Lecture co1 math 21-1

CALCULUS I( with Analytic Geometry)
MATH 21-1

Course Outcomes
1. Discuss comprehensively the fundamental concepts in Analytic
Geometry and use them to solve application problems and problems
involving lines.
2. Distinguish equations representing the circles and the conics; use the
properties of a particular geometry to sketch the graph in using the
rectangular or the polar coordinate system. Furthermore, to be able
to write the equation and to solve application problems involving a
particular geometry.
3. Discuss and apply comprehensively the concepts, properties and
theorems of functions, limits, continuity and the derivatives in
determining the derivatives of algebraic functions

4. Analyze correctly and solve properly application problems
concerning the derivatives to include writing equation of
tangent/normal line, curve tracing ( including all types of
algebraic curves and cusps), optimization problems, rate of
change and related-rates problems (time-rate problems).
5. Discuss comprehensively the concept and properties of the
transcendental functions ; to determine the derivatives and solve
application problems involving transcendental functions.

C
1 Mission and Vision of Mapua Institute of Technology
Orientation and Introduction to the Course
Discussion on COs, TLAs, and ATs of the course
Overview on student-centered learning and eclectic approaches to be used in the course.
Fundamental Concept of Analytic Geometry: Rectangular Coordinate System, Directed
Distance, Distance Formula
Division of Line Segment
Slope and Inclination of a Line
Angle Between Lines
Area of a Triangle/Polygon
Locus of a Moving Point
2
Normal Form of Equation of Line
Distance of Point from Line
Distance between Parallel Lines
Angle Bisector
Long Quiz 1 Coverage

Circle : Center at any point ( Include discussion on translation of axes)
3
CONICS: Properties and Application Involving the
Parabola, Ellipse and Hyperbola with
Vertex/ Center at any point with
Horizontal/Vertical/ Oblique Axis
4
Polar Curves and Parametric Curves; Sketching and Transformation to Rectangular forms of
equations

Limits: Definition and Concepts
Theorems One-Sided Limits
Limits of Functions
Infinite Limits and Limits at Infinity: Evaluation and Interpretation
Squeeze Theorem: Limits of Expression Involving Transcendental Functions
Continuity : Definition and Theorem
Types of Discontinuity;
Relationship between limits and
Discontinuity
The Derivative and Differentiability of a Function:
Definition and concept
Evaluation of the Derivative of a Function
based on Definition (Increment Method or
Four-Step Rule Method)
Derivatives of Algebraic Functions Using the Basic Theorems of Differentiation and the Chain Rule
Higher Order and Implicit Differentiation

Applications : Equations of Tangent and Normal Lines
Application of the Concepts of the Derivative and Continuity on Curve Tracing ( Include all types of the Algebraic
curves, cusps)
Optimization Problems: Applied Maxima/Minima Problems
Rate of Change Problems; Related-Rate Problems (Time-Rate Problems)

Derivatives of the Exponential and Logarithmic Functions with Applications
Derivatives of the Trigonometric and Inverse Trigonometric Functions with Application
The Hyperbolic and Inverse Hyperbolic Functions: Definition, Properties and Derivatives
Logarithmic Differentiation: Variable with Variable Exponent

Grading Matrix
Assessment Tasks Weight
(%)
Minimum Average for
Satisfactory
Performance (%)
CO 1
Diagnostic
Examination
10.0 7.00
CO 2
CPR 1 2.0 1.40
Classwork 1 1.0 0.70
Quiz 1 9.0 6.30
CO 3
CPR 2 2.0 1.40
Classwork 2 1.0 0.70
Quiz 2 10.0 7.00
CO 4
CPR 3 2.0 1.40
Class Presentation
and Critique 1
1.0 0.70
Quiz 3 10.0 7.00
CO 5
CPR 4 2.0 1.40
Class Presentation
and Critique 2
1.0 0.70
Project 5.0 3.50
Quiz 4 10.0 7.00
CO 6
CPR 5 2.0 1.40
Quiz 5 7.0 4.90
Summative Assessment:
Final Examination
25.0 17.5
TOTAL 100.0 70

CO1
Discuss comprehensively the fundamental
concepts in Analytic Geometry and use them to
solve application problems and problems
involving lines.

FUNDAMENTAL CONCEPTS OF ANALYTIC
GEOMETRY

Lesson 1: Rectangular Coordinate System,
Directed Distance, Distance Formula

OBJECTIVE:
At the end of the lesson, the students should be able to
illustrate properly and solve application problems involving
distance formula.

• Analytic Geometry – is the branch of mathematics, which
deals with the properties, behaviours, and solution of points,
lines, curves, angles, surfaces and solids by means of algebraic
methods in relation to a coordinate system(Quirino and
Mijares) .
• It is a unified algebra and geometry dealing with the study of
relationships between different geometric figures and
equations by means of the geometric properties and
processes of algebra in relation to a coordinate system (
Marquez, et al).
DEFINITION:
FUNDAMENTAL CONCEPTS

Two Parts of Analytic Geometry
1. Plane Analytic Geometry – deals with figures on a
plane surface (two-dimensional geometry, 2D).
2. Solid Analytic Geometry – deals with solid figures (
three-dimensional geometry, 3D).

Directed Line – a line in which one direction is chosen as
positive and the opposite direction as negative.
Directed Line Segment – portion of a line from one point
to another.
Directed Distance – the distance from one point to
another; may be positive or negative depending upon
which direction is denoted positive.
DEFINITION:

RECTANGULAR COORDINATES
A pair of number (x, y) in which x is the first and y the
second number is called an ordered pair. It defines the
position of a point on a plane by defining the directed
distances of the point from a vertical line and from a
horizontal line that meet at a point called the origin, O.
The x-coordinate of a point , known also as its abscissa, is
the directed distance of the point from the vertical axis, y-
axis; while the y-coordinate, also known as the ordinate, is
its directed distance from the horizontal axis, the x-axis.

DISTANCE BETWEEN TWO POINTS
The horizontal distance between any two points is the
difference between the abscissa (x-coordinate) of the
point on the right minus the abscissa (x-coordinate) of
the point on the left; that is,
Horizontal Distance Between Points
Distance, d=xright -xleft

Vertical Distance Between Any Two Points
The vertical distance between any two points is the
difference between the ordinate (y-coordinate) of the
upper point minus the ordinate (y-coordinate) of the
lower point; that is,
Distance d = yupper - ylower

Distance Between Any Two Points on a Plane
The distance between any two points on a plane
is the square root of the sum of the squares of the
difference of the abscissas and of the difference
of the ordinates of the points. That is, if
distance d = x2 - x1( )
2
+ y2 - y1( )
2
P1 x1, y1( ) and P2 x2, y2( ) are the points, then

SAMPLE PROBLEMS
1. By addition of line segments verify whether the points A ( - 3, 0 ) ,
B(-1, -1) and C(5, -4) lie on a straight line.
2. The vertices of the base of an isosceles triangle are at (1, 2) and
(4, -1). Find the ordinate of the third vertex if its abscissa is 6.
3. Find the radius of a circle with center at (4, 1), if a chord of length 4
is bisected at (7, 4).
4. Show that the points A(-2, 6), B(5, 3), C(-1, -11) and D(-8, -8) are the
vertices of a rectangle.
5. The ordinate of a point P is twice the abscissa. This point is
equidistant from (-3, 1) and (8, -2). Find the coordinates of P.
6. Find the point on the y-axis that is equidistant from (6, 1) and (-2, -3).

Lesson 2: DIVISION OF A LINE SEGMENT

OBJECTIVE:
At the end of the lesson, they students should be able
to illustrate properly and solve problems involving division of
line segments.

Let us consider a line segment bounded by the points
. This line segment can be subdivided
in some ratio and the point of division can be determined. It
is also possible to determine terminal point(s) whenever the
given line segment is extended beyond any of the given
endpoints or beyond both endpoints . If we consider the
point of division/ terminal point to be P (x, y ) and define the
ratio, r, to be
then the coordinates of point P are given by:
P1 x1, y1( ) and P2 x2, y2( )
r =
P1P
®
P1P2
®
x = x1 +r x2 - x1( )
y = y1 +r y2 - y1( )

If the line segment is divided into two equal parts, then the
point of division is called the midpoint. The ratio, r, is equal
to ½ and the coordinates of point P are given by:
or simply by:
x = x1 +
1
2
x2 - x1( )
y = y1 +
1
2
y2 - y1( )
x =
1
2
x1 + x2( )
y =
1
2
y1 + y2( )

SAMPLE PROBLEMS
1. Find the midpoint of the segment joining (7, -2) and (-3, 5).
2. The line segment joining (-5, -3) and (3, 4) is to be divided into five equal
parts. Find all points of division.
3. The line segment from (1, 4) to (2, 1) is extended a distance equal to twice
its length. Find the terminal point.
4. On the line joining (4, -5) to (-4, -2), find the point which is three-seventh
the distance from the first to the second point.
5. Find the trisection points of the line joining (-6, 2) and (3, 8).
6. Show that the points ( 0, -5), (3, -4), ( 8, 0) and ( 5, -1) are vertices of a
parallelogram.
7. What are the lengths of the segments into which the y-axis divided the
segment joining ( -6, -6) and (3, 6)?
8. The line segment joining a vertex of a triangle and the midpoint of the
opposite side is called the median of the triangle. Given a triangle whose
vertices are A(4,-4), B(10, 4) and C(2, 6), find the point on each median
that is two-thirds of the distance from the vertex to the midpoint of the
opposite side.

Lesson 3: INCLINATION AND SLOPE A LINE

OBJECTIVES:
At the end of the lesson, the students should be able
to use the concept of angle of inclination and slope of a line
to solve application problems.

INCLINATION AND SLOPE OF A LINE
The angle of inclination of the line L or simply
inclination , denoted by , is defined as the smallest
positive angle measured from the positive direction of
the x-axis to the line.
The slope of the line, denoted by m , is defined
as the tangent of the angle of inclination; that is,
And if two points are points on
the line L then the slope m can be defined as
a
m = tana
m = tana =
y2 - y1
x2 - x1
P1 x1, y1( ) and P2 x2, y2( )

PARALLEL AND PERPENDICULAR LINES
If two lines are parallel their slope are
equal. If two lines are perpendicular, the slope of
one 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
Sign Conventions:
Slope is positive (+), if the line is leaning to the right.
Slope is negative (-), if the line is leaning to the left.
Slope is zero (0), if the line is horizontal.
Slope is undefined , if the line is vertical.
m1m2 = -1.

SAMPLE PROBLEMS
1. Find the slope, m, and the angle of inclination of the
line through the points (8, -4) and (5, 9).
2. The line segment drawn from (x, 3) to (4, 1) is
perpendicular to the segment drawn from (-5, -6) to
(4, 1). Find the value of x.
3. Show that the triangle whose vertices are A(8, -4),
B(5, -1) and C(-2,-8) is a right triangle.
4. Find y if the slope of the line segment joining (3, -2)
to (4, y) is -3.
5. Show that the points A(-1, -1), B(-1, -5) and C(12, 4)
lie on a straight line.

ANGLE BETWEEN TWO INTERSECTING LINES


L1
L2
tanq =
m2 - m1
1+ m1m2
Where: m1 = slope of the initial side
m2 = slope of the terminal side
The angle between two intersecting lines is the positive angle
measured from one line (L1) to the other ( L2).
0
180:note  

Sample Problems
1. Find the angle from the line through the points (-1,
6) and (5, -2) to the line through (4, -4) and (1, 7).
2. The angle from the line through (x, -1) and (-3, -5)
to the line through (2, -5) and (4, 1) is 450 . Find x.
3. Two lines passing through (2, 3) make an angle of
450 with one another. If the slope of one of the lines
is 2, find the slope of the other.
4. Find the interior angles of the triangle whose
vertices are A (-3, -2), B (2, 5) and C (4, 2).

AREA OF A POLYGON BY COORDINATES
Consider the triangle whose vertices are P1(x1, y1), P2(x2, y2)
and P3(x3, y3) as shown below. The area of the triangle can
be determined on the basis of the coordinates of its
vertices.
o
y
x
 111 y,xP
 222 y,xP
 333 y,xP

Label the vertices counterclockwise and evaluate the area
of the triangle by:
1yx
1yx
1yx
2
1
A
33
22
11

The area is a directed area. Obtaining a negative value
will simply mean that the vertices were not named
counterclockwise. In general, the area of an n-sided
polygon can be determined by the formula :
1n54321
1n54321
yy..yyyyy
xx..xxxxx
2
1
A 

SAMPLE PROBLEMS
1. Find the area of the triangle whose vertices are (-
6, -4), (-1, 3) and (5, -3).
2. Find the area of a polygon whose vertices are (6, -
3), (3, 4), (-6, -2), (0, 5) and (-8, 1).
3. Find the area of a polygon whose vertices are (2, -
3), (6, -5), (-4, -2) and (4, 0).

OBJECTIVE:
At the end of the lesson, the students should be able to
determine the equation of a locus defining line, circle and
conics and other geometries defined by the given condition.

EQUATION OF A LOCUS
An equation involving the variables x and y is usually
satisfied by an infinite number of pairs of values of x and y,
and each pair of values corresponds to a point. These points
follow a pattern according to the given equation and form a
geometric figure called the locus of the equation.
Since an equation of a curve is a relationship satisfied by
the x and y coordinates of each point on the curve (but by
no other point), we need merely to consider an arbitrary
point (x,y) on the curve and give a description of the curve in
terms of x and y satisfying a given condition.

Sample Problems
1. Find an equation for the set of all points (x, y)
satisfying the given conditions.
2. It is equidistant from (5, 8) and (-2, 4).
3. The sum of its distances from (0, 4) and (0, -4) is 10.
4. It is equidistant from (-2, 4) and the y-axis
5. It is on the line having slope of 2 and containing the
point (-3, -2).
6. The difference of its distances from (3, 0) and (-3, 0) is
2.

Lesson 5: STRAIGHT LINES / FIRST DEGREE
EQUATIONS

OBJECTIVE:
At the end of the lesson, the students is
should be able to write the equation of a line in the
general form or in any of the standard forms; as well
as, illustrate properly and solve application
problems concerning the normal form of the line.

STRAIGHT LINE
A straight line is the locus of a point that
moves in a plane in a constant slope.
Equation of Vertical/ Horizontal Line
If a straight line is parallel to the y-axis (
vertical line ), its equation is x = k, where k is the
directed distance of the line from the y-axis.
Similarly, if a line is parallel to the x-axis (
horizontal line ), its equation is y = k, where k is
the directed distance of the line from the x-axis.

General Equation of a Line
A line which is neither vertical nor horizontal
is defined by the general linear equation
Ax + By + C=0,
where A and B are nonzeroes.
The line has y-intercept of and slope of .-
C
B
-
A
B

DIFFERENT STANDARD FORMS OF THE
EQUATION OF A STRAIGHT LINE
A. POINT-SLOPE FORM:
If the line passes through the points ( x , y) and (x1, y1), then
the slope of the line is .
Rewriting the equation we have
which is the standard equation of the point-slope form.
1
1
xx
yy
m



 11 xxmyy 

B. TWO-POINT FORM:
If a line passes through the points (x1, y1) and (x2, y2), then
the slope of the line is .
Substituting it in the point-slope formula will result to
the standard equation of the two-point form.
12
12
xx
yy
m



 1
12
12
1 xx
xx
yy
yy 









C. SLOPE-INTERCEPT FORM:
Consider a line containing the point P( x, y) and not parallel
to either of the coordinate axes. Let the slope of the line be m
and the y-intercept ( the intersection point with the y-axis) at
point (0, b), then the slope of the line is .
Rewriting the equation, we obtain
the standard equation of the slope-intercept form.
0x
by
m



bmxy 

D. INTERCEPT FORM:
Let the intercepts of a line be the points (a, 0), the x-
intercept, and (0, b), the y-intercept. Then the slope of the
line is defined by .
Using the Point-slope form, the equation is written as
or simply as
the standard equation of the Intercept Form.
a
b
m 
 0x
a
b
by 
1
b
y
a
x


E. NORMAL FORM:
Suppose a line L, whose equation is to be found, has its
distance from the origin to be equal to p. Let the angle of
inclination of p be .
Since p is perpendicular to L, then the slope of p is equal to the
negative reciprocal of the slope of L,
Substituting in the slope-intercept form y = mx + b , we obtain
or
the normal form of the straight line




 sin
cos
mor,cot
tan
1
m 


sin
p
x
sin
cos
y 
py sincosx  

Reduction of the General Form to the Normal Form
The slope of the line Ax+By+C=0 is . The slope of p which is
perpendicular to the line is therefore ; thus, .
From Trigonometry, we obtain the values
and .
If we divide the general equation of the straight line by
, we have
or
This form is comparable to the normal form .
Note: The radical takes on the sign of B.
B
A

A
B
A
B
tan 
22
BA
B
sin


22
BA
A
cos


22
BA  0
BA
C
y
BA
B
x
BA
A
222222






BA
C
y
BA
B
x
BA
A
222222






py sincosx  

PARALLEL AND PERPENDICULAR LINES
Given a line L whose equation is Ax + By + C = 0.
The line Ax + By + K = 0 , for any constant K not equal
to C, is parallel to L;
and the Bx – Ay + K = 0 is perpendicular to L.

DIRECTED DISTANCE FROM A LINE TO A POINT
The directed distance of the point P(x1, y1) from the
line Ax + By + C = 0 is ,
where the radical takes on the sign of B.
22
11
BA
CByAx
d




y
x


 111 y,xP
 222 y,xP
0CByAx 11 
0d1 
0d2 
linethebelowis
pointthe0,dif
linetheaboveis
pointthe0,dif
:note



Sample Problems
1. Determine the equation of the line passing through (2, -3) and
parallel to the line through (4,1) and (-2,2).
2. Find the equation of the line passing through point (-2,3) and
perpendicular to the line 2x – 3y + 6 = 0
3. Find the equation of the line, which is the perpendicular bisector
of the segment connecting points (-1,-2) and (7,4).
4. Find the equation of the line whose slope is 4 and passing through
the point of intersection of lines x + 6y – 4 = 0 and 3x – 4y + 2 = 0.
5. The points A(0, 0), B(6, 0) and C(4, 4) are vertices of triangles. Find:
a. the equations of the medians and their intersection point
b. the equations of the altitude and their intersection point
c.the equation of the perpendicular bisectors of the sides and their
intersection points

6. Find the distance from the line 5x = 2y + 6 to the points
a. (3, -5)
b. (-4, 1)
c. (9, 10)
7. Find the equation of the bisector of the acute angles and also the
bisector of the obtuse angles formed by the lines x + 2y – 3 = 0 and 2x + y –
4 = 0.
8. Determine the distance between the lines:
a. 2x + 5y -10 =0 ; 4x + 10y + 25 = 0
b. 3x – 4y + 25 =0 ; 3x – 4y + 45 = 0
9. Write the equation of the line a) parallel to b) perpendicular to 4x + 3y -
10 = 0 and is 3 units from the point ( 2, -1).
10. A line passes through ( 2, 2) and forms with the axes a triangle of area 9
sq. units. What is the equation of the line?

REFERENCES
Analytic Geometry, 6th Edition, by Douglas F. Riddle
Analytic Geometry, 7th Edition, by Gordon Fuller/Dalton Tarwater
Analytic Geometry, by Quirino and Mijares
Fundamentals of Analytic Geometry by Marquez, et al.
Algebra and Trigonometry, 7th ed by Aufmann, et al.

Lecture co1 math 21-1

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Lecture co1 math 21-1