Problem 94 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Since their introduction into the market in the late \(1990 \mathrm{~s}\), the sales of digital televisions, including high-definition television sets, have slowly gathered momentum. The model $$ S(t)=0.164 t^{2}+0.85 t+0.3 \quad(0 \leq t \leq 4) $$ describes the sales of digital television sets (in billions of dollars) between the beginning of \(1999(t=0)\) and the beginning of \(2003(t=4)\). a. Find \(S^{\prime}(t)\) and \(S^{\prime \prime}(t)\). b. Use the results of part (a) to conclude that the sales of digital TVs were increasing between 1999 and 2003 and that the sales were increasing at an increasing rate over that time interval.

Step-by-Step Solution

Answer

In conclusion, we found that \(S'(t) = 0.328t + 0.85\) and \(S''(t) = 0.328\). Since \(S'(t) > 0\) for the interval \(0 \leq t \leq 4\), we can conclude that the sales of digital TVs were increasing between 1999 and 2003. Since \(S''(t) > 0\) for the same interval, it indicates that the rate of sales was also increasing during that time. Thus, the sales of digital TVs were increasing at an increasing rate between 1999 and 2003.

Step 1: Find the first derivative \(S'(t)\)

To find the first derivative, we'll differentiate \(S(t)\) with respect to \(t\). We will use the power rule, which states that the derivative of \(t^n\) is \(nt^{n-1}\). \[S'(t) = \frac{d}{dt}(0.164t^2 + 0.85t + 0.3)\] \[S'(t) = 2(0.164)t^{(2-1)} + 1(0.85)t^{(1-1)}\] (Constant term 0.3 will vanish on differentiation) \[S'(t) = 0.328t + 0.85\] So, the first derivative is \(S'(t) = 0.328t + 0.85\)

Step 2: Find the second derivative \(S''(t)\)

To find the second derivative, we'll differentiate \(S'(t)\) with respect to \(t\) again. \[S''(t) = \frac{d}{dt}(0.328t + 0.85)\] \(S''(t) = 0.328\) (Constant term 0.85 will vanish on differentiation, and t term reduces to constant) So, the second derivative is \(S''(t) = 0.328\). Now that we've found the first and second derivatives, we can answer part (b):

Step 3: Determine if sales are increasing between 1999 and 2003 using \(S'(t)\)

If the first derivative, \(S'(t)\), is positive on the interval \(0 \leq t \leq 4\), then the sales are increasing during that time. We found: \[S'(t) = 0.328t + 0.85\] To determine if this function is positive over the given interval, we can check the sign of \(S'(t)\) at the end points, i.e., \(t = 0\) and \(t = 4\): \[S'(0) = 0.328(0) + 0.85 = 0.85 > 0\] \[S'(4) = 0.328(4) + 0.85 = 2.57 > 0\] Since \(S'(t)\) is positive for both endpoints of given interval, we can conclude that the sales of digital TVs were increasing between 1999 and 2003.

Step 4: Determine if the sales increase rate is also increasing using \(S''(t)\)

If the second derivative, \(S''(t)\), is positive on the interval \(0 \leq t \leq 4\), then the rate of sales is increasing during that time. We found: \[S''(t) = 0.328\] Since this is a constant value greater than 0, the second derivative is positive for all values of \(t\). This means the sales are increasing at an increasing rate between 1999 and 2003. In conclusion, the sales of digital TVs were increasing between 1999 and 2003, and the sales were increasing at an increasing rate over that time interval.

Key Concepts

Power Rule Differentiation

Understanding the power rule for differentiation is a critical skill in applied mathematics that simplifies the process of finding the rate of change of various functions. In the context of our example, dealing with the sales of digital televisions, the power rule is a quick and effective method of finding the first derivative of a polynomial function like
\( S(t) = 0.164t^2 + 0.85t + 0.3 \).

To apply the power rule, recall that for any term in the form of \( Ct^n \), where \( C \) is a constant and \( n \) is a positive integer, the derivative is given by \( nCt^{n-1} \). The constant term will drop out because it does not vary and thus its instantaneous rate of change is zero. The power rule is especially helpful as it streamlines the computation without the need for more complex differentiation techniques.

Example in the Sales Model

Let's apply the power rule to our function \( S(t) \).
The term \( 0.164t^2 \) becomes \( 2 \times 0.164t^{2-1} \), or \( 0.328t \).
The term \( 0.85t \) simplifies to \( 1 \times 0.85t^{1-1} \), or \( 0.85 \), since the power of \( t \) is reduced to zero, effectively removing \( t \) from the expression.
Hence, \( S'(t) = 0.328t + 0.85 \), which is the rate at which sales are changing over time.

First Derivative

When we talk about the first derivative of a function, we are essentially discussing the slope of the function at any given point or the rate at which the function's value is changing with respect to time or another variable. For businesses and economists, the first derivative can represent many things, like speed, growth rate, or as in our textbook example, the rate of sales of digital TVs.

The first derivative \( S'(t) \), calculated using power rule differentiation, tells us how sales change at a particular moment in time. If \( S'(t) > 0 \) for a range of time, sales are increasing during that period. Applying this to our model from 1999 to 2003, we can analyze the sales trend by examining the sign of \( S'(t) \) at various points. Since \( S'(t) = 0.328t + 0.85 \) is positive for the whole interval from \( t = 0 \) to \( t = 4 \), we can confidently state that sales were on the rise during this time.

Analyzing and verifying the behavior of the first derivative is not only crucial in understanding the present trend but also in forecasting future behavior based on existing patterns.

Second Derivative

Moving beyond the first derivative, the second derivative \( S''(t) \) in applied mathematics gives us a sense of the acceleration, or the rate of change of the rate of change. For the sales of digital TVs, a positive second derivative indicates that not only are sales increasing, they are increasing at an increasingly higher rate - an encouraging sign for market growth.

In our example, \( S''(t) = 0.328 \) which is a positive constant. The constancy comes from the fact that the second derivative of a linear function, which is what \( S'(t) \) is, will always be a constant. Because it’s positive, this constant second derivative indicates a consistent increase in the growth rate of digital TV sales between 1999 and 2003.

Understanding Acceleration

In the physical sciences, this concept of acceleration is directly related to motion — how quickly something speeds up or slows down. Likewise, in economics or business, it can signal how quickly an increase in sales is ramping up, which can have implications for production, inventory management, and forecasting. Recognizing and interpreting the significance of the second derivative is essential for strategizing and decision-making based on those predictions.