Problem 62
Question
Bacterial Growth Suppose that a bacterial colony grows in such a way that at time \(t\) the population size is $$ N(t)=N(0) 2^{t} $$ where \(N(0)\) is the population size at time \(0 .\) Find the per capita growth rate.
Step-by-Step Solution
Verified Answer
The per capita growth rate is \( \ln(2) \).
1Step 1: Understand the Equation
The given equation for population size is \(N(t) = N(0) \times 2^t\). This describes an exponential growth where the base of the exponent is 2, indicating that the population doubles every time unit.
2Step 2: Recall the Formula for Per Capita Growth Rate
The per capita growth rate, often denoted as \(r\), is defined as the rate at which the population grows per individual. For exponential growth, this can be expressed as \( r = \frac{1}{N(t)} \frac{dN}{dt}\).
3Step 3: Differentiate the Population Function
To find \( \frac{dN}{dt} \), differentiate the function \( N(t) = N(0) \times 2^t \) with respect to \( t \). Using the chain rule, \( \frac{dN}{dt} = N(0) \times 2^t \times \ln(2)\).
4Step 4: Substitute into the Per Capita Growth Rate Formula
Now, substitute \( \frac{dN}{dt} = N(0) \times 2^t \times \ln(2)\) and \(N(t) = N(0) \times 2^t\) into the per capita growth rate formula: \( r = \frac{ N(0) \times 2^t \times \ln(2)}{N(0) \times 2^t} = \ln(2) \).
5Step 5: Interpret the Result
The per capita growth rate \( r \) is a constant value of \( \ln(2) \). This indicates that the growth rate per individual remains constant over time for this type of exponential growth.
Key Concepts
Exponential GrowthPer Capita Growth RateDifferentiation
Exponential Growth
In biological systems, exponential growth is a process that shows how bacteria or other organisms reproduce rapidly over time. In our example, the population function \(N(t) = N(0) \times 2^t\) demonstrates exponential growth. Here, the base of the exponent, 2, represents that the bacterial population doubles every time unit.
Exponential growth is characterized by very rapid increases in population size. Initially, the growth may appear slow, but it escalates quickly. This type of growth is typical in environments where resources are abundant and there are minimal constraints, allowing the population to grow at its maximum potential.
Exponential growth is characterized by very rapid increases in population size. Initially, the growth may appear slow, but it escalates quickly. This type of growth is typical in environments where resources are abundant and there are minimal constraints, allowing the population to grow at its maximum potential.
- A key feature of exponential growth is its J-shaped curve, where the growth rate speeds up continuously as the population increases.
- Such growth cannot be sustained indefinitely, as environmental factors or resource limitations will eventually slow it down.
Per Capita Growth Rate
The per capita growth rate is a crucial metric in population studies. It gives us the rate at which each member of the population is contributing to the overall growth. In mathematical terms, for exponential growth, the per capita growth rate \(r\) is given by the formula: \[ r = \frac{1}{N(t)} \cdot \frac{dN}{dt} \]This formula calculates the change in population size per individual. It shows how much each individual contributes to the total increase over a small time step.
In our example, substituting the derivative and the population function back into the formula simply gives us \(r = \ln(2)\).
In our example, substituting the derivative and the population function back into the formula simply gives us \(r = \ln(2)\).
- Per capita growth rate does not depend on the size of the population—it remains constant even as the number of individuals grows.
- This feature makes it a powerful tool for comparing growth across different populations or time periods, offering insights about population dynamics.
Differentiation
Differentiation is a mathematical process used to find the rate at which a function changes at any point. In the context of bacterial growth, differentiation helps us understand how quickly the population is increasing at a given time.
When we differentiate the function \(N(t) = N(0) \times 2^t\), we use the chain rule, recognizing that it's a product of a constant \(N(0)\) and an exponential function \(2^t\). The result is:\[ \frac{dN}{dt} = N(0) \times 2^t \times \ln(2) \]
This derivative represents the instantaneous rate of growth of the bacterial population.
When we differentiate the function \(N(t) = N(0) \times 2^t\), we use the chain rule, recognizing that it's a product of a constant \(N(0)\) and an exponential function \(2^t\). The result is:\[ \frac{dN}{dt} = N(0) \times 2^t \times \ln(2) \]
This derivative represents the instantaneous rate of growth of the bacterial population.
- Differentiation allows us to calculate the change in population size with respect to time, essential for determining growth rates.
- The constant \(\ln(2)\) is derived from the derivative of the exponential function and is critical in solving for the per capita growth rate.
Other exercises in this chapter
Problem 62
Assume that \(x\) and \(y\) are differentiable functions of \(t\). Find \(\frac{d y}{d t}\) when \(y^{2}=x^{2}-x^{4}, \frac{d x}{d t}=1\) for \(x=\frac{1}{2}\),
View solution Problem 62
Use the quotient rule to show that $$ \frac{d}{d x} \cot x=-\csc ^{2} x $$
View solution Problem 62
Differentiate with respect to the independent variable. $$ f(x)=\frac{\sqrt{5 x\left(1+x^{2}\right)}}{\sqrt{2}} $$
View solution Problem 62
Assume that \(f(x)\) is differentiable with respect to \(x\). Show that $$\frac{d}{d x} \ln \left[\frac{f(x)}{x}\right]=\frac{f^{\prime}(x)}{f(x)}-\frac{1}{x}$$
View solution