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In an experiment, a certain type of bacteria was being added to a culture at the rate of \(e^{.03 t}+2\) thousand bacteria per hour. Suppose that the bacteria grow at a rate proportional to the size of the culture at time \(t\), with constant of proportionality \(k=.45 .\) Let \(P(t)\) denote the number of bacteria in the culture at time \(t\). Find a differential equation satisfied by \(P(t)\).

Short Answer

Expert verified
The differential equation is \( \frac{dP}{dt} = 0.45P(t) + (e^{0.03t} + 2) \).

Step by step solution

01

Identify the given information

The rate at which bacteria is being added to the culture is given by the function: \[ R(t) = e^{0.03t} + 2 \]The bacteria grow at a rate proportional to the size of the culture with a constant of proportionality \( k = 0.45 \). Let \( P(t) \) represent the number of bacteria at time \( t \).
02

Write the growth rate equation

The growth rate of the bacteria, at any time \( t \), can be written as: \[ \frac{dP}{dt} = kP(t) \]
03

Add the rate of new bacteria added to the culture

The differential equation also includes the rate at which new bacteria are being added:\[ \frac{dP}{dt} = kP(t) + (e^{0.03t} + 2) \]
04

Substitute the value of k

Substitute the constant of proportionality \( k = 0.45 \):\[ \frac{dP}{dt} = 0.45P(t) + (e^{0.03t} + 2) \]

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Key Concepts

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

bacterial growth
Bacterial growth refers to the increase in the number of bacteria in a culture over time. Bacteria reproduce through a process known as binary fission, where one cell divides into two. The growth of bacteria can be influenced by several factors, including nutrient availability, temperature, and the presence of inhibitory substances. When studying bacterial growth, scientists often look at the growth curve, which typically includes lag, exponential, stationary, and death phases.
The exponential phase is where we see rapid growth, and this behavior provides a great opportunity for mathematical modeling using differential equations.
proportional growth rate
The term 'proportional growth rate' implies that the rate at which something grows is directly proportional to its current size. In the context of bacterial growth, if we let \( P(t) \) represent the number of bacteria at time \( t \), then the growth rate can be represented as \[ \frac{dP}{dt} = kP(t) \]
Here, \( k \) is the constant of proportionality. This equation tells us that the rate of change of the bacterial population over time depends on the current population size. If \( P(t) \) is large, the population grows quickly; if it's small, the growth rate is slower. This kind of exponential growth is typical in biological systems during the early,
differential equation modeling
Differential equation modeling is a powerful tool for understanding complex systems, including biological processes. In this context, a differential equation can describe how the number of bacteria in a culture changes over time.
For our exercise, we use the following differential equation to model the system: \[ \frac{dP}{dt} = 0.45P(t) + (e^{0.03t} + 2) \]In this equation, \[ \frac{dP}{dt} \] represents the rate of change of the bacterial population, \[ 0.45P(t) \] accounts for the natural exponential growth, and \[ e^{0.03t} + 2 \] represents the external addition of bacteria to the culture. By solving this differential equation, we can predict the bacterial population at any point in time and understand how different factors influence growth.
constant of proportionality
The constant of proportionality, often denoted as \( k \), is a crucial parameter in our differential equation model. It determines the rate at which the bacteria grow relative to their current population size. In our exercise, the constant is given as \( k = 0.45 \).
This means that at any time, the rate of change in the bacterial population is 45% of the current population size. The constant of proportionality helps in understanding the intrinsic growth rate of the bacterial culture without external influences. In combination with the additional bacteria being added, it provides a comprehensive picture of the bacterial growth dynamics in the culture.

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Most popular questions from this chapter

One or more initial conditions are given for each differential equation in the following exercises. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solutions. Include a \(y z\) -graph if one is not already provided. Always indicate the constant solutions on the \(t y\) -graph whether they are mentioned or not. $$ y^{\prime}=\sin y, y(0)=-\pi / 6, y(0)=\pi / 6, y(0)=7 \pi / 4 $$

Suppose that \(f(t)\) satisfies the initial-value problem \(y^{\prime}=y^{2}+t y-7, y(0)=2\). Is the graph of \(f(t)\) increasing or decreasing at \(t=0\) ?

\(y^{\prime}=k y(M-y)\), where \(k>0, M>10\), and \(y(0)=1\)

A single dose of iodine is injected intravenously into a patient. The iodine mixes thoroughly in the blood before any is lost as a result of metabolic processes (ignore the time required for this mixing process). Iodine will leave the blood and enter the thyroid gland at a rate proportional to the amount of iodine in the blood. Also, iodine will leave the blood and pass into the urine at a (different) rate proportional to the amount of iodine in the blood. Suppose that the iodine enters the thyroid at the rate of \(4 \%\) per hour, and the iodine enters the urine at the rate of \(10 \%\) per hour. Let \(f(t)\) denote the amount of iodine in the blood at time \(t\). Write a differential equation satisfied by \(f(t)\).

When a certain liquid substance \(A\) is heated in a flask, it decomposes into a substance \(B\) at such a rate (measured in units of \(A\) per hour) that at any time \(t\) is proportional to the square of the amount of substance \(A\) present. Let \(y=f(t)\) be the amount of substance \(A\) present at time \(t .\) Construct and solve a differential equation that is satisfied by \(f(t)\).

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