Problem 9 - Calculus Concepts

Question

Electronics Sales (1990s) Annual U.S. factory sales of consumer electronics goods to dealers from 1990 through 2001 can be modeled as $$ s(t)=0.0388 t^{3}-0.495 t^{2}+5.698 t+43.6 $$ where output is measured in billion dollars and \(t\) is the number of years since \(1990 .\) (Sources: Based on data from Statistical Abstract, 2001 ; and Consumer Electronics Association a. Numerically estimate to the nearest tenth the derivative of \(s\) when \(t=10\) b. Interpret the answer to part \(a\).

Step-by-Step Solution

Answer

a. 6.744 b. Sales were increasing by 6.74 billion dollars per year in 2000.

Step 1: Evaluate the function value at t=10

To estimate the derivative numerically, we first need the function value. Plug \(t=10\) into the equation \(s(t)=0.0388 t^{3}-0.495 t^{2}+5.698 t+43.6\): \[s(10) = 0.0388(10)^3 - 0.495(10)^2 + 5.698(10) + 43.6\]Calculating each term: \(0.0388(1000) = 38.8\), \(-0.495(100) = -49.5\), \(5.698 (10) = 56.98\), then adding 43.6. The result is: \[s(10) = 38.8 - 49.5 + 56.98 + 43.6 = 89.88\] So \(s(10) = 89.88\).

Step 2: Evaluate the function value at t=10.1

Now, let's calculate the function value slightly after \(t=10\) for a small increment \(h = 0.1\). Plug \(t = 10.1\) into the function: \[s(10.1) = 0.0388(10.1)^3 - 0.495(10.1)^2 + 5.698(10.1) + 43.6\] Calculating approximations for each power: \(10.1^3 \approx 1030.301\) and \(10.1^2 \approx 102.01\). Substitute and calculate: \[s(10.1) \approx 0.0388(1030.301) - 0.495(102.01) + 5.698(10.1) + 43.6\]Simplifying each term: \(0.0388(1030.301) \approx 39.9796\), \(-0.495(102.01) \approx -50.495\), \(5.698(10.1) \approx 57.4698\). Then the result is: \[s(10.1) \approx 39.9796 - 50.495 + 57.4698 + 43.6 \approx 90.5544\]

Step 3: Calculate the numerical derivative

Use the difference quotient to estimate the derivative: \[s'(t) \approx \frac{s(t + h) - s(t)}{h} = \frac{s(10.1) - s(10)}{0.1}\]Substitute \(s(10)=89.88\) and \(s(10.1)=90.5544\) into the formula: \[s'(10) \approx \frac{90.5544 - 89.88}{0.1}\ = 6.744\]Thus, the approximate value of the derivative at \(t = 10\) is \(6.744\).

Step 4: Interpret the numerical derivative

The numerical derivative \(s'(10) \approx 6.744\) suggests that in the year 2000, the rate of change of electronics sales was approximately 6.74 billion dollars per year. This means that sales were increasing at a rate of 6.74 billion dollars annually at that time.

Key Concepts

Function Evaluation

To begin understanding numerical derivatives, it's essential to grasp the concept of function evaluation. Evaluating a function means finding the output or the value of the function at a specific input value. Imagine the function as a machine that transforms input values (let’s call them \( t \)) into output values (here, \( s(t) \)). For example, the function \( s(t) = 0.0388 t^3 - 0.495 t^2 + 5.698 t + 43.6 \) represents sales in billion dollars based on the number of years since 1990.
In the given context, evaluating the function at \( t = 10 \) helps us know the total electronics sales in 2000. We substitute \( t = 10 \) into the sales function to calculate \( s(10) \). By doing this, we derive the value of 89.88, meaning the sales were 89.88 billion dollars in that year. Understanding how to evaluate a function is critical, as it forms the basis for further calculations, such as finding derivatives.

Difference Quotient

The difference quotient is a central concept in finding the derivative of a function numerically. It helps us estimate how a function's value changes as the input changes slightly. Think of it as measuring how fast something is changing, like the speed of a car. By using the difference quotient, we can determine the slope of the tangent line to a curve at a point, which is precisely the derivative.
The formula for the difference quotient is:

• \( s'(t) \approx \frac{s(t + h) - s(t)}{h} \).

Here, \( h \) is a small number, and \( s(t + h) \) and \( s(t) \) are the function values before and after changing the input slightly. For instance, using \( t = 10 \) and \( h = 0.1 \), we calculated \( s(10.1) \) and \( s(10) \) then applied the formula to find that \( s'(10) \approx 6.744 \). This process allows us to approximate the rate of change of the function at any given point.

Rate of Change

The rate of change in mathematics refers to how a quantity changes relative to another quantity. It's like understanding how quickly things are happening. In this context, we examined how sales of electronics goods change over time. By estimating the derivative of the sales function at \( t = 10 \), we calculate the rate of change, which tells us the speed at which sales increased in the year 2000.
The rate of change, in this case, is calculated numerically as the derivative \( s'(10) \approx 6.744 \). This indicates that for each year during the period around 2000, the sales were increasing by approximately 6.74 billion dollars. Understanding the rate of change provides valuable insight into trends and allows businesses and analysts to make predictions and informed decisions.