} \right)}{1 + \frac{p_{0}}{M-p_{0}} e^{r t}} \).
07

Simplify the expression

Simplify the expression by multiplying the numerator and denominator by \( M-p_{0} \): \( p = \frac{M p_{0} e^{r t}}{M-p_{0} + p_{0} e^{r t}} \).
08

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Show that for $$ p(t)=\frac{M p_{0}}{p_{0}+\left(M-p_{0}\right) e^{-r t}} $$a. \(p(0)=p_{0}\) $$\text { b. } \lim _{t \rightarrow \infty} p(t)=M$$

Show that the variables are not separable in the equation $$\text { a. } \quad y^{\prime}(t)=\ln (t \times y) \quad \text { b. } \quad y^{\prime}(t)=\ln (t+y)$$

There is 'conventional wisdom' among SCUBA divers that if you are going to make a dive that involves two depths, 'do the deep part first'. This problem and the next explores rationale for that wisdom. To be concrete, assume that \(K=0.071 / \mathrm{min}\) which corresponds to approximately 10 minute half- life for the compartment \(((\ln 2) / 0.07=9.9 \mathrm{~min})\). a. Assume a diver (d1) descends immediately to 10 meters and stays there for 15 minutes, then descends to 30 meters and stays there for 10 minutes. Let $$d_{1}(t)=\left\\{\begin{array}{ll}10 & \text { for } 0 \leq t \leq 15 \\\30 & \text { for } 15

Some population scientists have argued that population density can get so low that reproduction will be less than natural attrition and the total population will be lost. Named the Allee effect. after W. C. Allee who wrote extensivelv about it \(^{6}\), this mav be a basis for arguing. for example that marine fishing of a certain species (Atlantic cod on Georges Bank, for example \({ }^{7}\) ee Paul Greenberg, "Four Fish, the Future of the Last Wild Food, The Penguin Press, 2010 . The model is a lot more complicated than we present here.) should be suspended, despite the presence of a small residual population. How should we modify the logistic differential equation, \(u^{\prime}=u(1-u),\) to incorporate such a threshold? Assume a fixed area and uniform density throughout the area and a threshold number, \(\epsilon .\) If the population number is less than \(\epsilon\) the population will decline; if the population number is more than \(\epsilon\) the population will increase. a. Modify the direction field for \(u^{\prime}=u(1-u)\) to account for the Allee effect. That is, a \(u, t\) plane, draw the line \(u=1\) and a threshold line \(u=\epsilon\) where \(\epsilon=0.1,\) say. Arrows below \(u=\epsilon\) should point downward; arrows between \(u=\epsilon\) and \(u=1\) should point upwards. Draw enough direction field arrows to indicate the paths of solutions for a threshold model. b. Draw the the logistic phase plane graph and the phase plane graphs for the following three candidates of a threshold logistic differential equation where \(\epsilon=0.1\). \(u^{\prime}=f(u)=u \times(1-u) \quad\) Logistic \(u^{\prime}=f_{1}(u)=u^{\frac{2}{3}} \times(u-\epsilon)^{\frac{1}{3}} \times(1-u) \quad\) Candidate 1 \(u^{\prime}=f_{2}(u)=u \times \frac{u-\epsilon}{u+\epsilon} \times(1-u)\) Candidate 2 \(u^{\prime}=f_{3}(u)=u \times(u-\epsilon) \times(1-u)\) Candidate 3

Write an equation that describes the temperature of an egg after it is uncovered (the adult bird leaves the nest to feed). Assume that the rate of change of the temperature of the egg is proportional to difference between the air temperature and the egg temperature.

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