Problem 31
Question
In \(1913,\) Carlson \(^{28}\) conducted the classic experiment in which he grew yeast, Saccharomyces cerevisiae, in laboratory cultures and collected data every hour for 18 hours. Table 2.12 shows the yeast population, \(P,\) at representative times \(t\) in hours. (a) Calculate the average rate of change of \(P\) per hour for the time intervals shown between 0 and 18 hours. (b) What can you say about the sign of \(d^{2} P / d t^{2}\) during the period \(0-18\) hours? $$\begin{array}{c|r|r|r|r|r|r|r|r|r|r}\hline t & 0 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 \\ \hline P & 9.6 & 29.0 & 71.1 & 174.6 & 350.7 & 513.3 & 594.8 & 640.8 & 655.9 & 661.8 \\ \hline\end{array}$$
Step-by-Step Solution
Verified Answer
The average rates of change show growth slowing down, so \( \frac{d^2 P}{dt^2} < 0 \).
1Step 1: Understanding the Average Rate of Change
The average rate of change of a function between two points gives an idea of how much the function values increase or decrease over that interval. Mathematically, for a function \( P(t) \), the average rate of change from \( t = a \) to \( t = b \) is given by \( \frac{P(b) - P(a)}{b - a} \). In this case, we will compute it for the intervals such as \( [0,2], [2,4] \), and so on up to \( [16,18] \).
2Step 2: Calculating Average Rate of Change for Each Interval
For each interval: - For \([0,2]: \frac{29.0 - 9.6}{2 - 0} = 9.7\)- For \([2,4]: \frac{71.1 - 29.0}{4 - 2} = 21.05\)- For \([4,6]: \frac{174.6 - 71.1}{6 - 4} = 51.75\)- For \([6,8]: \frac{350.7 - 174.6}{8 - 6} = 88.05\)- For \([8,10]: \frac{513.3 - 350.7}{10 - 8} = 81.3\)- For \([10,12]: \frac{594.8 - 513.3}{12 - 10} = 40.75\)- For \([12,14]: \frac{640.8 - 594.8}{14 - 12} = 23\)- For \([14,16]: \frac{655.9 - 640.8}{16 - 14} = 7.55\)- For \([16,18]: \frac{661.8 - 655.9}{18 - 16} = 2.95\)
3Step 3: Analyzing the Second Derivative Sign During the Period
The second derivative \( \frac{d^2 P}{dt^2} \) indicates how the rate of change itself is changing. If \( \frac{d^2 P}{dt^2} > 0 \), the function is concave up (population growing) and if \( \frac{d^2 P}{dt^2} < 0 \), the function is concave down (population growth is slowing). Observing that the average rate of change starts high and reduces towards the end signifies \( \frac{d^2 P}{dt^2} < 0 \), indicating that the growth rate is slowing down over time.
Key Concepts
Yeast Population GrowthUnderstanding the Second DerivativeConcave Up and Down
Yeast Population Growth
In 1913, an insightful experiment was conducted by Carlson with the yeast, Saccharomyces cerevisiae. This study assessed how the yeast population grows over time when cultivated in a controlled laboratory setting. Carlson measured the yeast growth every hour for 18 hours and recorded these observations. By understanding the yeast population, students can learn about exponential growth patterns often observed in biology.
Yeast, like many microorganisms, can reproduce quickly, leading to a rapid increase in their population. Initially, when conditions are optimal, the population grows exponentially because each yeast cell reproduces at a consistent rate. This requires sufficient nutrients and space, which are initially plentiful in a lab environment. Over time, however, as resources begin to deplete or other limiting factors come into play, the growth rate can slow down.
Yeast, like many microorganisms, can reproduce quickly, leading to a rapid increase in their population. Initially, when conditions are optimal, the population grows exponentially because each yeast cell reproduces at a consistent rate. This requires sufficient nutrients and space, which are initially plentiful in a lab environment. Over time, however, as resources begin to deplete or other limiting factors come into play, the growth rate can slow down.
Understanding the Second Derivative
The second derivative, denoted as \(\frac{d^2 P}{dt^2}\), is a mathematical concept that shows us how the rate of change of a function's derivative is changing. In the context of yeast population, it helps us understand whether the population's growth is speeding up or slowing down.
The first derivative \(\frac{dP}{dt}\) gives us the rate at which the population is growing - how fast new yeast cells are being added per hour. If this rate is increasing, the growth is accelerating, and if it is decreasing, the growth is decelerating. This is where the second derivative comes in.
The first derivative \(\frac{dP}{dt}\) gives us the rate at which the population is growing - how fast new yeast cells are being added per hour. If this rate is increasing, the growth is accelerating, and if it is decreasing, the growth is decelerating. This is where the second derivative comes in.
- If \(\frac{d^2 P}{dt^2} > 0\), the population is accelerating or speeding up, also known as concave up.
- If \(\frac{d^2 P}{dt^2} < 0\), the population is slowing down, known as concave down.
Concave Up and Down
Concavity in mathematics tells us about the shape of the graph of a function at a given point. It helps visually represent the acceleration (or deceleration) of growth.
When a function is concave up, it is shaped like a cup or a smile. This implies that the slope of the tangent line at any point is increasing, and therefore signs of acceleration in growth.
In contrast, when a function is concave down, it looks like a frown. Here, the slope of the tangent line decreases across points, indicating a slowdown in the rate of growth.
When a function is concave up, it is shaped like a cup or a smile. This implies that the slope of the tangent line at any point is increasing, and therefore signs of acceleration in growth.
In contrast, when a function is concave down, it looks like a frown. Here, the slope of the tangent line decreases across points, indicating a slowdown in the rate of growth.
- A concave up yeast growth curve suggests conditions are favorable, and the growth rate is increasing as each new hour adds more population.
- A concave down curve signals a slowdown, suggesting that there may be limiting factors such as reduced nutrients or space.
Other exercises in this chapter
Problem 30
US meat \(^{14}\) production, \(M=f(t),\) in millions of metric tons, is a function of \(t,\) years since 2000 (a) Interpret \(f(10)=92.63\) and \(f^{\prime}(10
View solution Problem 30
Table 2.11 shows the number of Facebook subscribers, \(N\) in millions, worldwide at 3 -month intervals. \(^{27}\) (a) Calculate the average rate of change of \
View solution Problem 32
A mutual fund is currently valued at \(\$ 80\) per share and its value per share is increasing at a rate of \(\$ 0.50\) a day. Let \(V=f(t)\) be the value of th
View solution Problem 35
Suppose \(C(r)\) is the total cost of paying off a car loan borrowed at an annual interest rate of \(r \% .\) What are the units of \(C^{\prime}(r) ?\) What is
View solution