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Chapter 7: Problem 3

A bacteria culture starts with 220 bacteria and grows at a rate proportional to its size. After 5 hours there will be 1100 bacteria. (a) Express the population after \(t\) hours as a function of \(t\). population: (function of (b) What will be the population after 5 hours? (c) How long will it take for the population to reach \(1510 ?\)

Short Answer

Expert verified
(a) Population function: \[ P(t) = 220 e^{\left(\frac{\text{ln}(5)}{5}\right)t} \] (b) The population after 8 hours is found by evaluating the function at \( t = 8 \). (c) Time to reach 1510 bacteria: \[ t = \frac{5\cdot\text{ln}\left(\frac{1510}{220}\right)}{\text{ln}(5)} \]

Step by step solution

01

- Determine the growth function

The bacteria grow at a rate proportional to their size, so the population can be modeled by the exponential growth function. The general form is \[ P(t) = P_0 e^{kt} \] where \( P_0 \) is the initial population and \( k \) is the growth rate constant. Given that the initial population \( P_0 \) is 220, we start with \[ P(t) = 220 e^{kt} \]
02

- Calculate the growth rate (\( k \))

After 5 hours, the population is 1100. Substitute \( P(5) = 1100 \) and \( t = 5 \) into the equation to find \( k \): \[ 1100 = 220 e^{5k} \] Divide both sides by 220: \[ 5 = e^{5k} \] Now take the natural logarithm of both sides: \[ \text{ln}(5) = 5k \] Solve for \( k \): \[ k = \frac{\text{ln}(5)}{5} \]
03

- Write the population function

Substitute the value of \( k \) back into the equation to get the population as a function of time:\[ P(t) = 220 e^{\left(\frac{\text{ln}(5)}{5}\right)t} \]
04

- Find the population after 8 hours

To find the population after 8 hours, substitute \( t = 8 \) into the population function:\[ P(8) = 220 e^{\left(\frac{\text{ln}(5)}{5}\right)8} \]Simplify the exponent:\[ P(8) = 220 e^{\left(\frac{8\text{ln}(5)}{5}\right)} \]
05

- Find the time to reach 1510 bacteria

Set the population to 1510 and solve for \( t \):\[ 1510 = 220 e^{\left(\frac{\text{ln}(5)}{5}\right)t} \]Divide both sides by 220:\[ \frac{1510}{220} = e^{\left(\frac{\text{ln}(5)}{5}\right)t} \]Take the natural logarithm of both sides:\[ \text{ln}\left(\frac{1510}{220}\right) = \left(\frac{\text{ln}(5)}{5}\right)t \]Solve for \( t \):\[ t = \frac{5\cdot\text{ln}\left(\frac{1510}{220}\right)}{\text{ln}(5)} \]

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Key Concepts

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

bacteria culture
A bacteria culture consists of a group of bacteria that grow and reproduce under controlled conditions. In this particular problem, we start with a colony of 220 bacteria, and we are interested in how their population increases over time.
Bacteria reproduce by binary fission, meaning one bacterium splits into two. When the environment is favorable (abundant food, optimal temperature, etc.), bacteria can reproduce rapidly.
This reproduction leads to exponential growth, which is a rapid increase in population where the growth rate is proportional to the current size of the population.
In our problem, the exponential growth function intuitively captures how bacteria reproduce at a rate consistent with their current number.
growth rate constant
The growth rate constant, denoted as \(k\), is a key parameter in the exponential growth equation. It indicates how quickly the population is growing.
In our example, after substituting the known values, we calculate \(k\) using the equation:
\[ k = \frac{\text{ln}(5)}{5} \]
Here, we used the fact that after 5 hours, the population increased from 220 to 1100 bacteria.
This growth rate constant is crucial because it tells us how fast the bacteria population doubles or increases by a certain factor.
The larger the value of \(k\), the faster the population grows.
population function
The population function \( P(t) = P_0 e^{kt} \) models how the population grows over time. In our problem, the function is:
\[ P(t) = 220 e^{\frac{\text{ln}(5)}{5} t} \]
We calculate this by using the initial population \( P_0 \) and the growth rate constant \( k \).
To find the population at any given time \( t \), simply plug \( t \) into the equation.
For example, to find the population after 8 hours, we set \( t = 8 \) in the function:
\[ P(8) = 220 e^{\frac{8 \text{ln}(5)}{5}} \]
This function gives us the flexibility to predict the population at any time point.
natural logarithm
The natural logarithm, abbreviated as ln, is a mathematical function that is the inverse of the exponential function. It is based on the constant \( e \), approximately equal to 2.718.
The natural logarithm is very useful in solving exponential growth problems. For example, we used it to solve for the growth rate constant \( k \):
\[ \text{ln}(5) = 5k \]
By taking the natural logarithm of both sides, we can isolate \( k \) and solve for it.
Similarly, to find the time \( t \) needed for the bacteria population to reach a certain number, we use:
\[ t = \frac{5 \text{ln}\frac{1510}{220}}{\text{ln}(5)} \]
The natural logarithm helps simplify these calculations and makes it easier to solve exponential equations.
Understanding how to work with natural logarithms is fundamental in dealing with exponential growth-related problems.

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Most popular questions from this chapter

Find the solution to the differential equation $$ \frac{d y}{d t}=0.5(y-250) $$ if \(y=70\) when \(t=0\).

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We have seen that the error in approximating the solution to an initial value problem is proportional to \(\Delta t .\) That is, if \(E_{\Delta t}\) is the Euler's method approximation to the solution to an initial value problem at \(\bar{t},\) then $$ y(\bar{t})-E_{\Delta t} \approx K \Delta t $$ for some constant of proportionality \(K .\) In this problem, we will see how to use this fact to improve our estimates, using an idea called accelerated convergence. a. We will create a new approximation by assuming the error is exactly proportional to \(\Delta t,\) according to the formula $$ y(\bar{t})-E_{\Delta t}=K \Delta t . $$ Using our earlier results from the initial value problem \(d y / d t=y\) and \(y(0)=1\) with \(\Delta t=0.2\) and \(\Delta t=0.1,\) we have $$ \begin{array}{l} y(1)-2.4883=0.2 \mathrm{~K} \\ y(1)-2.5937=0.1 \mathrm{~K} \end{array} $$ This is a system of two linear equations in the unknowns \(y(1)\) and \(K .\) Solve this system to find a new approximation for \(y(1)\). (You may remember that the exact value is \(y(1)=\) \(e=2.71828 \ldots . .)\) b. Use the other data, \(E_{0.05}=2.6533\) and \(E_{0.025}=2.6851\) to do similar work as in (a) to obtain another approximation. Which gives the better approximation? Why do you think this is? c. Let's now study the initial value problem $$ \frac{d y}{d t}=t-y, y(0)=0 $$ Approximate \(y(0.3)\) by applying Euler's method to find approximations \(E_{0.1}\) and \(E_{0.05}\). Now use the idea of accelerated convergence to obtain a better approximation. (For the sake of comparison, you want to note that the actual value is \(y(0.3)=0.0408 .\) )

Consider the solution of the differential equation \(y^{\prime}=-3 y\) passing through \(y(0)=1.5\). A. Sketch the slope field for this differential equation, and sketch the solution passing through the point (0,1.5) B. Use Euler's method with step size \(\Delta x=0.2\) to estimate the solution at \(x=0.2,0.4, \ldots, 1\), using these to fill in the following table. (Be sure not to round your answers at each step!) \begin{tabular}{|l|l|l|l|l|l|l|} \hline\(x=\) & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 \\ \hline\(y \approx\) & 1.5 & & & & & \\ \hline \end{tabular} C. Plot your estimated solution on your slope field. Compare the solution and the slope field. Is the estimated solution an over or under estimate for the actual solution? D. Check that \(y=1.5 e^{-3 x}\) is a solution to \(y^{\prime}=-3 y\) with \(y(0)=1.5\).

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