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Chapter 7: Problem 13

On Monday morning, Julia found that a colony of bacteria covered an area of \(100 \mathrm{cm}^{2}\) on the agar. After \(10 \mathrm{h},\) she found that the area had increased to \(200 \mathrm{cm}^{2}\). Assume that the growth is exponential. a) By Tuesday morning \((24 \mathrm{h}\) later), what area do the bacteria cover? b) Consider Earth to be a sphere with radius \(6378 \mathrm{km} .\) How long would these bacteria take to cover the surface of Earth?

Short Answer

Expert verified
a) The area covered by Tuesday morning is approximately 527.8 \mathrm{cm}^2.b) The bacteria would take around 567.67 hours to cover the surface of Earth.

Step by step solution

01

Understand exponential growth

Exponential growth can be modeled by the equation \[ A(t) = A_0 \times e^{kt} \] where \( A(t) \) is the area at time \( t \), \( A_0 \) is the initial area, \( k \) is the growth rate, and \( t \) is the time.
02

Use given data to find the growth rate

We know that \( A_0 = 100 \mathrm{cm}^{2} \) and after \(10 \mathrm{h} \), \( A(10) = 200 \mathrm{cm}^{2} \). Substitute these values into the exponential equation: \[ 200 = 100 \times e^{10k} \] Solve for \( k \): \[ 2 = e^{10k} \Rightarrow ln(2) = 10k \Rightarrow k = \frac{ln(2)}{10} \]
03

Calculate the area after 24 hours

Now use the value of \( k \) to find the area after 24 hours: \[ A(24) = 100 \times e^{24 \times \frac{ln(2)}{10}} \] Simplify: \[ A(24) = 100 \times e^{2.4 \times ln(2)} \] Since \( e^{ln(2)} = 2 \), \[ A(24) = 100 \times 2^{2.4} \] Compute \( 2^{2.4} \): \[ 2^{2.4} \approx 5.278 \] Therefore, \[ A(24) \approx 100 \times 5.278 = 527.8 \mathrm{cm}^{2} \]
04

Calculate how long it takes to cover Earth's surface

First, find the surface area of Earth using the radius \( R = 6378 \mathrm{km} \): \[ \text{Surface Area} = 4 \pi R^2 \Rightarrow 4 \pi (6378000 \mathrm{m})^2 \] \[ \text{Surface Area} \approx 5.1 \times 10^{14} \mathrm{m}^2 = 5.1 \times 10^{18} \mathrm{cm}^2 \] Use the exponential growth formula to find the time \( t \) when \( A(t) = 5.1 \times 10^{18} \mathrm{cm}^2 \): \[ 5.1 \times 10^{18} = 100 \times e^{kt} \] \[ \frac{5.1 \times 10^{18}}{100} = e^{kt} \] \[ ln(5.1 \times 10^{16}) = kt \] \[ t =\frac{ln(5.1 \times 10^{16})}{k} \] Substitute \( k \approx 0.0693 \): \[ t \approx \frac{ln(5.1 \times 10^{16})}{0.0693} \approx 567.67 \mathrm{hours} \]

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Key Concepts

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

Exponential Growth
Exponential growth describes a process where the quantity increases by a constant factor over equal time intervals. It's different from linear growth, where the increase is constant. Exponential growth happens in real-life scenarios like bacteria populations, where the number of bacteria doubles in regular intervals.
In mathematical terms, the exponential growth can be modeled using the formula: \[ A(t) = A_0 \times e^{kt} \] Here, A(t)represents the quantity at timet, A_0is the initial quantity, k is the growth rate, and tis the time. In our problem, A(t)is the area covered by bacteria, A_0 = 100 \text{cm}^2, and we need to find how fast the bacteria are growing to determine future growth.
Mathematical Modeling
Mathematical modeling involves creating a mathematical representation of a real-world scenario. In this exercise, we use an exponential growth model to predict how a bacteria colony grows over time.
Given data points help us find the growth rate. For instance, starting with \( A_0 = 100 \text{cm}^2 \)and knowing the bacteria cover \( 200 \text{cm}^2\) int = 10 \text{h}, we can plug these values into our exponential growth formula.
Solving for kgives us the growth rate specific to our problem. Once we have k, we can predict future growth by substituting any desired timet into the same formula.
Surface Area Calculation
Surface area calculation involves finding the total area that the bacteria can potentially cover. In our exercise, this includes both small-scale (like the initial petri dish) and large-scale areas (like the surface of Earth).
To calculate the Earth's surface area, we use the formula for the surface area of a sphere: \[ \text{Surface Area} = 4 \text{蟺} R^2\], where R is the radius of Earth, \( 6378 \text{km} \)
Converting the radius to centimeters, R = 6378 \times 10^5 \text{cm}and substituting in the formula, we get a very large surface area. This step is crucial for understanding how long it will take for the bacteria to cover such a vast amount of space.
Natural Logarithms
Natural logarithms (l) are mathematical functions that help us solve exponential equations. They are inverses of the exponential function. When working with exponential growth, natural logarithms allow us to find the growth rate and time.
In this exercise, we use natural logarithms to solve for the growth rate k. After substituting known values into the exponential growth equation, we get \[ 2 = e^{10k}\]. Applying the natural logarithm to both sides, \[ \text{ln}(2) = 10k\], we solve for k by dividing both sides by 10. This method simplifies the calculation, making it easy to find the time taken for bacteria to cover a large area like Earth by using the formula \( \text{t} = \frac{\text{ln(area ratio)}}{k} \) .

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Most popular questions from this chapter

Describe the transformations that must be applied to the graph of each exponential function \(f(x)\) to obtain the transformed function. Write each transformed function in the form \(y=a(c)^{b(x-h)}+k\). a) \(f(x)=\left(\frac{1}{2}\right)^{x}, y=f(x-2)+1\) b) \(f(x)=5^{x}, y=-0.5 f(x-3)\) c) \(f(x)=\left(\frac{1}{4}\right)^{x}, y=-f(3 x)+1\) d) \(f(x)=4^{x}, y=2 f\left(-\frac{1}{3}(x-1)\right)-5\)

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A flu virus is spreading through the student population of a school according to the function \(N=2^{t},\) where \(N\) is the number of people infected and \(t\) is the time, in days. a) Graph the function. Explain why the function is exponential. b) How many people have the virus at each time? i) at the start when \(t=0\) ii) after 1 day iii) after 4 days iv) after 10 days
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