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An experimenter reports that a certain strain of bacteria grows at a rate proportional to the square of the size of the population. Set up a differential equation that describes the growth of the population. Sketch a solution.

Short Answer

Expert verified
Differential equation: \( \frac{dP}{dt} = k \times [P(t)]^2 \). The population grows faster as it increases.

Step by step solution

01

Identify the growth rate

The growth rate is proportional to the square of the size of the population. This means if the population at time t is denoted by P(t), the rate of change of P(t), which is dP/dt, is proportional to [P(t)]^2.
02

Express the proportional relationship

If a quantity is proportional to another, it can be written as a constant times that quantity. Therefore, we can write \( \frac{dP}{dt} = k \times [P(t)]^2 \) where k is the proportionality constant.
03

Write the differential equation

We now have the differential equation describing the growth of the bacteria population: \( \frac{dP}{dt} = k \times [P(t)]^2 \).
04

Sketch the solution

The differential equation \( \frac{dP}{dt} = k \times [P(t)]^2 \) suggests that the population grows faster as its size increases. A graphical solution typically starts with an initial population size, increasing more rapidly over time.

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Key Concepts

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

Proportional Growth
Understanding proportional growth is crucial in solving differential equations related to population dynamics. Proportional growth means that the rate at which a population grows is related to the size of the population itself. In mathematical terms, if a population is growing proportionally, its rate of change can be expressed as a constant times the current population size. For example, if an experiment reports that bacterial growth rate is proportional to the square of its population, we can write this relationship as \( \frac{dP}{dt} = k \times [P(t)]^2 \) where \( P(t) \) is the population at time t and \( k \) is a constant. This kind of growth indicates that the larger the population, the faster it grows.
Population Dynamics
Population dynamics is the study of how and why populations change over time. Various factors such as birth rates, death rates, and environmental influences affect these dynamics. In the context of differential equations, understanding how populations evolve enables us to model and predict their future behavior. For bacterial growth, the differential equation \( \frac{dP}{dt} = k \times [P(t)]^2 \) states that as the population \( P(t) \) increases, its rate of growth \( \frac{dP}{dt} \) accelerates as well. This acceleration happens because the rate is proportional to the square of the population size. The concept can be applied to various biological populations, making it a fundamental aspect of studying biological systems.
Bacterial Growth
Bacterial growth is a fascinating application of differential equations. Bacteria often exhibit exponential growth under ideal conditions, where the growth rate is proportional to the population size. In more complex cases, bacteria may grow at a rate proportional to the square of the population size, as given by the differential equation \( \frac{dP}{dt} = k \times [P(t)]^2 \). This type of growth indicates an even faster increase in population as it grows larger. For instance, if the initial size of a bacterial culture is small, the growth starts slowly. However, as the bacteria multiply, the population expands more quickly, leading to exponential-like growth rates. This principle is vital for understanding how infections spread and can be controlled.
Differential Equation Solutions
Solving differential equations, like the one describing bacterial growth, is essential for predicting future population sizes. The equation \( \frac{dP}{dt} = k \times [P(t)]^2 \) might look complex but follows a straightforward method to reach a solution. First, we'll separate variables to solve it: $$ \int \frac{1}{P(t)^2} dP = \int k dt $$. Integrating both sides gives us: $$ - \frac{1}{P(t)} = kt + C $$. Solving for \( P(t) \), we're left with: $$ P(t) = \frac{1}{kt + C} $$. This solution shows that the population approaches infinity in a finite amount of time. By using initial conditions and calculations, one can determine specific behaviors and sizes of populations at given times. Differential equations thus provide powerful tools for understanding and managing population dynamics.

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Most popular questions from this chapter

Suppose that the Consumer Products Safety Commission issues new regulations that affect the toy-manufacturing industry. Every toy manufacturer will have to make certain changes in its manufacturing process. Let \(f(t)\) be the fraction of manufacturers that have complied with the regulations within \(t\) months. Note that \(0 \leq f(t) \leq 1\). Suppose that the rate at which new companies comply with the regulations is proportional to the fraction of companies who have not yet complied, with constant of proportionality \(k=.1\). (a) Construct a differential equation satisfied by \(f(t)\). (b) Use Euler's method with \(n=3\) to estimate the fraction of companies that comply with the regulations within the first 3 months. (c) Solve the differential equation in part (a) and compute \(f(3)\). (d) Compare the answers in parts (b) and (c) and approximate the error in using Euler's method.

Solve the given equation using an integrating factor. Take \(t>0\). $$ y^{\prime}+\frac{y}{10+t}=0 $$

Solve the initial-value problem. $$ t y^{\prime}-y=-1, y(1)=1, t>0 $$

According to the National Kidney Foundation, in 1997 more than 260,000 Americans suffered from chronic kidney failure and needed an artificial kidney (dialysis) to stay alive. (Source: The National Kidney Foundation, www.kidney.org.) When the kidneys fail, toxic waste products such as creatinine and urea build up in the blood. One way to remove these wastes is to use a process known as peritoneal dialysis, in which the patient's peritonium, or lining of the abdomen, is used as a filter. When the abdominal cavity is filled with a certain dialysate solution, the waste products in the blood filter through the peritonium into the solution. After a waiting period of several hours, the dialysate solution is drained out of the body along with the waste products. In one dialysis session, the abdomen of a patient with an elevated concentration of creatinine in the blood equal to 110 grams per liter was filled with two liters of a dialysate (containing no creatinine). Let \(f(t)\) denote the concentration of creatinine in the dialysate at time \(t\). The rate of change of \(f(t)\) is proportional to the difference between 110 (the maximum concentration that can be attained in the dialysate) and \(f(t)\). Thus, \(f(t)\) satisfies the differential equation $$ y^{\prime}=k(110-y) . $$ (a) Suppose that, at the end of a 4-hour dialysis session, the concentration in the dialysate was 75 grams per liter and it was rising at the rate of 10 grams per liter per hour. Find \(k\). (b) What is the rate of change of the concentration at the beginning of the dialysis session? By comparing with the rate at the end of the session, can you give a (simplistic) justification for draining and replacing the dialysate with a fresh solution after 4 hours of dialysis? [Hint: You do not need to solve the differential equation.]

Let \(f(t)\) be the solution of \(y^{\prime}=10-y, y(0)=1\). Use Euler's method with \(n=5\) to estimate \(f(1)\). Then, solve the differential equation and find the exact value of \(f(1)\).

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