Additional Mathematic 2011

7/31/2019 Additional Mathematic 2011

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PART 1


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Descartes and His Coordinate System

Every time you graph an equation on a Cartesian coordinate system, you are using the work of

Ren Descartes. Descartes, a French mathematician and philosopher, was born in La Haye,

France (now named in his honor) on March 31, 1596. His parents taught him at home until he

was 8 years old, when he entered the Jesuit college of La Flche. There he continued his studiesuntil he graduated at age 18.

Descartes was an outstanding student at La Flche, especially in mathematics. Because of his

delicate health, his teachers allowed him to stay in bed until late morning. Despite missing most

of his morning classes, Descartes was able to keep up with his studies. He would continue thehabit of staying late in bed for his entire adult life.

After graduating from La Flche, Descartes traveled to Paris and eventually enrolled at theUniversity of Poitiers. He graduated with a law degree in 1616 and then enlisted in a military

school. In 1619, he joined the Bavarian army and spent the next nine years as a soldier, touring

throughout much of Europe in between military campaigns. Descartes eventually settled inHolland, where he spent most of the rest of his life. There Descartes gave up a military career

and decided on a life of mathematics and philosophy.

Descartes attempted to provide a philosophical foundation for the new mechanistic physics that

was developing from the work of Copernicus and Galileo. He divided all things into twocategoriesmind and matterand developed a dualistic philosophical system in which,although mind is subject to the will and does not follow physical laws, all matter must obey the

same mechanistic laws.

The philosophical system that Descartes developed, known as Cartesian philosophy, was based

on skepticism and asserted that all reliable knowledge must be built up by the use of reasonthrough logical analysis. Cartesian philosophy was influential in the ultimate success of the

Scientific Revolution and provides the foundation upon which most subsequent philosophicalthought is grounded.

Descartes published various treatises about philosophy and mathematics. In 1637 Descartes

published his masterwork,Discourse on the Method of Reasoning Well and Seeking Truth in theSciences. InDiscourse, Descartes sought to explain everything in terms of matter and motion.

Discourse contained three appendices, one on optics, one on meteorology, and one titledLa

Gometrie (The Geometry). InLa Gometrie, Descartes described what is now known as the

system of Cartesian Coordinates, or coordinate geometry. In Descartes's system of coordinates,

geometry and algebra were united for the first time to create what is known as analytic

geometry.


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The Cartesian Coordinate System

Cartesian coordinates are used to locate a point in space by giving its relative distance from

perpendicular intersecting lines. In coordinate geometry, all points, lines, and figures are drawnin a coordinate plane. By reference to the two coordinate axes, any point, line, or figure may beprecisely located.

In Descartes's system, the first coordinate value (x-coordinate) describes where along thehorizontal axis (thex-axis) the point is located. The second coordinate value (y-coordinate)

locates the point in terms of the vertical axis (the y-axis). A point with coordinates (4, 2) islocated four units to the right of the intersection point of the two axes (point O, or the origin) and

then two units below the vertical position of the origin. In example (a) of the figure, point D is at

the coordinate location (4, 2). The coordinates for point A are (3, 2); for point B, (2, 4); and

for point C, (2, 5).

The coordinate system also makes it possible to exactly duplicate geometric figures. For

example, the triangle shown in (b) has coordinates A (3, 2), B (4, 5), and C (2, 4) that make it

possible to duplicate the triangle without reference to any drawing.

The triangle may be reproduced by using the coordinates to locate the position of the three

vertex points. The vertex points may then be connected with segments to replicate triangle ABC.More complex figures may likewise be described and duplicated with coordinates.

A straight line may also be represented on a coordinate grid. In the case of a straight line, everypoint on the line has coordinate values that must

satisfy a specific equation. The line in (c) may be expressed as y = 2x. The coordinates of everypoint on the line will satisfy the equationy = 2x, as for example, point A (1, 2) and point B (2, 4).

More complex equations are used to represent circles, ellipses, and curved lines.


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Example of Cartesian Cordinates :

Cartesian coordinate system with a circle of radius 2 centered at the origin marked in red. Theequation of a circle is (x - a)

2 + (y - b)2 = r

2 where a and b are the coordinates of the center (a, b)

and ris the radius.

The invention of Cartesian coordinates in the 17th century byRen Descartes(Latinizedname:

Cartesius) revolutionized mathematics by

Illustration of a Cartesian coordinate plane. Four points are marked and labeled with their

coordinates: (2, 3) in green, (3, 1) in red, (1.5, 2.5) in blue, and the origin (0, 0) in purple.

2,3
http://en.wikipedia.org/wiki/Ren%C3%A9_Descarteshttp://en.wikipedia.org/wiki/Ren%C3%A9_Descarteshttp://en.wikipedia.org/wiki/Ren%C3%A9_Descarteshttp://en.wikipedia.org/wiki/Latinisation_(literature)http://en.wikipedia.org/wiki/Latinisation_(literature)http://en.wikipedia.org/wiki/Latinisation_(literature)http://en.wikipedia.org/wiki/File:Cartesian-coordinate-system.svghttp://en.wikipedia.org/wiki/File:Cartesian-coordinate-system-with-circle.svghttp://en.wikipedia.org/wiki/File:Cartesian-coordinate-system.svghttp://en.wikipedia.org/wiki/File:Cartesian-coordinate-system-with-circle.svghttp://en.wikipedia.org/wiki/File:Cartesian-coordinate-system.svghttp://en.wikipedia.org/wiki/Latinisation_(literature)http://en.wikipedia.org/wiki/Ren%C3%A9_Descartes


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Other Contributions

La Gometrie made Descartes famous throughout Europe. He continued to publish hisphilosophy, detailing how to acquire accurate knowledge. His philosophy is sometimes summed

up in his statement, "I think, therefore I am."

Descartes also made a number of other contributions to mathematics. He discovered the Law of

Angular Deficiency for all polyhedrons and was the first to offer a quantifiable explanation ofrainbows. InLa Gometrie, Descartes introduced a familiar mathematics symbol, a raised

number to indicate an exponent. The expression 4 4 4 4 4 may be written as 45 using

Descartes's notation. He also instituted usingx,y, andz for unknowns in an equation.

In 1649, Descartes accepted an invitation from Queen Christina to travel to Sweden to be the

royal tutor. Unfortunately for Descartes, the queen expected to be tutored while she did her

exercises at 5:00 A.M. in an unheated library. Descartes had been used to a lifetime of sleepinglate, and the new routine was much too rigorous for him. After only a few weeks of this regimen,

Descartes contracted pneumonia and died on February 11, 1650.

GEOMETRY AND THE FLY

Some mathematics historians claim it may be that Descartes's inspiration for the coordinate

system was due to his lifelong habit of staying late in bed. According to some accounts, one

morning Descartes noticed a fly walking across the ceiling of his bedroom. As he watched the

fly, Descartes began to think of how the fly's path could be described without actually tracing its

path. His further reflections about describing a path by means of mathematics led to LaGometrie and Descartes's invention of coordinate geometry.

WHO USES COORDINATES?

The system of coordinates that Descartes invented is used in many modern applications. For

example, on any map the location of a country or a city is usually given as a set of coordinates.

The location of a ship at sea is determined by longitude and latitude, which is an application of

the coordinate system to the curved surface of Earth. Computer graphic artists create figures andcomputer animation by referencing coordinates on the screen.


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PART 2


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Malaysia with its warm tropical climate is rich in flora and fauna. Beautiful gardens are found all

over Malaysia. SMK Permata Decided to beautiful the school compound by getting the students

involved in the planting and maintenance of the greenery in the school compound as shown in

Diagram 1 . Each society is allocated a plot of land in various shapes and sizes to nurture

throughout the year. The Mathematics Society, English, Language Society and Malay LanguageSociety are allocated the region P, Q and R respectively as shown in Diagram 1.

a) Determine the area of region P , Q and R by using at least threedifferent methods including the use of calculus. Verify the answers obtained by using

computer software.

(Suggestions : GeoGebra, GSP, Graphing calculator etc )

Method 1 :

i) Divide area P into 3 segments, 2 triangle and 1 rectangle.Triangle 1 : Triangle 2 : Rectangle := x 2 x 1 = x 4 x3 = 2 x 4

= 1 = 6 = 8

P = ( 1+6+8)

P = 15 T 1T 2 R 1

Diagram 2


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ii) Divide area Q into 2 segments. 1 triangle and 1 rectangle.Triangle 1 : Rectangle 1 :

= x 4 x 3 = 2 x 3

= 6

= 6

Q = (6 + 6 )

Q = 12

iii) Divide area R into 3 segments. 1 triangle and 2 rectangles.Triangle 1 : Rectangle 1 : Rectangle 2 :

= x 1 x 2 = 2 x 3 = 4 x 2

= 1 = 6 = 8

R = ( 1 + 6 + 8 )

R = 15

R 1

R 1

T 1

R 2

T 1

Diagram 3

Diagram 4


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Method 2 :

Coordinate Geometry method.

Area P :

= | |

= I ( 0 +14 + 14 + 16 +0 ) ( 0 + 0 + 8 + 6 + 0 ) I

= I 30 I

= 15

(0, 0)

(4, 2)

(7, 2)

(0, 6) (3, 6)

(7, 6)

(3, 4)

(7, 0)0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8 9 10

QR

P

Graph 1


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Area Q :

= | |

= I ( 0 + 18 + 18 + 0 ) ( 0 + 12 + 0 +0 ) I

= I 24 I

= 12

(0, 0)

(4, 2)

(7, 2)

(0, 6) (3, 6)

(7, 6)

(3, 4)

(7, 0)0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8 9 10

P

QR

Graph 2


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Area R :

= | |= I ( 6 + 8 + 42 + 42 + 12) ( 18 + 18 + 14 + 14 + 16)

= I 30 I

= 15

(0, 0)

(4, 2)

(7, 2)

(0, 6) (3, 6)

(7, 6)

(3, 4)

(7, 0)0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8 9 10

P

Q R

Graph 3


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Area P

Graph 4

mAE = mED =

=

dx +

x + 10 dx +

x

mAE =

mED = -2 =

+ [ -x + 10] + [2x]

Equation AE: Equation ED: = ( 6 0 ) + ( 24 21 ) + ( 14 8 )

Y =x y = - 2x + 10 = 6 + 3 + 6

=15

A = (0, 0)

D = (4, 2)

C = (7, 2)

H = (0, 6) G = (3, 6)

F = (7, 6)

E = (3, 4)

B = (7, 0)0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8 9 10

3

0

4

3

7

4

p

QR


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Area Q

Graph 5

mAE =

Equation AE

Y=

X =

=

dy +

dy=

+ [3y]= 6 + 6

= 12

A = (0, 0)

D = (4, 2)

C = (7, 2)

H = (0, 6) G = (3, 6)

F = (7, 6)

E = (3, 4)

B = (7, 0)0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8 9 10

4

06

4

P

Q R


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Area R

Graph 6

= y -12-15

= [6x] - 27

= (42 0 ) 27

= 42 27

= 15

A = (0, 0)

D = (4, 2)

C = (7, 2)

H = (0, 6) G = (3, 6)

F = (7, 6)

E = (3, 4)

B = (7, 0)0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8 9 10

7

0

RQ

P


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Verification of answer by GeoGebra

Area P :

Graph 7

Area Q

E (6, 7)

D ( 7, 5)

C (10, 5)

H( 3, 9) G (6, 9) F (10, 9)

B( 10, 3)A (3,3)

e = 4

h = 2

E ( 6, 7)

D ( 7, 5)

C (10, 5)

H( 3, 9) G ( 6, 9) F (10, 9)

B(10, 3)A( 3,3)

e = 4

h = 2

f = 4

i = 5

d = 2.24

b = 2

a = 7

j = 6

g = 3

C = 3Poly = 15

b = 2

d = 2.24

i = 5

a =7

f = 4g = 3

j = 6

Poly =12

C = 3

P

Q R

Graph 8

Q

R

P


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Area R

Graph 9

E ( 6, 7)

D ( 7, 5)

C (10, 5)

H( 3, 9) G ( 6, 9) F (10, 9)

B(10, 3)A( 3,3)

e = 4

h = 2

b = 2

d= 2.24

i = 5

j = 6

a =7

g = 3f = 4

Poly = 15

C = 3

Q

P

R


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b) Suppose there is hedge along AB. The Mathematics Society wishes to fence up theremaining sides of the region P. Determine the length of fence required.

The remaining sides in the diagram is :

c) If a meter of fence costs RM 25.00, what is the total cost required by the MathematicsSociety to fence up region P ? It is possible for the society to carry out the fencing with

an allocation of RM 250.00. Explain your answer.

The society is impossible to carry out the fencing with an allocation of RM 250.00

The reason :

= RM 250.00 RM 25 per meter

= 10 m

RM 250.00 can only cover 10m of length.

Length of AEDCB = 12.24 m

The solution :

= 12. 24m x RM 25.00

RM 306 is needed to carry out the fencing plan

= + = 25AE = AE = 5 m

=[ ] + = + DE = DE = 2.24

CD = 3 m AB = 7cm

BC = ( 6 4 )

BC = 2 m

AEDCB = ( 2 + 3 + 2.24 + 5 ) m

AEDCB = 12.24 m


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d) During the mathematics Week, the society was given a single flag chain of length 9.20meters to be used completely. The President of the society wishes to tie the flag chain

continuously from A to B and then to another point along the hedge AB to crate

triangularshaped area.

i) Make conjecture about the number of points that the flag chain can be tied toalong AB. Support your conjecture with suitable calculations. Explain your

answer.

1 point for the flag chain to be tied at E

1 point for the flag chain to be tied at a point along the hedge AB.

2 points

ii)

Calculate the maximum area of the triangle obtained. Discuss.= 9.20 m AE

= 9.20 m 5 m

= 4.20 m

Divided the triangle obtained into 2 right triangle

Calculate the length.

Use Pythagoras Theorem.

Triangle 1 : Triangle 2 :

L = L = L = L = L = 3 L = 1.281 m

Area = x 3 x4 Area = x 4 x1.281

Area = 6 Area = 2.562

Solution of triangle method :

sin C

Angle EAL (

) = 53.13= [ 5 x ( 3+ 1.28) ] (sin 53.13)= 8.562


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Part 3


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The Mathematics society decided to build a pond in region P as shown in

Diagram 2. The pond is in the shape of a sector with centre E, radius ED and a

depth of 1 meter.

a) Calculate the angle AED, in radians, by using at least two different methods.Method 1:

By drawing a horizontal and vertical lines on GeoGebra .

A = (0, 0)

D = (4, 2)

C = (7, 2)

H = (0, 6) G = (3, 6)

F = (7, 6)

E = (3, 4)

B = (7, 0)0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8 9 10

b = 2

e = 4

i = 5

j = 6

h = 2

f = 4g = 3

c = 3

a = 7

Graph 10

P

Q R


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Next, calculate the angle on the left and right side of AED

Finally ,

AED = 180 ( 53.13 + 63.43 )

AED = 63.44

A = (0, 0)

D = (4, 2)

C = (7, 2)

H = (0, 6) G = (3, 6)

F = (7, 6)

E = (3, 4)

B = (7, 0)0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8 9 10

b = 2

e = 4

i= 5

j = 6

h = 2

f = 4g = 3

c = 3

= 53.13 = 63.43

A = (0, 0)

D = (4, 2)

C = (7, 2)

H = (0, 6) G = (3, 6)

F = (7, 6)

E = (3, 4)

B = (7, 0)0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8 9 10

b = 2

e = 4

i= 5

j = 6

h = 2

f = 4g = 3

c = 3= 63.44

a = 7

GeoGebras calculations:

So, AED is 63.44

Graph 11

Graph 12

Q

Q

R

R

P

P


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Method 2

By using the formula from the solution of triangle.

= + 2 bc = A

= + 5 2 ( 5 ) = A

20 = 5 + 25 - 2 ( 5 ) = A

10 = 2 ( 5 ) = A

= A

0.4472 = A

63.44 = A

b) Determine the volume of water that has to be pumped in to fill up 80 % of the pond.

So the depth of the pond is 1 m

=

= ( 2.24 ( 63.44 x )

= ( 2.24 ( 1.107 rad )

= 2.77 x 1m

= 2.77 x 80 %

= 2.216

c) If the water is pumped into the pond at a constant rate of 0.001 , calculatei) The rate of change of depth of the water,

= x = 0.001

=

x 0.001

= 0.000361


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ii) The depth of water after 10 minutes.= 0.000361 x 60 s x 10 min= 0.2166 m

iii) The minimum time taken, in minutes, before the water overflows.So, the maximum depth of the pond is 1 m. Ratio.

0.2166 m = 10 min

1m = ? min

=

0.2166 x = 10

X =

X = 46.17 min

iv) The minimum time taken, in minutes, before the water overflows, if the pond isshape is triangular AED and has a depth of 2 meters.

A = ab sin c

A = (5)( 2.4) sin 63.44

A =

V = x V =

The height of water after 10 mins

= 0.0002 x 60 s x 10 min= 0.012 m

Z =

Z = 166 .67 min

=

x

=

x 0.001

= 0.0002

Then , the ratio is

0.012m = 10 min

2m = ? min =

0.12 z = 20


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Map have been used for thousands of years to aid travelers during their journey from one place

to another. Map can also be used to estimate distance between places. In the year 2014, a

recreation park will be constructed in town marked X on the map of Malaysia in as shown

Diagram 3 . This town has the latitude of 5 41 N and has the same latitude as the city of

Malacca.

Explore and find the distance between these two places in kilometer by using

i) The map in diagram 3Since the scale is 60 km per square,

We can approximately the distance between X and the city of Malacca by

d = 60 km x 6 square

d = 360 km

Diagram 5


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ii) The formula given below :

1 nautical mile = 1.852 kilometers.

= difference in latitude in degrees.Malaccas latitude = 217NThus,

d = x 60 nautical milesd = (541 N - 217N ) x 60 ( 1. 852)

d = 3 24 x 111.12

d = 377.8 km

Yes, there is a different between the answers obtained. This is because the

calculation by using the scale given by a map is just an approximation method. The

answer is correct is correct but less accurate compared to the answer from the

calculation based on the formula given. By a using the latitude, the answer is very

accurate and significant.

a) Surf the internet and use the Google map to locate the position of your school and twonearby hospitals ? clinics. Print a copy of this Google map and mark the position of these

three places.

i) Solve the triangle obtained.ii) Calculate the shortest distance from your school to the line joining the two

hospitals / clinics.

Distance = x nautical miles


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RESOURCES

http://www.novelguide.com/a/discover/mmat_02/mmat_02_00096.html

http://en.wikipedia.org/wiki/Cartesian_coordinate_system
http://www.novelguide.com/a/discover/mmat_02/mmat_02_00096.htmlhttp://www.novelguide.com/a/discover/mmat_02/mmat_02_00096.htmlhttp://www.novelguide.com/a/discover/mmat_02/mmat_02_00096.html

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Additional Mathematic 2011