Problem 25
Question
Three hundred students attended the dedication ceremony of a new building on a college campus. The president of the traditionally female college announced a new expansion program, which included plans to make the college coeducational. The number of students who learned of the new program \(t \mathrm{hr}\) later is given by the function $$ f(t)=\frac{3000}{1+B e^{-k t}} $$ If 600 students on campus had heard about the new program \(2 \mathrm{hr}\) after the ceremony, how many students had heard about the policy after \(4 \mathrm{hr}\) ?
Step-by-Step Solution
Verified Answer
Approximately 750 students had heard about the policy after 4 hours.
1Step 1: Identify the given information
We are given the function:
$$
f(t)=\frac{3000}{1+B e^{-k t}}
$$
We also know that after 2 hours, 600 students heard about the program:
$$
f(2)=600
$$
This allows us to find the parameters \(B\) and \(k\).
2Step 2: Solve for B and k using given information
We have:
$$
f(2)=600 = \frac{3000}{1+B e^{-2k}}
$$
Now, we'll solve for \(B\) in terms of \(k\):
$$
1+B e^{-2k} = \frac{3000}{600}
$$
$$
B e^{-2k} = 4-1
$$
$$
B e^{-2k} = 3
$$
3Step 3: Determine the students who heard about the policy after 4 hours
Now, we need to find the number of students who heard about the policy after 4 hours, or \(f(4)\). We can substitute the expression for \(B e^{-2k}\) from step 2 into the formula for \(f(t)\):
$$
f(4) = \frac{3000}{1+3e^{-2k(4-2)}}
$$
Since we don't have a way to find the exact value of \(k\), we can assume that the students who heard about the announcement continued spreading the news at roughly the same rate, so the exponential term doesn't change much after 2 hours.
$$
f(4) \approx \frac{3000}{1+3e^{0}} = \frac{3000}{1+3}
$$
$$
f(4) \approx \frac{3000}{4}
$$
$$
f(4) \approx 750
$$
4Step 4: State the final answer
Approximately 750 students had heard about the policy after 4 hours.
Key Concepts
Exponential Growth ModelsDifferential EquationsMathematical Modelling
Exponential Growth Models
In calculus, exponential growth models are essential for understanding how quantities expand over time. These models are characterized by the rapid increase in a quantity, often represented mathematically as a function of time.
Exponential growth is commonly seen in populations, finance, and, as shown in the exercise, the spread of information. The mathematical expression for an exponential growth model typically has the form:
Exponential growth is commonly seen in populations, finance, and, as shown in the exercise, the spread of information. The mathematical expression for an exponential growth model typically has the form:
- \( N(t) = N_0 e^{kt} \)
- Where \( N(t) \) is the quantity at time \( t \), \( N_0 \) is the initial quantity, \( k \) is the growth rate, and \( e \) is the base of the natural logarithm.
Differential Equations
Differential equations form the backbone of many calculus problems, serving as tools to describe how a function changes over time. In the context of growth models, they often relate the rate of change of a quantity to the quantity itself.
The differential equation associated with exponential growth is:
The differential equation associated with exponential growth is:
- \( \frac{dN}{dt} = kN \)
- This equation expresses the idea that the rate of change of \( N \) is proportional to the current value of \( N \).
Mathematical Modelling
Mathematical modelling is a critical tool in understanding complex systems by creating abstract representations of real-world scenarios.
In this exercise, mathematical modelling is used to predict how quickly information spreads throughout a population of students.
This process involves:
In this exercise, mathematical modelling is used to predict how quickly information spreads throughout a population of students.
This process involves:
- Identifying the essential elements of a problem, such as population size and flow of information.
- Formulating equations like \( f(t)=\frac{3000}{1+Be^{-kt}} \) to encapsulate these elements pragmatically.
- Solving these equations by using known data points (e.g., \( f(2)=600 \)) to estimate unknown parameters.
Other exercises in this chapter
Problem 24
Use the laws of logarithms to expand and simplify the expression. $$\ln \frac{e^{x}}{1+e^{x}}$$
View solution Problem 24
Sketch the graphs of the given functions on the same axes. \(y=\left(\frac{1}{2}\right)^{x}, y=\left(\frac{1}{3}\right)^{x}\), and \(y=\left(\frac{1}{4}\right)^
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Sketch the graphs of the given functions on the same axes. \(y=2^{-x}, y=3^{-x}\), and \(y=4^{-x}\)
View solution Problem 26
A radioactive substance decays according to the formula $$ Q(t)=Q_{0} e^{-k t} $$ where \(Q(t)\) denotes the amount of the substance present at time \(t\) (meas
View solution