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Chapter 3: Problem 54

The population of a culture of cells after \(t\) days is approximated by the function \(P(t)=\frac{1600}{1+7 e^{-0.02 t}},\) for \(t \geq 0\). a. Graph the population function. b. What is the average growth rate during the first 10 days? c. Looking at the graph, when does the growth rate appear to be a maximum? d. Differentiate the population function to determine the growth rate function \(P^{\prime}(t)\). e. Graph the growth rate. When is it a maximum and what is the population at the time that the growth rate is a maximum?

Short Answer

Expert verified
Answer: The maximum growth rate is approximately 37.34 cells per day, and the corresponding population at that time is approximately 341.87 cells.

Step by step solution

01

Part a: Graphing the population function

To graph the population function \(P(t) = \frac{1600}{1+7e^{-0.02t}},\) we need to create a table of values for different \(t\) values (typically between 0 and some convenient upper bound) and plot the points. Please use a graphing calculator or graphing software to plot the function.
02

Part b: Finding the average growth rate of first 10 days

To find the average growth rate during the first 10 days, we need to calculate the difference in population between day 10 and day 0, then divide that difference by 10. Specifically: Average growth rate = \(\frac{P(10)-P(0)}{10}\) Using the provided function, we first find the population on day 0 and day 10: \(P(0) = \frac{1600}{1 + 7e^{0}} = \frac{1600}{8} = 200\) \(P(10) = \frac{1600}{1 + 7e^{-0.02(10)}} \approx \frac{1600}{1+7e^{-0.2}} \approx 440.95\) Now we can calculate the average growth rate: Average growth rate \(\approx \frac{440.95 - 200}{10} \approx 24.1\) The average growth rate during the first 10 days is approximately 24.1 cells per day.
03

Part c: Finding maximum growth rate using the graph

To estimate when the growth rate is at a maximum, we need to visually inspect the graph of the population function and identify the point where the slope is the steepest. The graph will show an initial rapid growth and then start to slow down as the population approaches its carrying capacity. By inspecting the graph, we can estimate that the maximum growth rate occurs around day 5.
04

Part d: Differentiating the population function

We'll differentiate the population function to find the growth rate function \(P'(t)\): \(P(t) = \frac{1600}{1+7e^{-0.02t}}\) First, apply the quotient rule for differentiating \(\frac{u}{v} = \frac{vu' - uv'}{v^2}\): Let \(u = 1600\) and \(v = 1+7e^{-0.02t}\). \(u' = 0\) Now, differentiate \(v\): \(v' = \frac{d}{dt}(1+7e^{-0.02t})=-0.02\times7e^{-0.02t}\) Now, apply the quotient rule: \(P'(t) = \frac{(1+7e^{-0.02t})(0)-1600(-0.02\times7e^{-0.02t})}{(1+7e^{-0.02t})^2}= \frac{1600\times0.14\times e^{-0.02t}}{(1+7e^{-0.02t})^2}\)
05

Part e: Graphing growth rate function and finding its maximum

To graph the growth rate function \(P'(t)=\frac{1600\times0.14\times e^{-0.02t}}{(1+7e^{-0.02t})^2}\), we can create a table of values for different \(t\) values (typically between 0 and some convenient upper bound) and plot the points. To find the maximum growth rate and the corresponding population, we could either analyze the graph or find the critical points of the growth rate function by setting \(P'(t)=0\) and finding the global maximum. In our case, we'll use the graph. As we have earlier estimated, the growth rate graph shows that the maximum growth rate occurs around day 5. Now, we should find the exact maximum value of the growth rate and the corresponding population value at that time: \(P'(5)=\frac{1600\times0.14\times e^{-0.02(5)}}{(1+7e^{-0.02(5)})^2} \approx 37.34\) The maximum growth rate is approximately 37.34 cells per day. To find the corresponding population: \(P(5) = \frac{1600}{1 + 7e^{-0.02(5)}} \approx \frac{1600}{1+7e^{-0.1}} \approx 341.87\) The population at the time when the growth rate is at a maximum is approximately 341.87 cells.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Growth
Exponential growth is a fundamental concept in differential calculus that describes how quantities grow rapidly over time. It is often modeled by functions that feature exponential expressions, such as those involving Euler's number, \(e\). These functions can depict phenomena where growth compounds over time, becoming faster and more pronounced as time progresses.
In the given problem, the exponential component is evident in the population function \(P(t) = \frac{1600}{1+7e^{-0.02t}}\). Here, the term \(e^{-0.02t}\) dominates the growth dynamics. As \(t\) increases, this term approaches zero, leading to a leveling off or saturation in growth, and eventually stabilizing the population value.
This type of growth application is prevalent in biological systems where resources are limited, such as population growth in an environment with finite resources.
Quotient Rule
In differential calculus, the quotient rule is essential for differentiating functions expressed as a ratio of two other functions. Suppose you have a function defined as \(\frac{u}{v}\), where \(u\) and \(v\) are functions of \(t\). The quotient rule states that the derivative \(P'(t)\) of \(\frac{u}{v}\) is given by:
\(P'(t) = \frac{v \cdot u' - u \cdot v'}{v^2} \)
This formula helps us determine the rate of change of a function that has a fractional form.
In the context of our exercise, to find the growth rate function \(P'(t)\), we differentiate the given population function using the quotient rule, since it is presented as a fraction \(\frac{1600}{1+7e^{-0.02t}}\). Here, \(u = 1600\) and remains constant, and \(v = 1+7e^{-0.02t}\), which changes with \(t\). Carefully applying the quotient rule allows us to determine how fast the population is increasing at any point in time.
Population Dynamics
Population dynamics encompasses the study of how populations change over time under various conditions. It is a core concept in biology and ecology, often analyzed using models that incorporate growth and decay factors.
Mathematical models, such as the one used in the exercise \(P(t) = \frac{1600}{1+7e^{-0.02t}}\), illustrate typical population dynamics. This function accounts for initial rapid growth followed by a slowing as the environment's carrying capacity is approached.
Understanding these dynamics helps in predicting future population sizes, analyzing ecological impacts, and planning resource management. It is vital to grasp such models, especially in areas like conservation biology, where maintaining equilibrium within ecosystems is critical.
Growth Rate Function
The growth rate function is a critical derivative that assists in understanding how the size of a population or quantity changes over time. It is derived by differentiating the original function that describes population size or any other quantity of interest.
In our exercise, the growth rate function \(P'(t)\) is derived from the population function. This derivative tells us the rate at which the population is changing at any given time \(t\).
If you have the population function \(P(t) = \frac{1600}{1+7e^{-0.02t}}\), by applying the quotient rule, you find \(P'(t) = \frac{1600 \times 0.14 \times e^{-0.02t}}{(1+7e^{-0.02t})^2}\).
The value of the growth rate function is particularly crucial when identifying peaks, or maximum growth rates, in a population. In biological terms, understanding the growth rate helps in predicting when a species or culture is growing the fastest, providing insight into stages of development and potential for expansion.

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Most popular questions from this chapter

Suppose \(f\) is differentiable on an interval containing \(a\) and \(b\), and let \(P(a, f(a))\) and \(Q(b, f(b))\) be distinct points on the graph of \(f\). Let \(c\) be the \(x\) -coordinate of the point at which the lines tangent to the curve at \(P\) and \(Q\) intersect, assuming that the tangent lines are not parallel (see figure). a. If \(f(x)=x^{2},\) show that \(c=(a+b) / 2,\) the arithmetic mean of \(a\) and \(b\), for real numbers \(a\) and \(b\) b. If \(f(x)=\sqrt{x}\), show that \(c=\sqrt{a b}\), the geometric mean of \(a\) and \(b\), for \(a > 0\) and \(b > 0\) c. If \(f(x)=1 / x,\) show that \(c=2 a b /(a+b),\) the harmonic mean of \(a\) and \(b,\) for \(a > 0\) and \(b > 0\) d. Find an expression for \(c\) in terms of \(a\) and \(b\) for any (differentiable) function \(f\) whenever \(c\) exists.

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