Applied mathematics

1.
𝑅 𝐹𝐷

𝑚𝑔𝑠𝑖𝑛𝜃
𝜇𝑅 𝑚𝑔𝑐𝑜𝑠𝜃
𝑚𝑔
𝜃
From Newton’s second law, 𝐹 = 𝑚𝑎
But 𝐹 = 𝐹𝐷 − (𝑚𝑔𝑠𝑖𝑛𝜃 + 𝜇𝑅)
1
𝑚𝑎 = 𝐹𝐷 − (1500𝑥9.8𝑥𝑠𝑖𝑛𝜃 + 𝑥1500𝑥9.8𝑥𝑐𝑜𝑠𝜃)
4
3 √7
But, 𝑠𝑖𝑛𝜃 = , 𝑐𝑜𝑠𝜃 =
4 4
At steady speed, acceleration, 𝑎 = 0𝑚𝑠 −1
3 1 √7
𝐹𝐷 = 1500𝑥9.8𝑥 + 𝑥1500𝑥9.8𝑥
4 4 4
𝐹𝐷 = 13455.7840𝑁
Therefore the driving force is 13455.7840N 5 marks
2
𝑥 𝑓 𝑓𝑥 𝑓𝑥 2
1 41 41 41
2 33 66 132
3 18 54 162
4 6 24 96
5 2 10 50
∑ 𝑓 = 100 ∑ 𝑓𝑥 = 195 ∑ 𝑓𝑥 2 = 481

(a) ∑ 𝑓𝑥
From mean, 𝑥̅ = ∑𝑓
195
𝑥̅ =
100
𝑥̅ = 1.95
(b) ∑ 𝑓𝑥 2
From variance,𝑣𝑎𝑟(𝑥) = ∑𝑓
− (𝑥̅ )2
481
𝑣𝑎𝑟(𝑥) = − (1.95)2
100
𝑣𝑎𝑟(𝑥) = 1.0075
5 marks
3 Since given is the number of ordinates, to get the number of sub-
intervals we subtract a one.
2−0 1 1
ℎ= = , and 𝑓(𝑥) = 2
6 3 3+4𝑥

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𝑥 1 1
𝑓 (𝑥 ) = 𝑓 (𝑥 ) =
3 + 4𝑥 2 3 + 4𝑥 2
0 0.3333
1 0.2903
3
2 0.2093
3
1 0.1429
4 0.0989
3
5 0.0707
3
2 0.0526
𝑠𝑢𝑚 0.3859 0.8121
2 1 1
From ∫0 𝑑𝑥 ≈ ℎ[(𝑓(𝑥)) + 2(𝑓(𝑥))]
3+4𝑥 2 2
2 1 1 1
∫0 3+4𝑥2 𝑑𝑥 ≈ 𝑥 [(0.3859) + 2(0.8121)]
2 3
2 1
∫0 3+4𝑥2 𝑑𝑥 ≈ 0.335 (3𝑑𝑝𝑠)
5 marks
4

𝑅 𝑇
𝑎
𝑇

5𝑔𝑠𝑖𝑛300
𝜇𝑅 5𝑔𝑐𝑜𝑠300 3𝑔
5𝑔
300
𝑇 = 3𝑔 = 3𝑥9.8 = 29.4𝑁
𝑇 = 3𝑔𝑠𝑖𝑛300 + 𝜇𝑅
But, 𝑅 = 𝑚𝑔𝑐𝑜𝑠300
𝑇 = 3𝑔𝑠𝑖𝑛300 + 𝜇𝑥𝑚𝑔𝑐𝑜𝑠300
29.4 = 3𝑥9.8𝑥𝑠𝑖𝑛300 + 𝜇𝑥5𝑥9.8𝑥𝑐𝑜𝑠300
𝜇 = 0.3464
Therefore the coefficient of friction between the two surfaces in
contact is 0.3464 5 marks
1 7 1
5 𝑃(𝐴) = , 𝑃 (𝐵) = , 𝑃(𝐴̅𝑛𝐵) =
2 12 2
𝑃(𝐵̅𝑛𝐴) = 𝑃 (𝐴) − 𝑃(𝐴𝑛𝐵)
But, 𝑃(𝐴̅𝑛𝐵 ) = 𝑃(𝐵) − 𝑃(𝐴𝑛𝐵)
1 7
= − 𝑃(𝐴𝑛𝐵)
2 12

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1 7 1
𝑃(𝐴𝑛𝐵) = − =
2 12 12
1 1 5
 Therefore, 𝑃(𝐵̅𝑛𝐴) = − = 5 marks
2 12 12
6 Extract,
97 105 𝑥
78 85 92
𝑥−97 108−97
=
92−78 85−78
𝑥 = 113
Therefore 113dollars are equivalent to 92 Euros
Extract,
79 85 97
64 𝑦 78
𝑦−64 78−64
=
85−79 97−79
𝑦 = 68.667
Therefore 69 Euros are equivalent to 85 dollars 5 marks
7

𝑣𝑟 = 3𝑚𝑠 −1
50𝑚
3𝑚𝑠 −1
𝑣𝑏 =? ?

(a) Velocity of the boat relative to the river,

𝑏𝑣𝑟 = 𝑣𝑏 − 𝑣𝑟
4 = 𝑣𝑏 − 3
𝑣𝑏 = 1𝑚𝑠 −1

(b) 𝑑 = 𝑣𝑥𝑡
50 = 1𝑥𝑡
𝑡 = 50𝑠𝑒𝑐𝑜𝑛𝑑𝑠 5 marks
8
(a) 𝑃(𝑅 𝑟𝑒𝑚𝑜𝑣𝑒𝑑 𝑓𝑟𝑜𝑚 𝐵) = 𝑃(𝑅1 𝑛𝑅2 ) + 𝑃(𝐵1 𝑛𝑅2 )
7 6 4 5
= 𝑥 + 𝑥
11 14 11 14
31
𝑃(𝑅 𝑟𝑒𝑚𝑜𝑣𝑒𝑑 𝑓𝑟𝑜𝑚 𝐵) =
77
4 5
𝐵 𝑃(𝐵1 𝑛𝑅) 𝑥 10
(b) 𝑃 ( 1⁄𝑅 ) = = 11 14
31 = 31
𝑃(𝑅)
77
5 marks

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9 (a)

63 in test 1 corresponds to 72 in test 2.

(b)(i)
𝑅𝑇1 𝑅𝑇1 𝑑 𝑑2
4 1 3 9
3 4 -1 1
5 6 -1 1
1.5 3 -1.5 2.25
7 8 -1 1
6 6 0 0
1.5 2 -0.5 0.25
8 6 2 4
∑ 𝑑 2 = 18.5
6 ∑ 𝑑2
From, 𝜌 = 1 −
8(82 −1)
6(18.5)
𝜌 =1− = 0.7798
8(82 −1)

(ii) Significant at 5%

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10 (a) 𝑟0 = (2𝑖 − 2𝑗 + 8𝑘 )𝑚
𝐹 = (4𝑡𝑖 + 𝑡 2 𝑗 + 5𝑘 )
From, 𝐹 = 𝑚𝑎
(4𝑡𝑖 + 𝑡 2 𝑗 + 5𝑘 ) = 4𝑎
1
𝑎 = (4𝑡𝑖 + 𝑡 2 𝑗 + 5𝑘 )
4
𝑡2 5
𝑎 = (𝑡𝑖 + 𝑗 + 𝑘) 𝑚𝑠 −2
4 4

(b) From, 𝑎 = 𝑑𝑣
𝑑𝑡
∫ 𝑑𝑣 = ∫ 𝑎𝑑𝑡
3
𝑣 = ∫0 𝑎𝑑𝑡
3 𝑡2 5
𝑣 = ∫0 (𝑡𝑖 + 𝑗 + 𝑘) 𝑑𝑡
4 4
𝑡2 𝑡3 5𝑡 3
𝑣 = ( 𝑖 + 𝑗 + 𝑘)|
2 12 4 0
32 33 5(3) 02 03 5(0)
𝑣=( 𝑖+ 𝑗+ 𝑘) − ( 𝑖 + 𝑗 + 𝑘)
2 12 4 2 12 4
9 27 15 −1
𝑣=( 𝑖+ 𝑗+ 𝑘) 𝑚𝑠
2 12 4

𝑑𝑟
(c) From, 𝑣 = 𝑑𝑡
𝑟 = ∫ 𝑣𝑑𝑡
𝑡2 𝑡3 5𝑡
𝑟(𝑡) = ∫ ( 𝑖 + 𝑗+ 𝑘) 𝑑𝑡
2 12 4
𝑡3 𝑡4 5𝑡 2
𝑟(𝑡) = ( 𝑖 + 𝑗 + 𝑘) + 𝑐
6 48 8
where 𝑐 is a constant of integration
But; at 𝑡 = 0, 𝑟𝑜 = 2𝑖 − 2𝑗 + 3𝑘, 𝑐 = 2𝑖 − 2𝑗 + 3𝑘
𝑡3 𝑡4 5𝑡 2
𝑟(𝑡) = ( 𝑖 + 𝑗+ 𝑘) + (2𝑖 − 2𝑗 + 3𝑘 )
6 48 8
𝑡 3 +2 𝑡 4 −2 5𝑡 2 +3
𝑟(𝑡) = ( 𝑖+ 𝑗+ 𝑘)|
6 48 8 𝑡=3
33 +2 34 −2 5(3)2 +3
𝑟(𝑡) = ( 𝑖+ 𝑗+ 𝑘)
6 48 8
13 5 69
𝑟(𝑡=3) = ( 𝑖 − 𝑗+ 𝑘)𝑚
2 16 8

𝑥
11 (a) Let,𝑚 = , if the 𝑀 is used to approximate 𝑚 with small change ∆𝑚
𝑦
(𝑋+∆𝑥)
then, (𝑀 + ∆𝑚) = (𝑌+∆𝑦)
𝑋+∆𝑥
∆𝑚 = −𝑀
𝑌+∆𝑦
𝑋+∆𝑥 𝑋
∆𝑚 = −
𝑌+∆𝑦 𝑌
𝑌(𝑋+∆𝑥)−𝑋(𝑌+∆𝑦)
∆𝑚 =
𝑌(𝑌+∆𝑦)
𝑌∆𝑥−𝑋∆𝑦
∆𝑚 = ∆𝑦
𝑌 2 (1+ )
𝑌

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∆𝑦
Since, ∆𝑦 ≪ 𝑦 then, ≈0
𝑌
𝑌∆𝑥−𝑋∆𝑦
∆𝑚 =
𝑌2
𝑌∆𝑥−𝑋∆𝑦
[ ∆𝑦 ]
∆𝑚 𝑌2 (1+ )
𝑌
= 𝑋
𝑀
𝑌
∆𝑚 𝑌∆𝑥−𝑋∆𝑦
=
𝑀 𝑌𝑋
∆𝑚 ∆𝑥 ∆𝑦
= −
𝑀 𝑋 𝑌
∆𝑚 ∆𝑥 ∆𝑦
| |=| − |
𝑀 𝑋 𝑌
∆𝑚 ∆𝑥 ∆𝑦
| |≤| |+| |
𝑀 𝑋 𝑌
𝑥 ∆𝑥 ∆𝑦
Therefore the relative error in approximating is | | + | |
𝑦 𝑋 𝑌

673.16
(b) From, 𝑇 =
40.345
Let 𝑥 = 673.16, 𝑦 = 40.345
then,
∆𝑥 = 0.5𝑥10−2 = 0.005, ∆𝑦 = 0.5𝑥10−3 = 0.0005
673.16+0.005
𝑢𝑝𝑝𝑒𝑟 𝑙𝑖𝑚𝑖𝑡 = = 16.6854
40.345−0.0005
673−0.005
𝑙𝑜𝑤𝑒𝑟 𝑙𝑖𝑚𝑖𝑡 = = 16.6848
40.345+0.0005
Therefore the interval within which the exact value of T can be
expected to lie is [16.6848, 16.6854]

12 (a) 𝑘𝑥 2 ; 𝑥 = 1,2,3
From 𝑓 (𝑥) = { 𝑘(7 − 𝑥)2 ; 𝑥 = 4,5,6
0; 𝑒𝑙𝑠𝑒 𝑤ℎ𝑒𝑟𝑒
(i) 𝑥 1 2 3 4 5 6
𝑃(𝑋 = 𝑥) 𝑘 4𝑘 9𝑘 9𝑘 4𝑘 𝑘

From, ∑𝑎𝑙𝑙 𝑥 𝑃 (𝑋 = 𝑥) = 1
(𝑘 + 4𝑘 + 9𝑘 ) + (9𝑘 + 4𝑘 + 𝑘 ) = 1
28𝑘 = 1
1
𝑘=
28

(ii) 𝑥 1 2 3 4 5 6
𝑃(𝑋 = 𝑥) 1 4 9 9 4 1
28 28 28 28 28 28

From, 𝐸 (𝑥) = ∑𝑎𝑙𝑙 𝑥 𝑥𝑃(𝑋 = 𝑥)

1 4 9 9 4 1
𝐸 (𝑥) = 1 ( ) + 2 ( ) + 3 ( ) + 4 ( ) + 5 ( ) + 6 ( ) = 3.5
28 28 28 28 28 28

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(iii) From, 𝑣𝑎𝑟(𝑥) = 𝐸 (𝑥 2 ) − ((𝐸(𝑥)2 )

But, 𝐸 (𝑥 2 ) = ∑𝑎𝑙𝑙 𝑥 𝑥 2 𝑃(𝑋 = 𝑥)
1 4 9 9 4 1
𝐸 (𝑥 2 ) = 1 ( ) + 4 ( ) + 9 ( ) + 16 ( ) + 25 ( ) + 36 ( )
28 28 28 28 28 28
𝐸(𝑥 2 ) = 13.5
𝑣𝑎𝑟(𝑥) = 13.5 − ((3.52 )
𝑣𝑎𝑟(𝑥) = 1.25

(b)
𝑥 1 2 3 4 5 6
𝑃(𝑋 = 𝑥) 1 4 9 9 4 1
28 28 28 28 28 28
𝐹 (𝑥) = 𝑃(𝑋 ≤ 𝑥) 1 5 14 23 27 1
28 28 28 28 28

𝐹(𝑥)

1
27
28
23
28
14
28
5
28
1
28

0 𝑥
0 1 2 3 4 5 6

13 (a)
D 3N C
√2N
4N 2N

2√2𝑁

A B
2N
2𝑚

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Resolving horizontally,
𝐹𝑥 = 2 + 3 + 2√2𝑐𝑜𝑠450 + √2𝑐𝑜𝑠450 = 8𝑁
Resolving vertically,
𝐹𝑦 = 4 + 2 + 2√2𝑠𝑖𝑛450 + √2𝑠𝑖𝑛450 = 7𝑁
8
𝐹⃗𝑅 = [ ]
7
7𝑁

𝐹𝑅

𝜃
8N
⃗⃗⃗⃗⃗
|𝐹 2
𝑅 | = √(𝐹𝑥 ) + (𝐹𝑦 )
2

⃗⃗⃗⃗⃗
|𝐹 2 2
𝑅 | = √(8) + (7) = 10.6301𝑁
𝐹𝑦 7
From, 𝜃 = 𝑡𝑎𝑛−1 ( ) = 𝑡𝑎𝑛−1 ( ) = 41.20
𝐹𝑥 8
Therefore the resultant force is 10.6301N and acts at 41.20 above the
positive x-axis.

(b) 𝐹𝑥 𝐹𝑦
From, | |=𝐺
𝑥 𝑦
8 7
| |=𝐺
𝑥 𝑦
√8
Taking moments about A. 𝐺 = 3𝑥2 − 2𝑥2 + (√2)𝑥 = 4𝑁𝑚
2
8 7
| |=4
𝑥 𝑦
8𝑦 − 7𝑥 = 4
Therefore the equation of line of action of the resultant force is; 8𝑦 −
7𝑥 = 4

14 (a) (ii) To calculate the tax paid (T) in dollars based on the amount (A)
earned after 200 iterations.

(b) N A T
1 1500 0
2 3500 70

3 9000 260

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(a) (i)
START

𝑁=1

READ, A

𝐼𝑠
YES
𝐴 < 2000? T=0

NO
𝐼𝑠
YES 2𝐴
𝐴 ≤ 5000? 𝑇=
100
𝑁 =𝑁+1

NO
5
𝑇 = 60 + (𝐴 − 5000)
100

NO 𝐼𝑠

𝑁 = 200?

YES

PRINT: T, A

STOP

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15 (a) 𝜇 = 600𝑔, 𝜎 = 20𝑔

7
𝑃 (𝑋 > 𝑥 ) =
100
𝑥−𝜇 0.43
𝑃 (𝑍 > ) = 0.07
𝜎
0.07

𝑥−600 𝜇=0 𝑥 − 600

= 1.476 20
20
𝑥 = 20𝑥 (1.476) + 600
𝑥 = 629.52𝑔

(b) 𝑛 = 1000
𝑃(𝑋 < 545)
545−600
𝑃 (𝑍 < )
20
𝑃(𝑍 < −2.75) = 2.98𝑥10−3
Number of packets that weighed less than 545g is;
2.98𝑥10−3 𝑥1000 = 2.98 ≈ 3packets

16 (a)
𝑦 49𝑚𝑠 −1
49𝑚𝑠 −1

ℎ𝑚
300
0
30
0
0 1 𝑑1 𝑥
From, 𝑠 = 𝑢𝑡 + 𝑎𝑡 2 𝑑2
2
1
For P, 𝑠 = (49𝑠𝑖𝑛300 )(𝑡 ) − 𝑥9.8𝑥𝑡 2
2
1
For Q, 𝑠 = (49𝑠𝑖𝑛30 𝑡 − 2) − 𝑥9.8𝑥(𝑡 − 2)2
0 )(
2
At the point they met, they had travelled the same distance, therefore;
1 1
(49𝑠𝑖𝑛300 )(𝑡 ) − 𝑥9.8𝑥𝑡 2 = (49𝑠𝑖𝑛300 )(𝑡 − 2) − 𝑥9.8𝑥(𝑡 − 2)2
2 2
68.6 = 19.6𝑡
𝑡 = 3.5𝑠𝑒𝑐𝑜𝑛𝑑𝑠
1
ℎ = (49𝑠𝑖𝑛300 )(3.5) − 𝑥9.8𝑥3.52 = 25.725𝑚
2
Therefore the two met at 25.725m from the start.

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(b) Distance between A and B is 𝑑 = 𝑑1 + 𝑑2

1
From, 𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
Horizontally there is no acceleration.
𝑑1 = (49𝑠𝑖𝑛300 )(3.5) = 85.75𝑚
𝑑2 = (49𝑠𝑖𝑛300 )(3.5 − 3) = 36.75𝑚
𝑑 = 85.75 + 36.75 = 122.5𝑚
Therefore the distance between A and B is 122.5m

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