Page 1 of 16
Modeling Questions
1
(a)
−
dy
∝ yt
dt
⇒
(b)
dy
=
− kyt where k is the constant of proportionality.
dt
dy
= − kyt
dt
∫ y dy= ∫ − kt dt
1
1
⇒ ln y =
− kt 2 + c
2
When t = 0, y = 50
⇒ ln 50 = c
1
∴ ln y =
− kt 2 + ln50
2
1
lny − ln50 =
− kt 2
2
1
 y 
ln   = − kt 2
2
 50 
1
− kt 2
y
=e 2
50
1
− kt 2
y = 50 e 2
(c)
When t = 10, y = 40
⇒ 40 = 50 e–50k
4
= e − 50 k
5
4
ln   = − 50k
5
k= −
1
4
ln  
50  5 
Unit 2 Answers: Modeling questions
© Macmillan Publishers Limited 2013
Page 2 of 16
1
∴y=
50 e100
4
t 2 ln  
5
When y = 25
⇒ 25 = 50 e
1 2 4
t In  
100
5
1 1 2 4
ln   =
t ln  
 2  100
5
 1
100 ln  
 2
= t2
 4
ln  
 5
t = 17.6 seconds
2
−
dθ
∝ (θ − θ )
dt
dθ
=
− k (θ − θ 0 ) where k is the constant of proportionality.
dt
dθ
=
− k (θ − 15)
dt
The temperature θ0 = 15˚
∫ θ − 15 dθ= ∫ −k dt
1
− kt + c
⇒ ln θ − 15 =
When t = 0, θ = 90
⇒ ln 75 = c
∴ ln (θ – 15) = –kt + ln 75
When t = 3, θ = 70
⇒ ln 55 = –3k + ln 75
3k = ln 75 – ln 55
k=
1  75  1  15 
ln   = ln  
3  55  3  11 
Unit 2 Answers: Modeling questions
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Page 3 of 16
1
 15 
∴ ln (θ – 15) = − t ln   + ln 75
3
 11 
Consider θ = 50˚C
1
 15 
⇒ ln 35 = − t ln   + ln 75
3
 11 
 1  15  
ln 35 – ln 75 =  − ln    t
 3  11  
 35 
ln  
 75 
=
t = 7.37
1  15 
− ln  
3  11 
It will take the body 7.37 minutes to cool to 50˚C
3
Q = Q1 (1 – e–αt).
(a)
Q
= 1 − e −αt
Q1
e −αt = 1 −
Q
Q1

Q
−α
=
t ln 1 − 
Q1 

1 
Q
t=
− ln 1 − 
α 
Q1 
(b)
Q = Q1 (1 − e −α t )
When t = 0, Q = 0,
When t → ∞, Q = Q 1
Unit 2 Answers: Modeling questions
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Page 4 of 16
(c)
When Q =
1
Q1 :
2
1
Q1 Q1 (1 − e −αt )
=
2
1
= 1 − e − αt
2
1
2
e −αt =
 1
−αt =
ln  
 2
t=
t=
4
−1
1
ln  
α 2
1
α
ln 2
−
t
θ= 10 + 90e 8
When t = 0, θ = 100
−
t
When t → ∞, e 8 → 0, θ → 10
Unit 2 Answers: Modeling questions
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Page 5 of 16
When θ = 50
−
t
50 = 10 + 90 e 8
t
−
4
=e 8
9
1
 4
ln   = − t
 9
8
4
t=
− 8ln   =
6.49
9
t = 6.49 minutes
5
Nokia
iPhone
Motorola
Male
50
30
10
Female
50
20
40
(a)
P(Nokia or iPhone)
=
(b)
P(female or owned iPhone)
=
(c)
100 50 150 3
+
=
=
200 200 200 4
110 50
20 140 7
+
−
=
= .
200 200 200 240 10
P(owned iPhone/female)
Unit 2 Answers: Modeling questions
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Page 6 of 16
P(owned iPhone ∩ female)
=
P(female)
=
(d)
P (male iPhone or female Nokia)
30
50
80 2
+
=
=
200 200 200 5
=
6
L
20
20
2
200
= =
110 110 11
200
di
+ Ri =
E (t )
dt
di
+ 10i =
12
dt
⇒ 0.5
di
+ 20i =
24
dt
P = 20, Q = 24
∫
=
e 20t
IF e=
20 dt
∫
i (e 20t ) = 24 e 20t dt
24 20t
e +c
20
i=
(e 20t )
When t = 0, i = 0
⇒ 0=
24
+c
20
c= −
24
20
∴ i (e 20t ) =
24 20t 24
e −
20
20
6 6
⇒ i = − e −20t
5 5
7
f1 = 1.04 f0 + 20
f1 = 1.04 (2000) + 20 = 2100
f2 = 1.04f1 + 20 = 1.04 (2100) + 20 = 2204
Unit 2 Answers: Modeling questions
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Page 7 of 16
f3 = 1.04f2 + 20 = 1.04 (2204) + 20 = 2312.2
After 3 months the population is approximately 2312 fish
8
j = price of a pair of jeans
s = price of ashirt
t = price of a T-shirt
2j + 2s + 4t = 90
j + 0s + 3t = 42.50
j + 3s + 2t = 62
In matrix form:
1

1
2

0
3
2
3   j   42.50 
  

2   s  =  62 
4   t   90 
The augmented matrix is:
1

1
2

0
3
2
3 42.50 

2 62 
4 90 
R2 → R2 – R1
R3 → R3 – 2R2
1

0
0

0
3
−4
3 42.50 

−119.50 
0 −34 
From row 2,
3s – t = 19.50
25.5 – t = 19.50
t=6
From row 1,
j + 3t = 42.50
j + 18 = 42.50
Unit 2 Answers: Modeling questions
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Page 8 of 16
j = 24.50
The prices are:
jeans: TT$ 24.50
shirt: TT$ 8.50
T-shirt: TT$ 6.00
9
I 1 – I2 – I3 = 0
6I1 + 3I3 = 0
6I1 + 6I2 = 36 ∴ I1 + I2 = 6
In matrix form:
1

6
1

−1
0
1
−1  I1   0 
   
3   I2  =  0 
0   I 3   6 
1

6
1

−1
0
1
−1 0 

3 0
0 6 
R2 → R2 – 6R1
R3 → R3 – R1
1

0
0

−1
6
2
−1 0 

9 0
1 6 
R 3 → 3R 3 − R 2
1

0
0

−1
2
0
−1 0 

3 0 
−6 18 
–6I3 = 18 ⟹ I3 = –3
2I2 + 3I3 = 0
2I2 – 9 = 0 ⟹ I2 = 4.5
I1 – I2 – I3 = 0
Unit 2 Answers: Modeling questions
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Page 9 of 16
I1 = 1.5
0.5
10
(a)
∫ 1 + x dx
3
0
0.5 − 0 1
=
6
12
=
Width of an interval
x
1+ x3
0
1
12
1
 4
 1
 3
 2
1+   =
1+   =
1+  
1+   =
 12 
 12 
 12 
 12 
2
12
3
12
3
∴
4
12
3
3
1.007782
1.000289
1.002312
∫
1 1 
  [(1 + 1.060660) +
2  12 
0.5
1 + x 3 dx ≈
0
1.018350
5
12
3
 5
1+  
 12 
1.035538
6
12
3
3
 6
1+   =
 12 
1.060660
2 (1.000289 + 1.002312 + 1.007782 + 1.018350 + 1.035538)]
= 0.508
(3 dp)
 1  1
   − 
1 3
2
2
3
1 + x =(1 + x ) =+
1
(x ) +
( x 3 )2 + 
2
2!
1
3 2
(b)
=
1+
1 3 1 6
x − x +
2
8
1
Since 1 + x 3 =+
∫
0.5
1 + x 3 dx =
0
1 3 1 6
x − x
2
8
1
1
up to x 6
∫ 1 + 2 x − 8 x dx
0.5
3
6
0
0.5
1
1 7

= x + x 4 −
x
8
56  0

=
0.5 +
1
1
(0.5)4 −
(0.5)7
8
56
= 0.508 (3 dp)
11
Let f (x ) = e − x
2
Unit 2 Answers: Modeling questions
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Page 10 of 16
f 1 ( x) = − 2 xe − x
2
2
2
2
f 2 ( x) =
− 2 x (−2 xe − x ) − 2 e − x =
4 x 2 e − x − 2e − x
2
2
2
2
2
–8x 3e − x + 12 xe − x
f 3 ( x) =
− 4 x 2 (−2 xe − x ) + 8 xe − x + 4 xe − x =
2
2
2
f 4 ( x) =
− 8 x3 (−2 xe − x ) − 24 x 2 e − x + 12e − x − 24 x 2 e − x
2
2
When x = 0, f (x) = e0 = 1
f 1 (x) = –2 (0) e0 = 0
f 2 (x) = 4 (0) e0 – 2e0 = –2
f 3 (x) = 0
f 4 (x) = 12
2
∴ e − x =1 + x (0) +
=
1 − x2 +
0.5
∫
2
e − x dx=
x2
x3
x4
(−2) +
(0) +
(12)
2!
3!
4!
1 4
x
2
1
∫ 1 − x + 2 x dx
0.5
2
4
0
0
0.5
1
1 5

= x − x3 +
x
3
10  0

=−
0.5
1
1
(0.5)3 +
(0.5)5
3
10
= 0.461 (3 dp)
12
d2 x
+ ω2x =
0
dt 2
AQE: m2 + ω 2 = 0
m=± ωi
General solution is x = A cos ω t + B sin ω t
When
2π
ω
, x=λ
Unit 2 Answers: Modeling questions
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Page 11 of 16
=
⇒ λ A cos 2π + B sin 2π
A=λ
∴ x = λ cos ω t + B sin ω t
dx
=
− ωλ sin ωt + ω B cos ωt
dt
=
t
When
2π dx
=
,
0
ω dt
⇒ 0 = − ωλ sin 2π + ω B cos 2π
B=0
∴ x = λ cos ω t
13
d2q
dq
+2
+2q=
50 cos t
2
dt
dt
q = CF + PI
CF:
d2q
dq
+2
+ 2q =
0
2
dt
dt
AEQ: m2 + 2m + 2 = 0
m=
=
−2 ± 4 − 8
2
−2 ± 2i
2
= –1 ± i
q = e–t (A cos t + B sin t)
P.I: Let q = a cos 2t + b sin 2t
dq
=
− 2a sin 2t + 2b cos 2t
dt
d2q
=
− 4a cos 2t − 4b sin 2t
dt 2
Substituting into the differential equation:
– 4a cos 2t – 4b sin 2t – 4a sin 2t + 4b cos 2t + 2a cos 2t + 2b sin 2t = 50 cos 2t
Unit 2 Answers: Modeling questions
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Page 12 of 16
Equating coefficients of cos 2t
–4a + 4b + 2a = 50 ⇒ –2a + 4b = 50
Equating coefficients of sin 2t
–4b – 4a + 2b = 0 ⇒ 2b = –4a ⇒ b = –2a
∴ –2a – 8a = 50
–10a = 50
a = –5
b = 10
∴ q = –5 cos 2t + 10 sin 2t
General solution is:
q = e–t (A cos t + B sin t) – 5 cos 2t + 10 sin 2t
When t = 0, q = 0
⇒ 0 = A – 5, A = 5
dq
=
e −t (− A sin t + B cos t ) − e −t ( A cos t + B sin t ) + 10 sin 2t + 20 cos 2t
dt
dq
=
t 0,= 0
When
dt
⇒ 0=
B − A + 20 ⇒ B =
− 15
Required solution is
q = e–t (5 cost – 15 sint) – 5 cos2t + 10 sin2t
14
No. to be chosen = 11
(a)
Choose the two spin bowlers and 9 from the other 18 players
Unit 2 Answers: Modeling questions
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Page 13 of 16
C2 × 18C9 = 48620
2
(b)
(c)
8
C4 × 3C1 × 9C6 = 17640
No. of choices with 1 spin bowler = 2C1 × 18C10 = 87516
No. of choices with 2 spin bowlers = 48620
No. of choices with at least 1 spin bowler = 87516 + 48620 = 136136
15
Q (K, L) = 100 K0.4 L0.6
(a)
Keep L constant and differentiate with respect to K:
∂Q
= 40 K − 0.6 L0.6
∂K
When K = 600, L = 825
∂Q
= 40 (600) −0.6 (825)0.6
∂K
= 48.42
Keep K constant and differentiate with respect to L:
∂Q
= 0.6 × 100 K 0.4 L−0.4
∂L
= 60 K0.4 L–0.4
When K = 600, L = 825
∂Q
= 60 (600)0.4 (825) −0.4
∂L
= 52.82
(b)
The gain for an increase in labour is slightly greater than that for an increase in capital,
so the manufacturer should increase the labour force.
16
Q(x, y) = 100x + 400y + x2y – x3 – y2
(a)
Differentiating wrt x while keeping y constant:
∂Q
=100 + 2 xy − 3x 2
∂x
When x = 21, y = 80
Unit 2 Answers: Modeling questions
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Page 14 of 16
⇒
∂Q
=100 + 2 (21) (80) − 3 (21) 2
∂x
= 2137
∴ Change in output = 2137 units
(b)
Differentiating wrt y keeping x constant:
∂Q
= 400 + x 2 − 2 y
∂y
When x = 20, y = 81
⇒
∂Q
= 400 + (20) 2 − 2 (81) = 638
∂y
Change in output = 638 units
17
Q = 2s3 + s2u + u3
When s = 40, u = 30
∴ Q = 2(40)3 + (40)2 (30) + (30)3
= 203000
Q must be maintained at 203000
∴ 2 s 3 + s 2u + u 3 =
203000
Differentiating wrt s:
6s 2 + 2su + s 2
∴
du
du
+ 3u 2
=
0
ds
ds
du −6s 2 − 2su
=2
ds
s + 3u 2
When s = 40, u = 30:
du −6 (40)2 − 2 (40) (30)
=
ds
(40)2 + 3 (30)2
= – 2.79
Unskilled labour must be decreased by 2.79 hours
18
(a)
For the useful life y = x
Unit 2 Answers: Modeling questions
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Page 15 of 16
4000 – 17t2 = 1500 + 8t2
25t2 = 2500
t2 = 100
t = 10
The useful life is 10 years
P (t ) = y (t ) – x (t )
(b)
= 4000 – 17t2 – (1500 + 8t2)
= 2500 – 25t2
10
25 3 

(2500 − 25t ) dt = 2500 t −
t
Net profit =
0
3  0

∫
= 2500 (10) −
10
2
25
(10)3
3
= 16666.67
∴ Net profit is TT$ 16666.67
19
For the first investment the amount of money deposited is 4000 e0.4t per year
r = 0.05, T = 10
Net value
=
∫ 4000 e
10
0.04 t
× e −0.05t dt − 2000
0
=
∫ 4000 e
10
−0.01t
dt − 2000
0
=
4000 −0.01t 10
e
 − 2000
0
−0.01 
= –400000 [e–0.1 – 1] – 2000
= TT$ 36065.03
For the second investment the money deposited is TT$ 6000 per year
r = 0.05, T = 10
=
Net value
∫ 6000 e
10
−0.05t
dt − 5000
0
Unit 2 Answers: Modeling questions
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Page 16 of 16
=
6000 − 0.05t 10
e
 0 − 5000
−0.05 
=
− 120000 [e −0.5 − 1] − 5000
= TT$ 42216.32
The net income generated by the second investment is better than the first.
20
V = πr2 h
∂V
= 2πrh
∂r
(a)
When h = 15 cm, r = 5 cm
∂V
=
2π (15) (5) =
π150
∂r
The change in volume for a 1 cm change in radius is 150π cm3
(b)
S = 2πr2 + 2πrh
∂S
(i) = 4πr + 2πh
∂r
When
=
h 15cm,
=
r 5 cm
∂S
= 4π (5) + 2 π (15)
∂r
= 50π
The change in surface area for a 1 cm change in radius is 50π cm2
(ii)
∂S
= 2πr
∂h
When r = 4 cm
∂S
=π
2 (4) =
8π
∂h
The change in surface area for a 1 cm change in height is 8π cm2
Unit 2 Answers: Modeling questions
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