Question

From 1996 through 2005, the numbers \(y\) (in millions) of employed people in the United States can be modeled by \(y=98.020+6.2472 t-0.24964 t^{2}+0.000002 e^{t}\) where \(t\) represents the year, with \(t=6\) corresponding to 1996. (a) Use a graphing utility to graph the model. (b) Use the graph to estimate the rates of change in the number of employed people in 1996,2000 , and 2005 . (c) Confirm the results of part (b) analytically.

Short answer

The graphical estimation in step 2 should provide approximate rates of change in employment for the years 1996, 2000, and 2005. In step 3, the analytical method will provide precise values of the rates of change for these years. Due to the exact values being dependent on how accurately the rates were estimated from the graph in step 2, they cannot be provided here.

Step-by-step solution

Graphing the Model

It would be great to start by sketching the model for the range from \(t=6\) (1996) to \(t=15\) (2005). This process involves plugging in the values of \(t\) into the model equation and plotting the corresponding \(y\) values. A graphing utility like Desmos or Graphmatica could be very useful for this task.

Estimating Rates from the Graph

From the previously plotted graph, estimate the rates of change for the years 1996 (\(t=6\)), 2000 (\(t=10\)), and 2005 (\(t=15\)). From 1996 through 2005, the curve is upward sloping indicating an increase in employment. The rates can be estimated from the slopes of the tangents at these points on the curves.

Confirming Analytically

The rate of change of the number of employed people can be found by taking the first derivative of the given function with respect to time \(t\). The derivative represents the rate of change at a given time. Therefore, compute the derivative of the given function \(y = 98.020 + 6.2472t - 0.24964t^2 + 0.000002e^t\) and substitute the values \(t=6\), \(t=10\), and \(t=15\) to find the respective rates in those years.

Key concepts

Understanding Graphing Utility

A graphing utility is a powerful tool utilized by students and professionals to visualize mathematical functions and their behavior. Applications like Desmos, Graphmatica, or even a graphic calculator enable users to input equations and instantly see the graph. This visual representation helps interpret complex functions, like the employment rate model provided in the exercise.

For the given task, plotting the employment model from 1996 to 2005 requires entering the function into the utility with values of 't' ranging from 6 to 15. This graph will show how employment numbers have changed over time. When creating such a graph, users can adjust the scale and range for clarity and also zoom in or out for a detailed view of specific areas. These features are particularly helpful when estimating rates of change, as they allow users to draw tangents at given points to assess the slope, which corresponds to the rate at which employment is increasing or decreasing.

Grasping Rates of Change

Rates of change are fundamental in understanding how a quantity varies with respect to another, such as time. In our employment rate model, the rates of change signify how employment numbers have evolved annually. To estimate these changes using the graph, one would need to look at the slope of the tangent lines at the points of interest on the graph for the years 1996, 2000, and 2005.

The steeper the slope, the greater the rate of employment change during that period. A slope of zero would indicate no change, while a negative slope would point to a decrease in employment. Estimating these slopes requires careful observation and sometimes drawing tangents at the points of interest, which then reflect the instantaneous rate of increase or decrease of employment within that specific year.

Demystifying Derivative Calculus

Derivative calculus is a branch of mathematics that deals with the rates at which things change. It's the mathematical way of finding the rate of change at any given point. The derivative of a function at a particular point is the slope of the tangent line at that point on the function's graph.

In the context of the employment model, taking the first derivative of the function with respect to time (t) shows us how the employment rate changes over the years. Mathematically, for the function \(y = 98.020 + 6.2472t - 0.24964t^2 + 0.000002e^t\), the first derivative would be \(y' = 6.2472 - 2*0.24964t + 0.000002e^t\), ignoring higher-order terms. By plugging in the values for 1996, 2000, and 2005, we get the rates of change for employment for these individual years. It's important to note that while graph estimates are useful, derivative calculus provides a precise and analytical approach to determining the exact rate of change at any given moment.