Problem 73 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Based on data obtained by polling automobile buyers, the number of subscribers of satellite radios is expected to grow at the rate of $$ r(t)=-0.375 t^{2}+2.1 t+2.45 \quad(0 \leq t \leq 5) $$ million subscribers/year between \(2003(t=0)\) and 2008 \((t=5)\). The number of satellite radio subscribers at the beginning of 2003 was \(1.5\) million. a. Find an expression giving the number of satellite radio subscribers in year \(t(0 \leq t \leq 5)\). b. Based on this model, what was the number of satellite radio subscribers in 2008 ?

Step-by-Step Solution

Answer

a. The number of satellite radio subscribers in year \(t\) can be expressed as \(s(t) = -\frac{1}{8}t^3 + \frac{2.1}{2}t^2 + 2.45t + 1.5\). b. Based on this model, there were approximately 24.375 million satellite radio subscribers in 2008.

Step 1: Integrate the rate function

We are given the rate of growth, \(r(t)\). To find the function that gives the number of subscribers, \(s(t)\), we integrate \(r(t)\): $$ s(t) = \int r(t) dt = \int (-0.375t^2 + 2.1t + 2.45) dt $$

Step 2: Evaluate the integral

Now we compute the integral: \[ s(t) = -\frac{1}{8}t^3 + \frac{2.1}{2}t^2 + 2.45t + C, \] where \(C\) is a constant that we need to determine from the initial condition that at the beginning of 2003 (\(t = 0\)), there were 1.5 million subscribers.

Step 3: Find the constant

Using the given initial condition of 1.5 million subscribers at \(t = 0\): \[ s(0) = 1.5 = -\frac{1}{8}(0)^3 + \frac{2.1}{2}(0)^2 + 2.45(0) + C \Rightarrow C = 1.5 \] Now, we know the entire function for the number of satellite radio subscribers as a function of time: $$ s(t) = -\frac{1}{8}t^3 + \frac{2.1}{2}t^2 + 2.45t + 1.5 $$

Step 4: Evaluate the function for 2008

To find the number of satellite radio subscribers in 2008, we need to evaluate the function at \(t = 5\): $$ s(5) = -\frac{1}{8}(5)^3 + \frac{2.1}{2}(5)^2 + 2.45(5) + 1.5 $$ Calculating the value: $$ s(5) = -\frac{1}{8}(125) + \frac{2.1}{2}(25) + 2.45(5) + 1.5 = \\ -15.625 + 26.25 + 12.25 + 1.5 = 24.375 $$ So, based on this model, there were approximately 24.375 million satellite radio subscribers in 2008.

Key Concepts

Definite Integrals

Definite integrals are a crucial concept in calculus that allow us to calculate the total accumulation of a particular quantity over a given interval. In this exercise, we use a definite integral to find the overall number of satellite radio subscribers over time. The formula for the number of subscribers is derived by integrating the rate of growth function, which tells us how fast subscribers are added each year.

• The integral of the rate function, \( r(t) \), provides us with the total number accumulated, \( s(t) \), which is the function we're ultimately interested in.
• When integrating, we add a constant \( C \) that accounts for the initial conditions. In this case, \( C \) ensures our function starts at the correct number of subscribers at the beginning of 2003.
• With definite integrals, the bounds \( [a, b] \) define the interval of accumulation. In this problem, it spans the beginning of 2003 to the end of 2008.

This exercise demonstrates how definite integrals are used to solve real-world problems, giving us a means to predict future quantities based on known rates.

Initial Value Problems

In calculus, initial value problems are solved using integration with known conditions at a specific point in time. Here, our task was to determine the number of satellite radio subscribers over time, given an initial value.

• An initial value problem requires us to find a function that not only fits the equation derived from the rate of change but also aligns with a known initial condition.
• The initial condition was the number of subscribers at \( t = 0 \) (beginning of 2003), which was given as 1.5 million.
• By substituting \( t = 0 \) into our integrated function \( s(t) \), we are able to find the constant \( C \) which ensures our solution fits the initial condition.

Solving initial value problems allows us to tailor general solutions to specific scenarios, providing predictions or descriptions that are accurate and meaningful in the real world.

Polynomial Functions

Polynomial functions appear often in calculus due to their flexibility and the ease with which they can be manipulated. In this exercise, both the rate function and the subscriber function are polynomials, making calculations straightforward.

• The rate of change of our subscriber function, \( r(t) = -0.375t^2 + 2.1t + 2.45 \), is a quadratic polynomial. This shape suggests certain traits about growth, such as acceleration and deceleration over time.
• When we integrate \( r(t) \), it results in the subscriber function \( s(t) = -\frac{1}{8}t^3 + \frac{2.1}{2}t^2 + 2.45t + C \), which is a cubic polynomial.
• Polynomials like these can be visually represented as curves, making interpretation easier — for instance, the cubic polynomial potentially indicates a shift in growth rates over the observed period.

Understanding polynomial functions enables us to effectively analyze and predict behaviors in data, especially in modeling contexts where rates of change vary over time.