Problem 8 - Calculus and its applications

Question

Pete Zah's, Inc., is selling franchises for pizza shops throughout the country. The marketing manager estimates that the number of franchises, \(N,\) will increase at the rate of \(10 \%\) per year, that is, \(\frac{d N}{d t}=0.10 \mathrm{~N}\) a) Find the function that satisfies this equation. Assume that the number of franchises at \(t=0\) is 50 b) How many franchises will there be in 20 yr? c) In what period of time will the initial number of 50 franchises double?

Step-by-Step Solution

Answer

a) \(N(t) = 50e^{0.10t}\). b) Approximately 369 franchises. c) Approximately 6.93 years to double.

Step 1: Set up the Differential Equation

The first step is to understand the given differential equation: \( \frac{dN}{dt} = 0.10N \). This is a first-order separable differential equation, which represents exponential growth.

Step 2: Integrate the Differential Equation

To solve the differential equation, we separate variables and integrate: \[ \frac{1}{N} dN = 0.10 \, dt \]Integrating both sides gives:\[ \ln|N| = 0.10t + C \]where \(C\) is the integration constant.

Step 3: Solve for N(t)

To express \(N\) explicitly, we exponentiate both sides:\[ N = e^{(0.10t + C)} = e^C \cdot e^{0.10t} \]Let \(e^C = N_0\), the initial number of franchises. Thus, \(N(t) = N_0e^{0.10t}\).

Step 4: Find the Initial Condition

We are given that at \(t=0\), \(N = 50\). Using the equation \(N(0) = N_0e^{0} = N_0\), we have:\[ N_0 = 50 \]So, \(N(t) = 50e^{0.10t}\).

Step 5: Calculate N(20)

To find the number of franchises in 20 years, substitute \(t=20\) into \(N(t)\):\[ N(20) = 50e^{0.10 imes 20} = 50e^{2} \]Calculate this to find the number of franchises after 20 years.

Step 6: Solve for Doubling Time

To find when the franchises will double from 50 to 100, set \(N(t) = 100\):\[ 100 = 50e^{0.10t} \]Divide both sides by 50:\[ 2 = e^{0.10t} \]Take the natural logarithm of both sides:\[ \ln(2) = 0.10t \]Solve for \(t\):\[ t = \frac{\ln(2)}{0.10} \]

Step 7: Calculate Doubling Time

Substitute \( \ln(2) \approx 0.693\) into the equation from Step 6:\[ t \approx \frac{0.693}{0.10} = 6.93 \]Thus, it will take approximately 6.93 years for the franchises to double from 50 to 100.

Key Concepts

Differential Equations

The concept of differential equations is a cornerstone of understanding how quantities change over time. In this exercise, we tackled a specific type of differential equation, which is "separable." These types of equations can be written in the form \( \frac{dy}{dx} = g(x)h(y) \). This allows us to separate the variables \(x\) and \(y\) to opposite sides of the equation, enabling us to integrate both sides independently.
In the given problem, the equation \( \frac{dN}{dt} = 0.10N \) is a classic example of exponential growth. It tells us that the rate of change of \( N \) with respect to time \( t \) is proportional to \( N \) itself. The solutions to such equations are functions that grow exponentially, and they are crucial for modeling real-world systems where growth accelerates over time. For Pete Zah's franchises, this indicates an increase in growth that is consistently speeding up, given that a constant 10% growth rate applies at all times.

Initial Conditions

Initial conditions are essential in solving differential equations as they pin down a specific solution from a family of possible solutions. Once we solve the differential equation generally, we need this information to identify the precise behavior of the system at the starting point.
In this exercise, we were given an initial condition stating that Pete Zah's franchises numbered 50 at time \(t=0\). Knowing this starting point allows us to find the constant of integration that comes up during the process of solving the differential equation.
By substituting this value into the equation \( N(t) = N_0e^{0.10t} \) and setting \( t=0 \), we learned that \( N_0 = 50 \). This information is crucial for predicting future growth. Without it, our solution would lack any specific information about the actual number of franchises at any given time. Initial conditions help convert a general solution into a model tailored to the situation.

Doubling Time

Doubling time refers to the period needed for a quantity to double in size or value, assuming a constant growth rate. It is a straightforward concept, yet plays a crucial role in fields such as finance, biology, and business. In exponential growth contexts, like that of Pete Zah's franchises, it provides a quick way to estimate how long the initial number will take to multiply by two.
To calculate the doubling time, we set up the equation where the final number of entities is twice the initial number. In this case, doubling means going from 50 franchises to 100. We found this by setting up the equation \( 100 = 50e^{0.10t} \) and solving for \(t\).
The results show that \( t \approx 6.93 \) years are required for the franchises to double when growing at a constant exponential rate of 10% annually. Understanding doubling time helps businesses and other entities to plan strategically for the future, giving insights into how fast resources or investments will grow.