Problem 1 - Differential Equations and Linear Algebra

Question

The number of bacteria in a culture grows at a rate that is proportional to the number present. Initially there were 10 bacteria in the culture. If the doubling time of the culture is 3 hours, find the number of bacteria that were present after 24 hours.

Step-by-Step Solution

Answer

After 24 hours, there were approximately 16384 bacteria in the culture.

Step 1: Setting up a differential equation

Since the rate of growth is proportional to the number of bacteria present, we can write it as follows: \( \frac{dB}{dt} = k B \) Where B is the number of bacteria, t is the time (in hours), and k is the constant of proportionality (positive since the growth is positive).

Step 2: Solving the differential equation

To find the number of bacteria as a function of time, we need to solve the differential equation: \( \frac{dB}{dt} = k B \) Dividing both sides by B: \( \frac{dB}{B} = k dt \) Now, we integrate both sides with respect to the variables B and t, respectively: \( \int \frac{dB}{B} = \int k dt \) Applying integrals, we obtain: \( ln(|B|) = kt + C_1 \) To find the amount of bacteria on the timescale we are observing, let's solve for B: \[ B(t) = e^{kt + C_1} = e^{kt} \cdot e^{C_1} \] Let's denote the constant \( e^{C_1} \) as C, and rewrite the equation: \[ B(t) = C \cdot e^{kt} \]

Step 3: Applying initial conditions

To find the values of k and C, we apply the initial conditions given in the problem: Initially, there were 10 bacteria: \( B(0) = 10 \) So, \( C*e^{k*0} = 10 \) Which simplifies to: \[ C = 10 \] Now we know that the doubling time is 3 hours, so we have: \( B(3) = 2 * 10 = 20 \) Plugging this into our equation: \( 10*e^{3k} = 20 \) Now, let's solve for k: \( e^{3k} = \frac{20}{10} = 2 \) Obtaining: \[ k = \frac{ln(2)}{3} \]

Step 4: Finding the number of bacteria after 24 hours

Now that we have solved for k and C, we can substitute the values back into the initial equation for B(t): \[ B(t) = 10 * e^{\frac{ln(2)}{3}t} \] Now we can find the number of bacteria after 24 hours by simply plugging t=24 into the equation: \[ B(24) = 10 * e^{\frac{ln(2)}{3} \cdot 24} \] This gives us the number of bacteria after 24 hours: \[ B(24) \approx 16384 \] So, after 24 hours, there were approximately 16384 bacteria in the culture.

Key Concepts

Exponential Growth

Exponential growth describes a situation where the rate of increase of a quantity is directly proportional to the current amount of the quantity itself. It's a fundamental concept in understanding how populations of organisms, including bacteria, grow over time. In our example, the culture of bacteria expands in such a way that the larger the number of bacteria present, the faster the population grows.

This growth can be mathematically represented by the equation \( B(t) = C \times e^{kt} \), where \( B(t) \) represents the number of bacteria at any given time \( t \), \( C \) is the initial quantity of bacteria, \( e \) is the base of the natural logarithm, and \( k \) is a constant that represents the rate of growth. This mathematical model is very powerful as it can be used to predict how large a population will become after a certain period, assuming that conditions remain constant for that growth to continue.

Solving Differential Equations

A differential equation is a mathematical equation that involves unknown functions and their derivatives. Solving these equations often helps in understanding how a system changes over time. In the context of bacterial growth, the differential equation \( \frac{dB}{dt} = k B \) encapsulates the idea of exponential growth, with \( \frac{dB}{dt} \) representing the rate of change of the bacteria population over time.

To solve this differential equation, we must separate the variables and integrate. This yields the general solution \( B(t) = C \times e^{kt} \), which includes a constant \( C \) that can be determined using initial conditions of the system. This process converts the abstract representation of the growth rate into a specific formula we can use to make actual predictions about the bacteria population at any given time.

Initial Value Problem

An initial value problem in the context of differential equations is a problem in which we are given the rate of change of a quantity and we’re asked to find the quantity itself, given its initial value. This is vital because it anchors the general solution of the differential equation to a specific context, ensuring the solution is applicable to the real-world scenario in question.

In our bacterial growth example, the initial value problem is solved by finding the constants \( C \) and \( k \) in the equation, which represent the initial quantity of bacteria and the growth rate, respectively. We were given that the initial number of bacteria is 10, and the doubling time is 3 hours. This information allows us to calculate the exact values of \( C \) and \( k \), making the equation specific to the initial conditions of our bacteria culture. Once this is achieved, we can predict the number of bacteria at any future time, such as after 24 hours, through the equation \( B(t) \).