Additional Mathematics Project 2014 Selangor Sample Answers

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PART 1
Gottfried
Leibniz
Thoughtof the
variables x and y as
ranging over
sequences of
infinitely close values
Knew that dy/dx
gives the tangent but
he did not use it as a
defining property
His notation was
better suited to
generalizing calculus
to multiple variables
and in addition it
highlighted the
operator aspectof
the derivative and
integral
Introduced dx and dy
as differences
between successive
values of these
sequences

PART 2
a) From the graph,find
i. the acceleration of the car in the first hour,
o equation: Y= mX + c
o PQ= RS
o same gradient= -160
Area under the graph= displacement
= 1/2 (20 + 80)(1)
= 1(50)(1)
= 50 km
Velocity= displacement/ time
= 50/ 1
= 50kmh-1
Acceleration= velocity/ time
= 50/ 1
= 50kmh-2
ii. the average speed of the car in the first two hours.
Average speed= total distance travelled/ total time taken
= [area of trapezium + area of rectangle + area of triangle] ÷ 2
= [(1/2)(20+80)(1)] + [0.5 x 80] + [(1/2)(0.5)(80)] ÷ 2
= (50 + 40 +20) ÷ 2
=110/2
= 55kmhˉ¹
b) What is the significance of the position of the graph
i. above the t-axis
 A positive slope (starting point at low time and low position) moves away from
the base point.
ii. below the t-axis
 a negative slope (starting at low time but high position) will move back towards
the base position.

c) Using two different methods, find the total distance travelled by the car.
 Method 1 : Area under the graph
Distance= [area of trapezium + area of rectangle + area of triangle] + [area of triangle +
area of rectangle + area of triangle]
= [(1/2)(20+80)(1) + (0.5 x 80) + (1/2 x 0.5 x 80)] + [(1/2 x 0.5 x 80) + (0.5 x 80)
+ (1/2 x 0.5 x 80)]
= [50 + 40 + 20] + [20 + 40 + 20]
= 190km
 Method 2 : Integration
ʃ1
0 60ʈ+ 20 ᶑʈ= [ 60ʈ2
/2+ 20ʈ]0
1
=[ 30ʈ2+ 20ʈ]
= [30( 1)2+ 20( 1)]- [ 30(0)2+ 20( 0)]
= 30+ 20 -0
= 50
ʃ1.5
80 ᶑʈ = [ 80ʈ]1.5
= [ 80( 1.5)]- [80(1)]
= [ 120]- [80]
= 40
∫²-160t + 320dt = [-160t²/2 + 320t]²
= [-80t² + 320t]²
= [-80(2)² + 320(2)] - [-80(1.5)² + 320(1.5)]
= 320 – 300
= 20
equation : Y=mX + c
-80= -160(3)+ c
= -480 + c
c = 400
Y= -160X + 400
∫ -160t +400ᶑʈ= [-160t²/2 + 400t]
= [-80t² + 400t]²
= [-80(3)² + 400(3)] – [-80(2.5)² + 400(2.5)]
= [-720 + 1200] – [-500 +1000]
= 480-[500]
= -20
Y=-80

∫[-80]ᶑʈ = [-80t]
= [-80(3.5)] + [-80(3)]
= -280 – (-240)
= -40
equation: Y= mX + c
0= 160(4) + c
c= -640
y= 160X – 640
∫ 160t- 640 ᶑʈ= [160t²/2 – 640t]
= [1280 – 2560] – [980- 2240]
= -1280 – [-1260]
= -20
Distance= [50 + 40 + 20] – [-20 – 40 – 20]
= 110 – [-80]
= 190
d) Based on the above graph,write an interesting story of the journey in not more than 100
words.
 Ali was driving his car at 20km/h on the highway from PJ to Subang. After he
started to time his journey, he drove with an acceleration of 60 km/h for one hour,
then he drove at 80km/h for 30 minutes,then he decelerated with a deceleration
of 160km/h for 30 miuntes until he reached Puchong where he found that he
missed the exit at Subang. Dismayed,he spent 30 minutes to calm down at
Puchong. He drove back to Subang with an deceleration of 160km/h for 30
minutes. Then, he drove at 80km/h for another 30 minutes and completed his
journey with an acceleration of 160km/h for 30 minutes.

PART 3
Diagram 2 shows a parabolic satallite disc which is symmetrical at the y-axis.Given that the
diameter of the disc is 8m and the depth is 1m.
(a) Find the equation of the curve y=f(x)
y=ax² + c
y-axis cuts at the point (0 , 4). So y-intercept = 4
y=ax² + 4
y-coordinate = (4 ,5)
substitute point (4 ,5) into y=ax² + 4
5=a(4)² + 4
5=a(16) + 4
5=16a + 4
a=1/16
y=x²/16 + 4
(b) To find the approximate area under a curve, we can divide the region into several
vertical strips, then we add up all the areas of all the strips.
Using a scientific calculator or any suitable computer software,estimate the area
bounded by the curve y=f(x) at (a),x-axis, x = 0 and x = 4.

When x = 0, f(0)=0²/16 + 4
=4
When x = 0.5, f(0.5)=0.5²/16 + 4
=4.016
When x = 1, f(1)=1²/16 + 4
=4.063
When x = 1.5, f(1.5)=1.5²/16 + 4
=4.141
When x = 2, f(2)=2²/16 + 4
=4.250
When x = 2.5, f(2.5)=2.5²/16 + 4
=4.391
When x = 3, f(3)=3²/16 + 4
=4.563
When x = 3.5, f(3.5)=3.5²/16 + 4
=4.766
b (i) Area of first strip = 4 x 0.5
=2
Area of second strip = 4.016 x 0.5
= 2.008
Area of third strip = 4.063 x 0.5
= 2.032
Area of fourth strip = 4.141 x 0.5
= 2.071
Area of fifth strip = 4.250 x 0.5
= 2.125
Area of sixth strip = 4.391 x 0.5
= 2.196
Area of seventh strip = 4.563 x 0.5
= 2.282
Area of eighth strip = 4.766 x 0.5
= 2.383
Total area = 17.097

b (ii) Area of first strip = 4.016 x 0.5
= 2.008
Area of second strip = 4.063 x 0.5
= 2.032
Area of third strip = 4.141 x 0.5
= 2.071
Area of fourth strip = 4.250 x 0.5
= 2.125
Area of fifth strip = 4.391 x 0.5
= 2.196
Area of sixth strip = 4.563 x 0.5
= 2.282
Area of seventh strip = 4.766 x 0.5
= 2.383
Area of eighth strip = 5 x 0.5
=2.500
Total area = 17.597
b (iii) Area of first and second strip = 4.016 x 1
= 4.016
Area of third and fouth strip = 4.141 x 1
= 4.141
Area of fifth and sixth strip = 4.391 x 1
= 4.391
Area of seventh and eighth strip = 4.766 x 1
= 4.766
Total area = 17.314 m²
c (i) Calculate the area under the curve using integration.
Area = ∫ y dx
= ∫ (x²/16 + 4) dx

= [ x³/`16(3) + 4x ]
= 4³/48 + 4(4) – 0
= 17.33 m²
(ii) Compare your answer in c (i) with the values obtained in (b). Hence, discuss
which diagram gives the best approximate area.
 The diagram 3 (iii) gives the best approximate area, which is 17.314 m²
(iii) Explain how you can improve the value in c (ii)
 We can improve the value in c (ii) by having more strips from x = 0 to x = 4
d Calculate the volume of the satellite disc.
y = x²/16 + 4
x² = 16 (y – 4)
= 16y – 64
volume = π∫ x² dy
= π∫ (16y – 64) dy
= π [16y²/2 – 64y]
= π [ 8y² – 64y]
= 8π m³

FURTHER EXPLORATION
A gold ring in Diagram 4 (a) has the same volume as the solid of revolution obtained when the
shaaded region in Diagram 4 (b) is rotated 360˚ about the x-axis
Find
(a) the volume of gold needed,
Let f(x) = y
y = 1.2 – 5x²
When x = 0.2 , y = 1.2 – 5(0.2)²
y = 1 y = 1
y = 1.2 – 5x² y²= 1²
y² = (1.2 – 5x²)² y²= 1
y² = 1.44 – 25x – 12x²
volume of gold needed
= π∫ (1.44 + 25x – 12x²) dx – π∫ 1 dx
= π [1.44x + 25x /5 – 12x³/3] – π [x]
= π [1.44x + 5x – 4x³] – π [x]
= 0.5152 π – 0.4 π
= 0.36191 cm³
Gold density = 19.3gcmˉ³
Let k be the weight of the gold in g
1cmˉ³ ---------19.3 g
0.36191cmˉ³---------k g
So, k = 6.9849
(b) the cost of gold needed for the ring
RM130 x 6.9849 = RM908.04

REFLECTION Calculus by Sarah Glaz
I tell my students the story of Newton versus Leibniz,
the war of symbols, lasting five generations,
between The Continent and British Isles,
involving deeply hurt sensibilities,
and grievous blows to national pride;
on such weighty issues as publication priority
and working systems of logical notation:
whether the derivative must be denoted by a "prime,"
an apostrophe atop the right hand corner of a function,
evaluated by Newton's fluxions method, Δy/Δx;
or by a formal quotient of differentials dy/dx,
intimating future possibilities,
terminology that guides the mind.
The genius of both men lies in grasping simplicity
out of the swirl of ideas guarded by Chaos,
becoming channels, through which her light poured clarity
on the relation binding slope of tangent line
to area of planar region lying below a curve,
The Fundamental Theorem of Calculus,
basis of modern mathematics, claims nothing more.
While Leibniz―suave, debonair, philosopher and politician,
published his proof to jubilant cheers of continental followers,
the Isles seethed unnerved,
they knew of Newton's secret files,

locked in deep secret drawers—
for fear of theft and stranger paranoid delusions,
hiding an earlier version of the same result.
The battle escalated to public accusation,
charges of blatant plagiarism,
excommunication from The Royal Math. Society,
a few blackened eyes,
(no duels);
and raged for long after both men were buried,
splitting Isles from Continent, barring unified progress,
till black bile drained and turbulent spirits becalmed.
Calculus―Latin for small stones,
primitive means of calculation; evolving to abaci;
later to principles of enumeration advanced by widespread use
of the Hindu-Arabic numeral system employed to this day,
as practiced by algebristas―barbers and bone setters in Medieval Spain;
before Calculus came the Σ (sigma) notion―
sums of infinite yet countable series;
and culminating in addition of uncountable many dimensionless line segments―
the integral integral―snake,
first to thirst for knowledge, at any price.
That abstract concepts, applicable―at start,
merely to the unseen unsensed objects: orbits of distant stars,
could generate intense earthly passions,

is inconceivable today;
when Mathematics is considered a dry discipline,
depleted of life sap, devoid of emotion,
alive only in convoluted brain cells of weird scientific minds.

Additional Mathematics Project 2014 Selangor Sample Answers

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Additional Mathematics Project 2014 Selangor Sample Answers