Problem 28
Question
Bacteria Growth The number \(N\) of bacteria in a culture is given by the model \(N=250 e^{k t}\), where \(t\) is the time (in hours), with \(t=0\) corresponding to the time when \(N=250\). When \(t=10\), there are 320 bacteria. How long does it take the bacteria population to double in size? To triple in size?
Step-by-Step Solution
Verified Answer
To find the time it takes for the bacteria population to double and triple, calculate the growth rate constant \(k\) using the given data, then solve for \(t\) when \(N = 500\) (double) and \(N = 750\) (triple) respectively in the formula \(N=250 e^{k t}\).
1Step 1: Solve for the growth rate \(k\)
First, plug in \(N = 320\) and \(t = 10\) into the given formula \(N=250 e^{k t}\) and solve for \(k\). Doing so, we get the equation 320 = 250 \(e^{10k}\). Solving this equation for \(k\), we get \(k = \ln(320/250) / 10\).
2Step 2: Calculate the time it takes for the population to double
With the found value for \(k\), we can now solve for the time it takes for the bacteria to double in size. We set \(N = 500\) in our model formula, and solve for \(t\) in the equation 500 = 250 \(e^{k t}\). This gives \(t = \ln(500/250) / k\).
3Step 3: Calculate the time it takes for the population to triple
We follow the same procedure as in step 2, but now we set \(N = 750\). Solving the equation 750 = 250 \(e^{k t}\) for \(t\), we get \(t = \ln(750/250) / k\).
Key Concepts
Bacteria GrowthDoubling TimeTripling TimeGrowth Rate Calculation
Bacteria Growth
Bacteria growth often follows an exponential pattern, which means the number of bacteria increases at a rate proportional to its current size. This concept is crucial for understanding how quickly bacteria populations can explode under optimal conditions.
For example, if a culture starts with 250 bacteria and grows according to the model \( N = 250 e^{kt} \), where \( k \) is the growth rate and \( t \) is time in hours, we see that this model predicts substantial growth as time passes.
Using exponential models allows scientists to forecast bacteria populations in various scenarios, from medical research to food safety and beyond. It's important to note that conditions such as nutrient availability, temperature, and environmental factors can affect the real-life application of these models.
For example, if a culture starts with 250 bacteria and grows according to the model \( N = 250 e^{kt} \), where \( k \) is the growth rate and \( t \) is time in hours, we see that this model predicts substantial growth as time passes.
Using exponential models allows scientists to forecast bacteria populations in various scenarios, from medical research to food safety and beyond. It's important to note that conditions such as nutrient availability, temperature, and environmental factors can affect the real-life application of these models.
Doubling Time
The doubling time is a crucial metric in understanding exponential growth. It represents the time required for a population to double in size.
To find the doubling time, we use the formula \( N = 500 = 250 e^{kt} \) which simplifies into \( t = \frac{\ln(2)}{k} \).
Hence, once the growth rate \( k \) is known, calculating the doubling time becomes straightforward.
Knowing the doubling time allows us to predict future population sizes, manage resources efficiently, and prepare for potential impacts on ecosystems or health situations. It is an essential measure for microbiologists, ecologists, and anyone involved in growth studies.
To find the doubling time, we use the formula \( N = 500 = 250 e^{kt} \) which simplifies into \( t = \frac{\ln(2)}{k} \).
Hence, once the growth rate \( k \) is known, calculating the doubling time becomes straightforward.
Knowing the doubling time allows us to predict future population sizes, manage resources efficiently, and prepare for potential impacts on ecosystems or health situations. It is an essential measure for microbiologists, ecologists, and anyone involved in growth studies.
Tripling Time
Tripling time indicates the period required for a population to triple. Much like doubling time, it gives insight into the dynamics of exponential growth.
Using the bacteria model \( N = 250 e^{kt} \), to find tripling time, we substitute \( N = 750 \) into the formula, leading to the equation \( 750 = 250 e^{kt} \). This simplifies to \( t = \frac{\ln(3)}{k} \).
Tripling time is essential for understanding longer-term forecasts of population size in micro and macro ecological contexts. It proves beneficial in evaluating scenarios over time, such as understanding bacteria's impact on an environment or planning accordingly to ensure sustainable resources.
Using the bacteria model \( N = 250 e^{kt} \), to find tripling time, we substitute \( N = 750 \) into the formula, leading to the equation \( 750 = 250 e^{kt} \). This simplifies to \( t = \frac{\ln(3)}{k} \).
Tripling time is essential for understanding longer-term forecasts of population size in micro and macro ecological contexts. It proves beneficial in evaluating scenarios over time, such as understanding bacteria's impact on an environment or planning accordingly to ensure sustainable resources.
Growth Rate Calculation
To calculate the growth rate \( k \), we need initial data points of the population size at known times.
Given the formula \( N = 250 e^{kt} \), inserting values \( N = 320 \) and \( t = 10 \) helps determine \( k \). Solving the equation \( 320 = 250 e^{10k} \), we derive \( k = \frac{\ln(\frac{320}{250})}{10} \).
Accurate growth rate calculations are vital, as they affect predictions and analyses based on exponential models. Once established, \( k \) provides insights into how fast a population increases under the given conditions, influencing decisions across many scientific and practical applications.
Given the formula \( N = 250 e^{kt} \), inserting values \( N = 320 \) and \( t = 10 \) helps determine \( k \). Solving the equation \( 320 = 250 e^{10k} \), we derive \( k = \frac{\ln(\frac{320}{250})}{10} \).
Accurate growth rate calculations are vital, as they affect predictions and analyses based on exponential models. Once established, \( k \) provides insights into how fast a population increases under the given conditions, influencing decisions across many scientific and practical applications.
Other exercises in this chapter
Problem 27
Evaluate the expression without using a calculator.\(\log _{3} 9\)
View solution Problem 27
Sketch the graph of the function.\(y=2^{-x^{2}}\)
View solution Problem 28
Solve the exponential equation algebraically. Approximate the result to three decimal places.\(6^{x}+10=47\)
View solution Problem 28
Evaluate the logarithm. Round your result to three decimal places.\(\log _{19} 42\)
View solution