Problem 25
Question
Bacterium population When a bactericide was added to a nutrient broth in which bacteria were growing, the bacterium population continued to grow for a while, but then stopped growing and began to decline. The size of the population at time \(t\) (hours) was \(b=10^{6}+10^{4} t-10^{3} t^{2} .\) Find the growth rates at a. \(t=0\) hours. b. \(t=5\) hours. c. \(t=10\) hours.
Step-by-Step Solution
Verified Answer
Growth rates: (a) 10^4; (b) 0; (c) -10^4 bacteria per hour.
1Step 1: Differentiate the Population Function
To find the growth rate of the bacterium population at any time \(t\), we need to differentiate the function \(b(t)\). The function given is \(b(t) = 10^6 + 10^4 t - 10^3 t^2\). The derivative, \(b'(t)\), represents the rate of change of the bacterium population with respect to time. Differentiating the function, we get:\[b'(t) = \frac{d}{dt}[10^6 + 10^4 t - 10^3 t^2] = 10^4 - 2 \times 10^3 t.\]
2Step 2: Calculate Growth Rate at t=0
We calculate the rate of growth at \(t = 0\) by substituting \(t = 0\) into the derivative \(b'(t)\):\[b'(0) = 10^4 - 2 \times 10^3 \times 0 = 10^4.\]Thus, the growth rate at \(t = 0\) hours is \(10^4\) bacteria per hour.
3Step 3: Calculate Growth Rate at t=5
To find the growth rate at \(t = 5\), substitute \(t = 5\) into the derivative \(b'(t)\):\[b'(5) = 10^4 - 2 \times 10^3 \times 5 = 10^4 - 10^4 = 0.\]Thus, the growth rate at \(t = 5\) hours is 0 bacteria per hour.
4Step 4: Calculate Growth Rate at t=10
Now, calculate the growth rate at \(t = 10\) by substituting \(t = 10\) into \(b'(t)\):\[b'(10) = 10^4 - 2 \times 10^3 \times 10 = 10^4 - 2 \times 10^4 = -10^4.\]Thus, the growth rate at \(t = 10\) hours is \(-10^{4}\) bacteria per hour.
Key Concepts
DifferentiationRate of ChangePopulation GrowthExponential Functions
Differentiation
Differentiation is a key concept in calculus that helps us understand how a function changes at any given point. It gives us the derivative of a function, which represents the rate at which one quantity changes with respect to another. In simple terms, differentiation tells us how fast something is changing.
For the bacterium population given by the function \(b(t) = 10^6 + 10^4 t - 10^3 t^2\), differentiating this function gives us its derivative: \(b'(t) = 10^4 - 2 \times 10^3 t\).
Here, \(b'(t)\) tells us the rate of change of the bacterium population over time. By plugging in different values of \(t\) into \(b'(t)\), we can find out how quickly the population is increasing or decreasing at those specific times.
For the bacterium population given by the function \(b(t) = 10^6 + 10^4 t - 10^3 t^2\), differentiating this function gives us its derivative: \(b'(t) = 10^4 - 2 \times 10^3 t\).
Here, \(b'(t)\) tells us the rate of change of the bacterium population over time. By plugging in different values of \(t\) into \(b'(t)\), we can find out how quickly the population is increasing or decreasing at those specific times.
Rate of Change
The rate of change is an important concept that shows us how a certain quantity is changing over time. In the context of the bacterium population, it is crucial for understanding how quickly the population is growing or declining.
When we compute \(b'(t)\), we're really finding the rate at which the bacteria count changes as time progresses. For example, at \(t=0\), the rate of change \(b'(0)\) is \(10^4\), meaning the population is increasing by 10,000 bacteria per hour initially.
As time progresses, this rate can change. At \(t = 5\), the rate is 0, indicating the population has stopped growing temporarily. By \(t = 10\), the rate becomes \(-10^4\), implying the population is decreasing by 10,000 bacteria per hour.
When we compute \(b'(t)\), we're really finding the rate at which the bacteria count changes as time progresses. For example, at \(t=0\), the rate of change \(b'(0)\) is \(10^4\), meaning the population is increasing by 10,000 bacteria per hour initially.
As time progresses, this rate can change. At \(t = 5\), the rate is 0, indicating the population has stopped growing temporarily. By \(t = 10\), the rate becomes \(-10^4\), implying the population is decreasing by 10,000 bacteria per hour.
Population Growth
Population growth describes how a population size changes over time. It can be influenced by birth rates, death rates, and external factors like a bactericide. In this case, the growth of bacteria is modeled by a mathematical function.
Initially, the bacterial population grows due to a positive growth rate. As seen in the example, at the start \(t = 0\), the bacteria are rapidly multiplying. This is typical in population growth scenarios where resources are abundant.
However, as time goes on, the growth rate decreases, showing that the bacteria begin to consume the resources or are affected by inhibitors like the bactericide. Eventually, the population size begins to decline, marking a shift from growth to decay. Understanding these shifts helps in predicting long-term population trends.
Initially, the bacterial population grows due to a positive growth rate. As seen in the example, at the start \(t = 0\), the bacteria are rapidly multiplying. This is typical in population growth scenarios where resources are abundant.
However, as time goes on, the growth rate decreases, showing that the bacteria begin to consume the resources or are affected by inhibitors like the bactericide. Eventually, the population size begins to decline, marking a shift from growth to decay. Understanding these shifts helps in predicting long-term population trends.
Exponential Functions
Exponential functions are frequently used to model populations, especially when dealing with bacterial growth. These functions are characterized by their rapid rate of increase. While the given bacterium population function is quadratic, the initial part of the growth behaves similarly to an exponential function when the growth rate is high.
In exponential growth, populations can grow unchecked due to constant reproduction rates, resulting in a characteristic J-shaped curve. This doesn't apply indefinitely in real-life scenarios due to resource limitations or added inhibitors like bactericides, which convert exponential growth into a more complex model like the one described here.
The initial rapid growth phase in our bacteria example mimics this exponential behavior until external factors curtail it, leading to a decline as seen by the negative growth rate at \(t=10\). Understanding exponential functions is key in explaining the initial phases of population growth before limiting factors take effect.
In exponential growth, populations can grow unchecked due to constant reproduction rates, resulting in a characteristic J-shaped curve. This doesn't apply indefinitely in real-life scenarios due to resource limitations or added inhibitors like bactericides, which convert exponential growth into a more complex model like the one described here.
The initial rapid growth phase in our bacteria example mimics this exponential behavior until external factors curtail it, leading to a decline as seen by the negative growth rate at \(t=10\). Understanding exponential functions is key in explaining the initial phases of population growth before limiting factors take effect.
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