Problem 62
Question
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. $$ \frac{1}{N} \frac{d N}{d t} $$
Step-by-Step Solution
Verified Answer
The per capita growth rate is \(\ln(2)\), a constant.
1Step 1: Understand the Given Function
The bacterial population at time \(t\) is given by the function \(N(t) = N_0 2^t\). \(N_0\) is the initial population size at time \(t=0\). This exponential function indicates the population doubles every unit of time.
2Step 2: Differentiate the Population Function
To find the rate of change of population, we need to differentiate \(N(t) = N_0 2^t\) with respect to \(t\). The derivative of an exponential function \(a^t\) is \(a^t \ln(a)\). Thus, \(\frac{dN}{dt} = N_0 \cdot 2^t \cdot \ln(2)\).
3Step 3: Apply the Per Capita Growth Rate Formula
The per capita growth rate is given by \(\frac{1}{N} \frac{dN}{dt}\). Substituting \(N = N_0 2^t\) and \(\frac{dN}{dt} = N_0 2^t \ln(2)\), we get: \[\frac{1}{N} \frac{dN}{dt} = \frac{1}{N_0 2^t} (N_0 2^t \ln(2))\].
4Step 4: Simplify the Equation
After substitution, simplify the expression: \[\frac{1}{N_0 2^t} (N_0 2^t \ln(2)) = \ln(2)\]. This shows that the per capita growth rate is constant and equals \(\ln(2)\).
Key Concepts
Exponential GrowthDifferentiationBacterial Population Dynamics
Exponential Growth
Exponential growth describes a situation where the rate of change of a quantity is proportional to the current amount of that quantity. This results in the quantity growing at an ever-accelerating pace. In natural settings, such as bacterial populations or compound interest, exponential growth models how rapidly something can expand under optimal conditions.
For a function such as \(N(t) = N_0 2^t\), where \(N_0\) represents the initial size of the population, exponential growth indicates that the population doubles for each unit increase in time \(t\). Since the base of the exponent is 2, this implies consistent doubling of the population with each time interval. Such predictable and consistent growth is characteristic of exponential processes, representing perfect conditions where resources are unlimited and growth can proceed unchecked.
For a function such as \(N(t) = N_0 2^t\), where \(N_0\) represents the initial size of the population, exponential growth indicates that the population doubles for each unit increase in time \(t\). Since the base of the exponent is 2, this implies consistent doubling of the population with each time interval. Such predictable and consistent growth is characteristic of exponential processes, representing perfect conditions where resources are unlimited and growth can proceed unchecked.
- Base greater than one: means growth (e.g., population doubling).
- Resources are assumed unlimited: allows continued growth.
- Compound nature: emphasizes the multiplicative growth pattern.
Differentiation
Differentiation is a fundamental concept in calculus, used to determine how a function changes at any point. In the context of a bacterial population growing over time, differentiation allows us to find the rate at which the population is increasing at any given moment.
When differentiating an exponential function like \(N(t) = N_0 2^t\), we use the chain rule for differentiation. If \(a^t\) is an exponential function, its derivative with respect to \(t\) is \(a^t \ln(a)\). Therefore, the rate of change of the population, \(\frac{dN}{dt}\), is given by:\[\frac{dN}{dt} = N_0 \cdot 2^t \cdot \ln(2)\]This expression tells us how fast the bacterial population is growing at any time \(t\). The derivative reflects the interplay of the exponential base and its natural logarithm, providing vital insights on acceleration and instantaneous changes in the system.
When differentiating an exponential function like \(N(t) = N_0 2^t\), we use the chain rule for differentiation. If \(a^t\) is an exponential function, its derivative with respect to \(t\) is \(a^t \ln(a)\). Therefore, the rate of change of the population, \(\frac{dN}{dt}\), is given by:\[\frac{dN}{dt} = N_0 \cdot 2^t \cdot \ln(2)\]This expression tells us how fast the bacterial population is growing at any time \(t\). The derivative reflects the interplay of the exponential base and its natural logarithm, providing vital insights on acceleration and instantaneous changes in the system.
Bacterial Population Dynamics
Bacterial population dynamics explores how populations of bacteria change over time, influenced by factors like growth rates, environmental conditions, and resource availability. Mathematically, it is often modeled using exponential functions due to the rapid replication rates observed in bacteria under ideal conditions.
For the population function \(N(t) = N_0 2^t\), the term \(N_0\) represents the initial population, while \(2^t\) models the doubling behavior of bacteria. The per capita growth rate, expressed as \(\frac{1}{N}\frac{dN}{dt}\), provides insight into how each individual bacterium contributes to the overall population growth. By substituting \[N = N_0 2^t\] and \[\frac{dN}{dt} = N_0 2^t \ln(2)\], we obtain the simplified per capita growth rate:\[\frac{1}{N} \frac{dN}{dt} = \ln(2)\].This result underscores that each bacterium contributes equally to a constant relative growth rate of \(\ln(2)\), independent of the population size. Such insights are crucial for predicting bacterial behavior in diverse environments, affecting areas like medicine and environmental science.
For the population function \(N(t) = N_0 2^t\), the term \(N_0\) represents the initial population, while \(2^t\) models the doubling behavior of bacteria. The per capita growth rate, expressed as \(\frac{1}{N}\frac{dN}{dt}\), provides insight into how each individual bacterium contributes to the overall population growth. By substituting \[N = N_0 2^t\] and \[\frac{dN}{dt} = N_0 2^t \ln(2)\], we obtain the simplified per capita growth rate:\[\frac{1}{N} \frac{dN}{dt} = \ln(2)\].This result underscores that each bacterium contributes equally to a constant relative growth rate of \(\ln(2)\), independent of the population size. Such insights are crucial for predicting bacterial behavior in diverse environments, affecting areas like medicine and environmental science.
Other exercises in this chapter
Problem 61
Find the normal line to $$ f(x)=\frac{a x^{2}}{a+1} $$ at \(x=2 .\) Assume that \(a\) is a positive constant.
View solution Problem 62
Use the quotient rule to show that $$ \frac{d}{d x} \cot x=-\csc ^{2} x $$
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 Problem 62
Differentiate with respect to the independent variable. \(f(x)=\frac{\sqrt{5 x}\left(1+x^{2}\right)}{\sqrt{2}}\)
View solution