Problem 24 - Precalculus

Question

Yeast is often used in biological experiments. A research technician estimates that a sample of yeast suspension contains 2.5 million organisms per cubic centimeter (cc). Two hours later, she estimates the population density to be 6 million organisms per cc. Let \(t\) be the time elapsed since the first observation, measured in hours. Assume that the yeast growth follows the Law of Uninhibited Growth \(N(t)=N_{0} e^{k t}\). (a) Find the growth constant \(k\). Round your answer to four decimal places. (b) Find a function which gives the number of yeast (in millions) per cc \(N(t)\) after \(t\) hours. (c) What is the doubling time for this strain of yeast?

Step-by-Step Solution

Answer

(a) \(k \approx 0.4378\); (b) \(N(t) = 2.5e^{0.4378t}\); (c) Doubling time is approximately 1.5838 hours.

Step 1: Understand the Problem

We need to find the growth constant \(k\), derive a function \(N(t)\) for the yeast population, and compute the doubling time. The problem follows the uninhibited growth formula \(N(t) = N_0 e^{kt}\), with the given information that \(N_0 = 2.5\) million organisms per cc and after 2 hours, \(N(2) = 6\) million organisms per cc.

Step 2: Set Up the Equation for Growth Constant

We know that initially \(N_0 = 2.5\) million and after 2 hours \(N(2) = 6\) million. Using these values in the growth formula \(N(t) = N_0 e^{kt}\), we set up the equation for \(t = 2\): \[ 6 = 2.5e^{2k} \]

Step 3: Solve for the Growth Constant \(k\)

Rearrange the equation from Step 2 to solve for \(k\):\[ e^{2k} = \frac{6}{2.5} \]\[ e^{2k} = 2.4 \]Take the natural logarithm on both sides:\[ 2k = \ln(2.4) \]\[ k = \frac{\ln(2.4)}{2} \]Using a calculator, \[ \ln(2.4) \approx 0.8755 \] Thus, \[ k \approx \frac{0.8755}{2} \approx 0.4378 \]

Step 4: Write the Function for Yeast Population

With \(k\) found, the function for yeast after \(t\) hours becomes:\[ N(t) = 2.5e^{0.4378t} \]

Step 5: Calculate Doubling Time

For doubling time, set \(N(t) = 2N_0\). Since \(N_0 = 2.5\), doubling means \(N(t) = 5\). Use the equation:\[ 5 = 2.5e^{0.4378t_d} \]Solving for \(t_d\), we get:\[ e^{0.4378t_d} = 2 \]\[ 0.4378t_d = \ln(2) \]\[ t_d = \frac{\ln(2)}{0.4378} \]\[ t_d \approx \frac{0.6931}{0.4378} \approx 1.5838 \] hours.

Key Concepts

Growth Constant

The growth constant, often denoted as \(k\), is a crucial parameter in understanding exponential growth. It reflects the rate at which a population expands over time. In the context of uninhibited growth, like our yeast experiment, the growth constant provides a direct measure of how fast the organisms multiply. The formula given, \(N(t) = N_0 e^{kt}\), relies heavily on finding an accurate \(k\). To find \(k\) for our yeast sample, we observed that the initial population \(N_0\) was 2.5 million organisms per cubic centimeter (cc), and after 2 hours, the population increased to 6 million organisms per cc. Using this data, the equation \(N(t) = N_0 e^{kt}\) is used, where:\[6 = 2.5e^{2k}\]Solving for \(k\) involves rearranging and simplifying the equation with logarithmic calculations:

• First, we divide by 2.5: \(e^{2k} = 2.4\)
• Take the natural logarithm: \(2k = \ln(2.4)\)
• Finally, calculate \(k\): \(k = \frac{\ln(2.4)}{2} \approx 0.4378\)

This value demonstrates how rapidly the yeast population grows during the experiment.

Doubling Time

Doubling time is an important aspect of understanding exponential growth, representing the period it takes for a population to double in size. In our yeast experiment, given that the growth follows \(N(t) = N_0 e^{kt}\), it's essential to calculate the doubling time to predict future growth accurately.To find the doubling time, denoted as \(t_d\), we need the population to double from its initial size, which means setting \(N(t) = 2N_0\). Thus, if \(N_0 = 2.5\), we want \(N(t) = 5\):

• Start by substituting into the formula: \(5 = 2.5e^{0.4378t_d}\)
• Move the terms around to find \(t_d\): \(e^{0.4378t_d} = 2\)
• Using logarithms: \(0.4378t_d = \ln(2)\)
• Finally, compute \(t_d\): \(t_d = \frac{\ln(2)}{0.4378} \approx 1.5838\) hours

This result shows that the yeast population doubles roughly every 1.58 hours under uninhibited growth conditions. It's a valuable measure when planning further experiments or predicting colony sizes.

Uninhibited Growth

Uninhibited growth refers to a scenario where a population grows at a constant relative rate unrestrained by external factors, resulting in exponential growth. This type of growth is ideal for theoretical models because it simplifies calculations and predictions about future population sizes.For organisms like yeast, uninhibited growth is modeled using the formula \(N(t) = N_0 e^{kt}\), whereby the initial number \(N_0\) can exponentially increase over time \(t\) depending on the growth constant \(k\). In the example of our yeast population, \(N_0\) was set at 2.5 million, and we used our calculated \(k\) to describe the growth:

• With \(k \approx 0.4378\), our function becomes: \(N(t) = 2.5e^{0.4378t}\)
• This formula predicts how the population grows without limitations such as resource scarcity or environmental resistance.
• It helps researchers in estimating how a population might expand under optimal conditions, essential for various practical applications in biology and industry.

Uninhibited growth offers a simplified view but is powerful for preliminary analysis and helps in conceptualizing larger studies where other factors might come into play later.