Problem 54
Question
Suppose that the number of bacteria in a culture at time \(t\) is given by \(N=5000\left(25+t e^{-t / 20}\right) .\) (a) Find the largest and smallest number of bacteria in the culture during the time interval \(0 \leq t \leq 100\). (b) At what time during the time interval in part (a) is the number of bacteria decreasing most rapidly?
Step-by-Step Solution
Verified Answer
(a) Largest: \(136840\) at \(t=20\); Smallest: \(125000\) at \(t=0\). (b) Decreasing most rapidly around \(t=20\).
1Step 1: Understand the Function
The function given is \[ N(t) = 5000 \left(25 + t e^{-t/20}\right) \] where \(N\) represents the number of bacteria, and \(t\) is the time.
2Step 2: Find the Derivative
To find when the number of bacteria is increasing or decreasing, we first need the derivative of \(N(t)\). Use the product rule for differentiation: \[ N'(t) = 5000\left( 1 \cdot e^{-t/20} + t \cdot \left(\frac{-1}{20}e^{-t/20}\right)\right) = 5000 \left( e^{-t/20} - \frac{t}{20} e^{-t/20} \right).\]
3Step 3: Simplify the Derivative Expression
Factor out \(e^{-t/20}\) from the derivative:\[ N'(t) = 5000 e^{-t/20} \left( 1 - \frac{t}{20} \right). \]
4Step 4: Find Critical Points
Set the derivative equal to zero to find critical points:\[ 0 = 5000 e^{-t/20} \left(1 - \frac{t}{20}\right). \]e^{-t/20} is always positive, so set \(1 - \frac{t}{20} = 0\). Solving gives \( t = 20 \).
5Step 5: Analyze Critical Points and Endpoints
Evaluate \(N(t)\) at the critical point and the endpoints of the interval:- At \(t = 0\): \[ N(0) = 5000(25 + 0) = 125000.\]- At \(t = 20\): \[ N(20) = 5000(25 + 20e^{-1}) \approx 5000 (25 + 20 \times 0.368) \approx 136840.\]- At \(t = 100\): \[ N(100) = 5000(25 + 100e^{-5}) \approx 5000(25 + 100 \times 0.0067) \approx 126565.\]Thus, the maximum is at \(t = 20\) and the minimum is at \(t = 0\).
6Step 6: Analyze the Rate of Decrease
To find when the bacteria decrease most rapidly, analyze when \(N'(t)\) is minimum. The most rapid decrease occurs not with sign changes in \(N'(t)\) but in large negative values. Hence consider \(N''(t)\):\[ N''(t) = 5000e^{-t/20}\left( \frac{-1}{20} + \frac{t}{400} \right). \]Set \[ N''(t) = 0 \] which simplifies to \[ \frac{-1}{20} + \frac{t}{400} = 0 \Rightarrow t = 20. \]However, this is a zero point, for decreasing rate \(N'(t)\) remains decreasing as \(t\) goes to 100 as negative eigenvalues take place there.
Key Concepts
Bacteria Culture ModelCritical PointsDerivative AnalysisRate of Change
Bacteria Culture Model
The Bacteria Culture Model is a mathematical representation that describes how a population of bacteria grows over time. In this exercise, the function given is \[ N(t) = 5000 \left(25 + t e^{-t/20}\right), \]where \(N(t)\) represents the number of bacteria at time \(t\), and the task is to analyze the population dynamics within a specific time frame. Such models are essential in microbiology to predict the growth patterns of bacterial cultures and to understand under what conditions the bacteria thrive or decline.
Understanding the biological context behind this model is crucial too. Bacteria typically follow a growth curve that includes phases of lag, exponential growth, stationary, and decline. The mathematical model captures how the population may increase initially and then decrease as resources become limited.
This particular model combines a constant term and an exponential decay factor, indicating that initial growth may peak before declining due to the negative exponential factor. This allows the model to reflect situations where bacteria expand rapidly, but resource limitations or increased waste accumulation eventually slow the growth.
Understanding the biological context behind this model is crucial too. Bacteria typically follow a growth curve that includes phases of lag, exponential growth, stationary, and decline. The mathematical model captures how the population may increase initially and then decrease as resources become limited.
This particular model combines a constant term and an exponential decay factor, indicating that initial growth may peak before declining due to the negative exponential factor. This allows the model to reflect situations where bacteria expand rapidly, but resource limitations or increased waste accumulation eventually slow the growth.
Critical Points
Critical points in a function are values of \(t\) where the derivative of the function is zero or undefined. These points are significant because they can determine where a function reaches its maximum or minimum, marking crucial shifts in the behavior of the bacteria population.
In our exercise, the derivative of the number of bacteria function is found as \[ N'(t) = 5000 e^{-t/20} \left( 1 - \frac{t}{20} \right). \] The critical points occur where this derivative equals zero. Given that \(e^{-t/20}\) is always positive, we set\[ 1 - \frac{t}{20} = 0, \]which simplifies to finding \(t = 20\).
These critical points, determined through setting the derivative to zero, help pinpoint where maximum or minimum population numbers occur. Finding and understanding these points is key in optimization problems since they tell us where the bacteria stop increasing and start decreasing in quantity.
In our exercise, the derivative of the number of bacteria function is found as \[ N'(t) = 5000 e^{-t/20} \left( 1 - \frac{t}{20} \right). \] The critical points occur where this derivative equals zero. Given that \(e^{-t/20}\) is always positive, we set\[ 1 - \frac{t}{20} = 0, \]which simplifies to finding \(t = 20\).
These critical points, determined through setting the derivative to zero, help pinpoint where maximum or minimum population numbers occur. Finding and understanding these points is key in optimization problems since they tell us where the bacteria stop increasing and start decreasing in quantity.
Derivative Analysis
Derivative Analysis involves studying the derivatives of functions to understand how these functions change, which is pivotal in optimization tasks like this one.
By calculating the first derivative, \( N'(t) \), we gain insights into where the function increases or decreases. The derivative \[ N'(t) = 5000 e^{-t/20} \left( 1 - \frac{t}{20} \right) \]indicates that changes in the population growth rate occur due to the interplay between population-gain from birth and population-loss due to environmental limitations reflected by the decay factor.
To find out when the bacteria population is changing most rapidly, we also look at the second derivative, \( N''(t) \), which gives information about the concavity of the function. This can highlight points of inflection where the rate of growth changes most steeply, which is crucial for understanding different growth phases in a bacteria culture model.
By calculating the first derivative, \( N'(t) \), we gain insights into where the function increases or decreases. The derivative \[ N'(t) = 5000 e^{-t/20} \left( 1 - \frac{t}{20} \right) \]indicates that changes in the population growth rate occur due to the interplay between population-gain from birth and population-loss due to environmental limitations reflected by the decay factor.
To find out when the bacteria population is changing most rapidly, we also look at the second derivative, \( N''(t) \), which gives information about the concavity of the function. This can highlight points of inflection where the rate of growth changes most steeply, which is crucial for understanding different growth phases in a bacteria culture model.
Rate of Change
The Rate of Change, as described by the derivative \( N'(t) \), tells us how quickly the population of bacteria is growing or declining at any point in time. In the context of this exercise, identifying when the bacteria population decreases the fastest can be essential for scenarios such as resource management or predicting population crashes.
A rapid decrease typically means the bacteria are facing unfavorable conditions. By setting the second derivative \( N''(t) \) to zero, we find points where the rate of change shifts in its dynamics, showing us times of maximum rapid decrease. For this exercise, analyzing:\[ N''(t) = 5000e^{-t/20}\left( \frac{-1}{20} + \frac{t}{400} \right), \]shows us how the derivative continues to decay deeper into the negative values as \(t\) increases.
This concept helps manage and predict the behavior of bacteria under varying conditions over time.
A rapid decrease typically means the bacteria are facing unfavorable conditions. By setting the second derivative \( N''(t) \) to zero, we find points where the rate of change shifts in its dynamics, showing us times of maximum rapid decrease. For this exercise, analyzing:\[ N''(t) = 5000e^{-t/20}\left( \frac{-1}{20} + \frac{t}{400} \right), \]shows us how the derivative continues to decay deeper into the negative values as \(t\) increases.
This concept helps manage and predict the behavior of bacteria under varying conditions over time.
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