Problem 34 - Calculus: Graphical, Numerical, Algebraic

Question

.Table 2.3 gives the amount of federal spending in billions of dollars for agriculture for several years. \(\begin{array}{ll}{\text { Year }} & {\text { Agriculture Spending(dollar billion) }} \\ {1990} & {12.0} \\ {1995} & {9.0} \\ {1999} & {23.0} \\\ {2000} & {26.6} \\ {2001} & {26.4} \\ {2002} & {22.0} \\ {2003} & {2003}\end{array}\) (a) Let \(x=0\) represent \(1990, x=1\) represent \(1991,\) and so forth. Make a scatter plot of the data. (b) Let \(P\) represent the point corresponding to \(2003, Q_{1}\) the point corresponding to \(2000, Q_{2}\) the point corresponding to \(2001,\) and \(Q_{3}\) the point corresponding to \(2002 .\) Find the slope of the secant line \(P Q_{i}\) for \(i=1,2,3 .\)

Step-by-Step Solution

Answer

The slopes of the lines \( PQ_1 \), \( PQ_2 \), and \( PQ_3 \) are 658.8, 988.3, and 1981, respectively.

Step 1: Make a Scatter Plot

First, you will plot these data points on a scatter plot. The year (with 1990 represented as 0, 1991 as 1, and so forth) will be the x-values and the Agriculture Spending in billions of dollars will be the y-values.

Step 2: Identify relevant points

We have to find the points referred to for each of \( Q_1 \), \( Q_2 \), \( Q_3 \), and \( P \). \nFor \( Q_1 \), corresponding to the year 2000 (or \( x = 10 \)), we find the point \( (10,26.6) \). \nSimilarly, for \( Q_2 \), corresponding to 2001 (or \( x = 11 \)), we find \( (11,26.4) \), and for \( Q_3 \) corresponding to 2002 (or \( x = 12 \)), we find \( (12,22.0) \).\nFinally, for \( P \), corresponding to 2003 (or \( x = 13 \)), we find \((13,2003) \).

Step 3: Calculate the Slope

Calculate the slope between P and each Qi for i = 1,2,3, using the formula for the slope between two points (x1, y1) and (x2, y2), which is \((y2-y1)/(x2-x1)\). \nFor the line \( PQ_1 \): slope = \( (2003 - 26.6) / (13 - 10) = 658.8 \) \nFor the line \( PQ_2 \): slope = \( (2003 - 26.4) / (13 - 11) = 988.3 \) \nFor the line \( PQ_3 \): slope = \( (2003 - 22) / (13 - 12) = 1981 \)