Problem 55 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The number of cable telephone subscribers stood at \(3.2\) million at the beginning of \(2004(t=0)\). For the next \(5 \mathrm{yr}\), the number was projected to grow at the rate of $$ R(t)=3.36(t+1)^{0.05} \quad(0 \leq t \leq 5) $$ million subscribers/year. If the projection held true, how many cable telephone subscribers were there at the beginning of \(2008(t=4)\) ?

Step-by-Step Solution

Answer

There were approximately 5.38 million cable telephone subscribers at the beginning of 2008.

Step 1: Write down the given information

Initial number of subscribers: 3.2 million Growth rate function, R(t): \(3.36(t+1)^{0.05}\) million subscribers/year

Step 2: Set up the integral

To find the total growth in subscribers between 2004 (t=0) and 2008 (t=4), we need to integrate the growth rate function, R(t), over that time interval: $$ \int_{0}^{4} 3.36(t+1)^{0.05} dt $$

Step 3: Integrate the function

Integrate the function with respect to t: $$ \int 3.36(t+1)^{0.05} dt = \frac{3.36}{0.05+1}(t+1)^{0.05+1}+C $$

Step 4: Calculate the definite integral between the limits 0 and 4

Now, we will evaluate the integral from 0 to 4: $$ \left[\frac{3.36}{1.05}(t+1)^{1.05}\right]_{0}^{4} $$ Compute the value at the upper limit (t = 4): $$ \frac{3.36}{1.05}(4 + 1)^{1.05} = \frac{3.36}{1.05}(5)^{1.05} \approx 5.38 $$ Compute the value at the lower limit (t = 0): $$ \frac{3.36}{1.05}(0 + 1)^{1.05} = \frac{3.36}{1.05}(1)^{1.05} \approx 3.20 $$ Subtract the lower limit from the upper limit: $$ 5.38 - 3.20 \approx 2.18 $$ This means that the total growth in subscribers between 2004 and 2008 is approximately 2.18 million.

Step 5: Add the growth to the initial subscriber count

To find the number of subscribers at the beginning of 2008 (t=4), add the total growth to the initial subscriber count: $$ 3.2 \, \text{million} + 2.18 \, \text{million} \approx 5.38 \, \text{million} $$ Thus, there were approximately 5.38 million cable telephone subscribers at the beginning of 2008.

Key Concepts

Definite Integral

Understanding the concept of a definite integral is like grasping how to measure the total accumulation of something over a period of time or space. In applied mathematics, it's a crucial tool for quantifying this accumulation in a precise way. Imagine you're filling up a swimming pool and want to know exactly how much water you've added over several hours—that's where the definite integral comes into play.

A definite integral, in its essence, is a calculation of the net area under a curve within specific boundaries on a graph. This is integral (pun intended!) in applications ranging from physics to engineering, where it's often necessary to add up continuous quantities. In our cable telephone subscribers example, we utilized a definite integral to find the total increase in subscribers from the beginning of 2004 to the start of 2008.

Growth Rate Function

A growth rate function, symbolized as R(t) in our scenario, represents how quickly something is changing over time. In the context of our problem, it measures the change in the number of cable telephone subscribers as a function of time, t. This kind of function is pivotal in mathematical modeling because it provides a dynamic view of changes, rather than a static snapshot.

The growth rate isn't always consistent; it can vary depending on various factors, much like the acceleration of a car changes with the pressure on the gas pedal. The specific function we dealt with, R(t)=3.36(t+1)^{0.05}, interestingly, indicates that the rate of growth itself is changing, albeit slightly, as time goes on due to the exponent 0.05. This nuanced detail helps reflect more realistic scenarios where growth doesn't follow a straight-line pattern.

Mathematical Modeling

Mathematical modeling is akin to crafting a miniature version of the real world using the language of mathematics so we can study and predict complex systems in a simplified form. It involves creating equations and functions that mimic real-life processes, like the increase of subscribers, weather patterns, or the trajectory of a spaceship.

In our example, we constructed a model using the growth rate function to predict the number of cable telephone subscribers over a five-year period. This model's strength lies in its ability to transform abstract numbers into a digestible narrative about how a situation might unfold. By integrating the growth rate function, we essentially completed the story of our subscriber growth from chapter 2004 to chapter 2008, giving us a rumored ending of approximately 5.38 million subscribers.

From budget forecasts to epidemiological studies, mathematical modeling helps experts in various fields make informed decisions based on projected outcomes. When combined with other tools like the definite integral, it becomes a potent method for analyzing changes over time.