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Bacteria Growth The number of bacteria in a culture is increasing according to the law of exponential growth. There are 125 bacteria in the culture after 2 hours and 350 bacteria after 4 hours. (a) Find the initial population. (b) Write an exponential growth model for the bacteria population. Let \(t\) represent time in hours. (c) Use the model to determine the number of bacteria after 8 hours.

Short Answer

Expert verified
The initial population (N0) is found to be approximately 48. The exponential growth model for the bacteria population is \(N = 48 \times e^{0.546t}\). After 8 hours, the number of bacteria will be approximately \(N = 48 \times e^{0.546\times 8}\).

Step by step solution

01

- Calculate the Growth Rate

We will first find the growth rate and to do that we need to solve for \(r\) in the formula. We know that \(N=350\), \(N0=125\) after 2 hours, and \(t=4-2=2\). So, \(350 = 125 \times e^{2r}\). Solving this equation, the growth rate \(r\) is found to be \(ln(350/125)/2\). Calculate this value using a calculator to get approx 0.546.
02

- Find Initial Population

Now we can find the initial population of the bacteria. We use the exponential growth formula again, this time inserting \(N=125\), \(r=0.546\), and \(t=2\). The resulting equation is \(125 = N0 \times e^{0.546\times 2}\). Solving this equation for \(N0\), we get \(N0 = 125 / e^{0.546 \times 2}\). Calculating the value, we get approx 48.
03

- Write the Exponential Growth Model

Next, we'll write the exponential growth model. Substitute the calculated values into the formula \(N = N0 \times e^{rt}\) to get the growth model as \(N = 48 \times e^{0.546t}\). This equation can be used to predict the bacteria population at any given time.
04

- Predict the Bacteria Population

Finally, use the model to predict the number of bacteria after 8 hours. Substitute \(t=8\) into the growth model, we get \(N = 48 \times e^{0.546\times 8}\). By calculating this, the number of bacteria can be predicted.

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Key Concepts

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

Growth Rate Calculation
When studying populations, such as bacteria in a culture, it's crucial to understand their rate of increase over time. This increase is often described as exponential growth, characterized by a constant proportionality factor known as the growth rate. To calculate this growth rate, one must have data points at different times.

In our case, we had a population of 125 bacteria after 2 hours and 350 bacteria after 4 hours. Using the exponential growth formula, we set up an equation involving the known quantities and the unknown growth rate, which we denote as 'r'. The formula for one data point is given by: \[N = N_0 \times e^{rt}\] where the variable 'r' is the growth rate, 'N_0' is the initial population size, and 't' is the time passed. By substituting our data values, we were able to solve for 'r' algebraically. This growth rate is essential for the exponential growth model and can be used to predict future populations, as it represents the rate at which the bacteria multiply.
Initial Population Estimation
Determining the initial population size, or 'N_0', is like setting the starting point of our exponential journey. Without an accurate estimation of this starting point, predictions for future population sizes will be off.

Given the population at a known time, we can rearrange the exponential growth formula to solve for 'N_0'. In our bacteria culture, we utilized the known size at 2 hours and the previously calculated growth rate 'r'. By isolating 'N_0' on one side of the equation, we found the initial number of bacteria in the culture. This value is not only important for record-keeping and understanding past conditions but is also the foundation from which we build our predictive growth model.
Bacteria Population Prediction
Predicting future populations provides valuable insight into how a culture of bacteria will behave over time. With the exponential growth model in hand, we use the formula \[N = N_0 \times e^{rt}\] where 'N' is the future population size after time 't'.

By plugging in the initial population size 'N_0', the growth rate 'r', and the desired future time, we get an estimate of the population size at that future moment. Predictions like these are used in many fields, from biology to environmental science, and even economics, to prepare for and manage future scenarios, like resource needs or containment strategies.
Differential Equations
Behind the scenes of our exponential growth calculations are differential equations. In our context, these are mathematical equations that describe the rate of change of the bacteria population over time. The link between exponential growth and differential equations is intrinsic; the exponential model is actually a solution to a certain type of differential equation known as a growth equation.

For any continuously growing process, the differential equation often takes the form \[\frac{dN}{dt} = rN\] where \(\frac{dN}{dt}\) is the derivative of the population with respect to time, indicating how quickly the population changes. 'N' is the population size, and 'r' is the growth rate. Solving this differential equation under appropriate initial conditions gives us the exponential growth model that we've been using to make our bacteria population predictions.

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Most popular questions from this chapter

Mixture In Exercises \(35-38\) , consider a tank that at time \(t=0\) contains \(v_{0}\) gallons of a solution of which, by weight, \(q_{0}\) pounds is soluble concentrate. Another solution containing \(q_{1}\) pounds of the concentrate per gallon is running into the tank at the rate of \(r_{1}\) gallons per minute. The solution in the tank is kept well stirred and is withdrawn at the rate of \(r_{2}\) gallons per minute. Let \(Q\) be the amount of concentrate in the solution at any time t. Show that $$ \frac{d Q}{d t}+\frac{r_{2} Q}{v_{0}+\left(r_{1}-r_{2}\right) t}=q_{1} r_{1} $$

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Using a Logistic Equation In Exercises 51 and 52 , the logistic equation models the growth of a population. Use the equation to (a) find the value of \(k,\) (b) find the carrying capacity, (c) find the initial population, (d) determine when the population will reach 50\(\%\) of its carrying capacity, and (e) write a logistic differential equation that has the solution \(P(t)\) . $$ P(t)=\frac{5000}{1+39 e^{-0.2 t}} $$

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