The book The Closing o/the American Mind byAllan Bloom was published in 1987 and spent many weeks on the bestseller list. In the book, Mr. Bloom recalls being in a restau
rant in France and overhearing a waiter call another waiter a "Cartesian." He goes on to say that French people today define themselves in terms of the philosophy of either Rene Descartes (1595-1650) or Blaise Pascal ( 1623-1662). Followers of Descartes are sometimes referred to as Cartesians. As a philosopher, Descartes is responsible for the statement "I think, therefore I am." In mathematics, Descartes is credited with, among other things, the invention of the rectangular coordinate system, which we sometimes call the Cartesian coordinate system. Until Descartes invented his coordinate system in 1637, algebra and geometry were treated as separate subjects. The rectangular coordi
nate system allows us to connect algebra and geometry by associating geometric shapes with algebraic equations. For example, every nonvertical straight line (a geometric con
cept) can be paired with an equation of the form y = rnx + b (an algebraic concept), where m and b are real numbers, andx and yare variables that we associate with the axes of a coordinate system. In this section we wi1l review some of the concepts developed around the rectangular coordinate system and graphing in two dimensions.
The rectangular (or Cartesian) coordinate system is shown in Figure 1. The axes di
vide the plane into four quadrants that are numbered I through IV in a counterclockwise direction. Looking at Figure 1, we see that any point in quadrant I will have both coor
dinates positive; that is, ( +, +). In quadrant II, the form is ( -, +). In quadrant ill, the form is ( -, -), and in quadrant IV it is ( +, -). Also, any point on the x-axis will have
y
Figure 1
- -
•
NOTE Example 1 illustrates the connection between algebra and geometry that we mentioned in the introduction to this section. The rectangular coordinate system allows us to associate the equation
y=
(an algebraic concept) with a specific straight line (a geometric concept). The study of the relationship between equations in algebra and their associated geometric figures is called analytic geometry and is based on the coordinate system eredited to Descartes.
Section 1.2 The Rectangular Coordinate System
a y-coordinate of 0 (it ha.'l no vertical displacement), and any point on the y-axis will have an x-coordinate of 0 (no horizontal displacement).
Graphing Lines
Graph the line y = ~x.
SOLUTION Because the equation of the line is written in slope-intercept form, we see that the slope of the line is ~ 1.5 and the y-intercept is O. To graph the line, we begin at the origin and use the slope to locate a second point. For every unit we tnl
verse to the right, the line will rise 1.5 units. Ifwe trnverse 2 units to the right, the line will rise 3 units, giving us the point (2, 3). Or, if we traverse 3 units to the right, the line will rise 4.5 units yielding the point (3, 4.5). The graph of the line is shown in Figure 2.
y
OIl' I : ' ! ' ' . x
Figure 2 •
Notice in Example 1 that the points (2, 3) and (3, 4.5) create two similar right tri
angles whose corresponding sides are in proportion. That is, 4.5 3
3 2
In general, for any point (x, y) on the line other than the origin, the ratio y/x will always be equal to the ratio ~, which is the slope of the line.
USING :~
We can use a graphing calculator to verify that for any point (other than the origin) on the graph of the line y ~x, the ratio of the y-coordinate to the x-coordinate will always be equivalent to the slope of ~, or 1.5 as a decimal.
Define this function as Y 1 3X/2. To match the graph shown in Figure 3, set the window variables so that -6 x::; 6 and -6::; y ::; 6. (By this, we mean that Xmin = -6, Xmax 6, Ymin = -6, and Ymax 6. We will assume that the scales for both axes, Xsc1 and Ysc1, are set to I unless noted otherwise.) Use the TRACE feature to move the cursor to any point on the line other than the origin itself
I Chapter 1 The Six Trigonometric Functions
NOTE Although quadratic functions are not pertinent to a study of trigonometry, this example introdnces the Human Cannonball theme that runs throughout the text and lays the foundation for problems that will follow in later sections, where trigonometric concepts are used.
Figure 3 Figure 4
(Figure 3). The current coordinates are stored in the variables x and y. Ifwe check the ratio ylx, the result should be the slope of 1.5 as shown in Figure 4. Try this for several different points to see that the
Graphing Parabolas
Recall from your algebra classes that any parabola that opens up or down can be de
scribed by an equation of the form
y = a(x h)2 + k
Likewise, any equation of this form will have a graph that is a parabola. The highest or lowest point on the parabola is called the vertex. The coordinates of the vertex are (h, k). The value of a determines how wide or narrow the parabola will be and whether it opens upward or downward.
At the 1997 Washington County Fair in Oregon, David Smith, Jr., The Bullet, was shot from a cannon. As a human cannonball, he reached a height of 70 feet before landing in a net 160 feet from the cannon. Sketch the graph of his path, and then find the equation of the graph. .
SOLUTION We assume that the path taken by the human cannonball is a parabola.
Ifthe origin of the coordinate system is at the opening of the cannon, then the net that catches him will be at 160 on the x-axis. Figure 5 shows a graph of this path.
y
Figure 5
Because the curve is a parabola we know that the equation will have the form y a(x h)2 + k
•
Section 1.2 The Rectangular Coordinate System
Because the vertex of the parabola is at (80, 70), we can fill in two of the three con
stants in our equation, giving us
y = a(x 80)2 + 70
To find a we note that the landing point will be (160, 0). Substituting the coordinates of this point into the equation, we solve for a.
0= a(l60 - 80? + 70
o= a(80)2 + 70
o 6400a + 70
70 7
a 6400 640
The equation that describes the path of the human cannonball is
7 •
Y = - 640 (x - 80)2 + 70 for 0 :'S X s 160
, -\
USIN<I'''~ >#
80 To verify that the equation from Example 2 is correct, we can graph the parabola and 'check the vertex and the x-intercepts. Graph the equation using the window settings
shown below.
o s X :'S 180, scale = 20; 0 :'S Y :'S 80, scale 10
Use the appropriate command on your calculator to find the maximum point on the graph (Figure 6), which is the vertex. Then evaluate the function at x 0 and again
180 at x 160 to verify the x-intercepts (Figure 7).
Note There are many different models of graphing calculators, and each model has its own set of commands. For example, to perform the previous steps on a TI-84 we would press I2nd II CALC Iand use the maximum and value commands. On a TI-86, it is the FMAX and EVAL commands found in the IGRAPH Imenu. Because we have no way of knowing which model of calculator you are working with, we will generally avoid providing specific key icons or command names throughout the remainder of this book. Check your calculator manual to find the appropriate command for your particular model.
180 Figure 6
Figure 7
The Distance Formula
Our next definition gives us a formula for finding the distance between any two points on the coordinate system.
THE DISTANCE FORMULA
The distance between any two points (x}, Yl) and (X2, Y2) in a rectangular coor
dinate system is given by the formula
r v'(X2 - Xl)2 + (yz yd2
c
I Chapter 1
[]
y
c
[
~~---~X
r =.J(X2 - Xl)2 + (Y2 - YJ)2
Figure 8
y
-I'---'---x Figure 10
y
Figure 11
The Six Trigonometric Functions
The distance formula can be derived by applying the Pythagorean Theorem to the right triangle in Figure 8. Because r is a distance, r 2:: O.
Find the distance between the points ( - I, 5)ã and (2, 1).
SOLUTION It makes no difference which of the points we call (Xl. YI) and which we call (X2, yz) because this distance will be the same between the two points regard
less (Figure 9).
r = Y'"(2---(--1-))-2-+-(-1---5-)Z y
V9+16
=Y2s
=5 x
Figure 9 •
Find the distance from the origin to the point (x, y).
SOLUTION The coordinates of the origin are (0, 0). As shown in Figure 10, applying the distance formula, we have
r = V(x - 0)2 + (y - 0)2
Vx2 + y2 •
Circles
Because of their perfect symmetry, circles have been used for thousands of years in many disciplines, including art, science, and religion. One example is Stonehenge, a 4,500-year-old site in England. The arrangement of the stones is based on a circular plan that is thought to have both religious and astronomical significance.
A circle is defined as the set of all points in the plane that are a fixed distance from a given fixed point. The fixed distance is the radius of the circle, and the fixed point is called the center. Ifwe let r> 0 be the radius, (h, k) the center, and (x,y) rep
resent any point on the circle, then (x, y) is r units from (h, k) as Figure 11 illustrates.
Applying the distance formula, we have
Squaring both sides of this equation gives the formula for a circle.
•
e
I.
~
Section 1.2 The Rectangular Coordinate System
EQUATION OF CIRCLE
The equation of a circle with center (h, k) and radius r > 0 is given by the formula
(x - hf + (y kf = il-
Ifthe center is at the origin so that (h, k) = (0, 0), this simplifies to .x2 + y2 = il-
Verify that the points ( ~, ) and (-~, -i)
both lie on a circle of radius 1 centered at the origin.
SOLUTION Because r = 1, the equation of the circle is + l 1. We check
each point by showing that the coordinates satisfy the equation.
1 _1_ v3 1
If x= v2 andy v2 If x= - 2 andy -2"
1)2 (_1)2 2
Then .x2+ l =(v2 + v2 Then x + y2 ( - ~ r+ ( -i r
1 1 3 1
=2"+2" 4+4