Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

A bacterial population grows at a rate proportional to its size. Initially, it is 10,000 , and after 10 days it is 20,000 . What is the population after 25 days?

Step-by-Step Solution

Answer

The bacterial population after 25 days is approximately 56,560.

Step 1: Understand the Growth Model

The bacterial population grows at a rate proportional to its size, which can be modeled with the equation \( P(t) = P_0 e^{kt} \), where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, and \( k \) is a constant growth rate.

Step 2: Establish Known Values

We know that \( P_0 = 10000 \) and after 10 days, \( P(10) = 20000 \). We need to find \( P(25) \).

Step 3: Calculate the Growth Rate Constant

Set up the equation \( 20000 = 10000 e^{10k} \). This simplifies to \( 2 = e^{10k} \). Take the natural logarithm of both sides to solve for \( k \):\[ \ln(2) = 10k \]\[ k = \frac{\ln(2)}{10} \]

Step 4: Solve for the Population at 25 Days

Now substitute \( k \) into the original equation to find \( P(25) \):\[ P(25) = 10000 e^{25 \left(\frac{\ln(2)}{10}\right)} \]\[ P(25) = 10000 e^{2.5 \ln(2)} \]Simplify using the property \( e^{a\ln(b)} = b^a \):\[ P(25) = 10000 \cdot 2^{2.5} \]\[ P(25) = 10000 \cdot 5.656 \approx 56560 \]

Step 5: Final Answer

After 25 days, the bacterial population is approximately 56,560.

Key Concepts

Population Growth

In the context of the exponential growth model, population growth refers to how a population increases over time in a way that is proportionate to its current size. This reflects real-world situations like bacterial growth, where the growth rate is continually increasing as the population itself grows. Such growth can be modeled with the equation \[P(t) = P_0 e^{kt}\] where:

• \(P(t)\) is the population at time \(t\)
• \(P_0\) is the initial population size
• \(k\) represents the growth rate constant

The key aspect of population growth in this context is that it's exponential, meaning that the population grows by a fixed percentage every period, rather than a fixed amount. This causes the population to rise sharply as time goes on, assuming that the growth rate remains constant. In practical terms, this model can predict the size of populations in the future, helping guide planning in areas ranging from ecology to public health.

Natural Logarithm

The natural logarithm is a concept used to solve equations where the variable is in the exponent, as seen in exponential growth models. It is denoted as \(\ln(x)\) and is based on the constant \(e\), approximately 2.718. The natural logarithm helps to "undo" the exponential function, allowing us to solve for variables like the growth rate constant \(k\). For example, in finding the growth rate constant in an exponential equation like: \[e^{10k} = 2\] Taking the natural logarithm of both sides simplifies to: \[\ln(e^{10k}) = \ln(2)\] Using the property that \(\ln(e^x) = x\), you can simplify this to: \[10k = \ln(2)\] This step is crucial in solving exponential equations because it allows for the algebraic manipulation necessary to isolate \(k\) or any other variable exponent. Understanding the properties of the natural logarithm is essential when working with natural exponential functions.

Growth Rate Constant

The growth rate constant \(k\) is an essential part of the exponential growth model, influencing how rapidly a population grows over time. It is a positive number that indicates the proportional rate of growth of the population, defined as how quickly the population doubles in size under continuous growth.To find the growth rate constant \(k\), you can use known values in the growth model equation and the natural logarithm. Consider this equation: \[20000 = 10000 e^{10k}\] Simplifying to: \[2 = e^{10k}\] By taking the natural logarithm of both sides, you have: \[\ln(2) = 10k\] This gives: \[k = \frac{\ln(2)}{10}\] Thus, \(k\) is computed by dividing the natural logarithm of 2 by 10. The value of \(k\) gives insight into how quickly the initial population will grow when time \(t\) is in integer units, like days. A larger \(k\) value means faster population growth. Understanding \(k\) is critical since it directly affects the projections and informs whether resources might become constrained.