CS 001 - PowerPoint Presentation

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CS

001

CALCULUS AND ANALYTIC GEOMETRY

LECTURE 01Slide2

CS

001

CALCULUS AND ANALYTIC GEOMETRY

Class Presence

10%

Quiz 10%Assignment 20%Midterm 20%Final 40%Total 100%

Instructor : Somchai ThangsathityangkulYou can download lecture note at http://kbucomsci.weebly.com/

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Rectangular Coordinate Systems

We shall now show how to assign an ordered pair (a, b) of real numbers to each point in a plane.

Although we have also used the notation (a, b) to denote an open interval, there is little chance for confusion, since it should always be clear from our discussion whether (a, b) represents a point or an interval.3Slide4

We introduce a

rectangular, or Cartesian, coordinate system in a plane by means of two perpendicular coordinate lines, called coordinate axes, that intersect at the origin O, as shown in Figure 1.

Figure 1Rectangular Coordinate Systems4Slide5

We often refer to the horizontal line as the

x-axis and the vertical line as the y-axis and label them x and y, respectively.The plane is then a coordinate plane, or an xy-plane. The coordinate axes divide the plane into four parts called the first, second, third, and

fourth quadrants, labeled I, II, III, and IV, respectively (see Figure 1). Points on the axes do not belong to any quadrant. Each point P in an xy-plane may be assigned an ordered pair (a, b), as shown in Figure 1.

Rectangular Coordinate Systems

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We call

a the x-coordinate (or abscissa) of P, and b they-coordinate (or ordinate). We say that P has coordinates (a, b) and refer

to the point (a, b) or the point P (a, b).Conversely, every orderedpair (a

,

b

) determines a point P with coordinates a and b. Weplot a point by using a dot, as illustrated in Figure 2.Figure 2Rectangular Coordinate Systems6Slide7

Distance between two points

The length of a horizontal line segment is the abscissa (x coordinate) of the point on the right minus the abscissa (x coordinate) of the point on the left. Distance D = X2 – X1

Rectangular Coordinate Systems7Slide8

Distance between two points

The length of a vertical line segment is the ordinate (y coordinate) of the upper point minus the ordinate (y coordinate) of the lower point. Distance D = y2 – y1

Rectangular Coordinate Systems8Slide9

To determine the distance between two points of a slant line segment add the square of the difference of the abscissa to the square of the difference of the ordinates and take the positive square root of the sum.

Rectangular Coordinate Systems

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Rectangular Coordinate Systems10Slide11

Example 1 –

Finding the distance between pointsPlot the points A(–3, 6) and B(5, 1), and find the distance d (A, B).

Solution:The points are plotted in Figure 4.Figure 411Slide12

By the distance formula,

d (A, B) = = =

=  9.43.

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The inclination of the line, L, (not parallel to the x-axis) is defined as the smallest positive angle measured from the positive direction of the x-axis or the counterclockwise direction to L. The slope of the line is defined as the tangent of the angle of inclination.

INCLINATION AND SLOPE OF A LINESlide14
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If two lines are parallel their slope are equal. If two lines are perpendicular the slope of one of the line is the negative reciprocal of the slope of the other line.

If m1 is the slope of L1 and m2

is the slope of L2 then, or m1m2 = -1.

PARALLEL AND PERPENDICULAR LINESSlide17

Example 1 –

Finding slopesSketch the line through each pair of points, and find its slope m:

(a) A(–1, 4) and B(3, 2) (b) A(2, 5) and B(–2, –1)(c) A(4, 3) and B(–2, 3)

(d)

A(4, –1) and B(4, 4)17Slide18

The lines are sketched in Figure 3. We use the definition

of slope to find the slope of each line.Figure 3(a)Figure 3(b)

m = m =

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Figure 3(c)

Figure 3(d)m = 0 m undefined

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(a)

(b)(c)

(d) The slope is undefined because the line is parallel to the y-axis. Note that if the formula for m is used, the denominator is zero.

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LINES

Let us next find an equation of a line l through a point

P1(x1, y1) with slope m. If P (x, y) is any point with x  x1 (see Figure 7), then P is on

l

if and only if the slope of the line through

P1 and P is m—that is, ifFigure 721Slide22

This equation may be written in the form

y – y1 = m(x – x1).Note that (

x1, y1) is a solution of the last equation, and hence the points on l are precisely the points that correspond to the solutions. This equation for l is referred to as the point-slope form.

LINES

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Example 4 –

Finding an equation of a line through two pointsFind an equation of the line through A(1, 7) and B

(–3, 2).Solution:The line is sketched in Figure 8.Figure 823Slide24

The formula for the slope

m gives usWe may use the coordinates of either A or B for (x1, y

1) in the point-slope form. Using A(1, 7) gives us the following: y – 7 = (x – 1) 4(y – 7) = 5(x – 1)

point-slope form

multiply by 4

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4

y – 28 = 5x – 5 –5x + 4y = 23 5x – 4y

= –23The last equation is one of the desired forms for an equation of a line. Another is 5x – 4y + 23 = 0.multiply by –1

subtract 5

x

and add 28multiply factors25Slide26

The point-slope form for the equation of a line may be rewritten as

y = mx – mx1 + y1, which is of the form y = mx + b

with b = –mx1 + y1. The real number b is the y-intercept of the graph, as indicated in Figure 9. Figure 9

LINES

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Example 5 –

Expressing an equation in slope-intercept formExpress the equation 2x – 5y

= 8 in slope-intercept form.Solution:Our goal is to solve the given equation for y to obtain the form y = mx + b.

We may proceed as follows:

2

x – 5y = 8given27Slide28

–5

y = –2x + 8 y =

y =The last equation is the slope-intercept form y = mx + b with slope m = and y-intercept b =

.

equivalent equationdivide by –5subtract 2x28