Mathematical Modeling and its Applications in Various Fields

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Contents
LO1..................................................................................................................................................2
LO2..................................................................................................................................................6
LO3..................................................................................................................................................8
LO4................................................................................................................................................12

LO1
a) Equate the dimensions on both side of the given equation then result which will get is:
[L] / [T] = ([M] / [T] 2 [L]) x ([M] / [L] 3) y [L3] z
The comparison will be made on both sides of exponents, so the expression that will obtain:
1 = -x -3y +3z
0 = x + y
-1 = -2x
On solving, the values which will obtain are:
x= ½, y= -1/2, z= 0,
Therefore,
p= ([M] / [T] 2 [L]) x and ( p) = ([M] / [L] 3) y [L3] z
Answer:
V = C√ p /( p).
b) Let us consider, F ∝ ra ,
F ∝ vb
F ∝ nc or F ∝ ravbnc
Or, F = kravbnc ........................................ (i)
In the above equation k is constant.
Now substitute the dimensional formula in equation (i)
[M1 L1 T-2] = [M0L1T0] a [M0L1T-1] b [M1 L-1 T-1] c
Or, [M1 L1 T-2] = [Mc La+b-c T-b-c]
Now equate the powers on both the sides:
c = 1
a + b – c = 1
-b – c = -2
On solving, a = 1, b = 1, c = 1,
Now substitute the values of a, b, c in equation (i),
Answer: F = k r v n

c) The equation of the formula used is Sn=n/2 (2a + (n – 1 ) d) for solving d and by the
question the equation which is given is: s8 = 2 (s5) and given that a = 3, therefore,
8/2(2(3)) + (8 - 1) d) = 2(5/2(2(3) + (5 – 1) d))
4(6 + 7d) = 5(6 + 4d)
24 + 28d = 30 + 20d
24 – 30 = 20d - 28d
-6 = -8d
d = ¾
Sum of the given series can be calculated by:
Sn = a1 (1 - rn) / (1- r)
S5 = 8(1 – (3/4)5) / (1 – (3/4))
= 8(1- (243/1024) / (4-3) / 4)
= 24.40
d) To calculate h, first calculate d by using time and speed and consider (1 minute = 1/60 hr)
d = 600 * (1 / 60) = 10
Now, explain the tangent of the given elevation angles:
tan (20o) = h / (d + x) and tan (60o) = h / x
By eliminating x from the above two equations:
h = d / [1 / tan (20o) – 1 / tan (60o)]
= 4.6 miles
e) (i) The graph is given below:

0 1 2 3 4 5 6 7 8 9 10 11
0
5
10
15
20
25 Radioactivity per week
Radioactivity
(ii) The formula is:
X (sub t) = X (sub 0) x (1 / 2) time/ half life where,
X (sub t) is the remaining radioactivity amount at time t and X (sub 0) = the initial
radioactivity amount.
(iii) We know: X (sub 0) = 20
Time (t) = 3 x 7 days = 21 days
Half life will be = 7 days
X (sub t) = X (sub 0) x (1 / 2) 21/ 7 = 20 x (1 / 2) 3 = 20/ 8 = 2.5 counts/ sec

(f)
(i) The exponential equation needed is in the form: y = c .ax
Now let’s assume,
x = number of people own the cell phones (in thousands)
y = years since 2000
a = (375 / 225) (1 / 2)
= 1.29
y = c.(1.29)8
c = take (8, 1725), sub x = 8, y = 1725 into y = c. (1.29) x
1725 = c. (1.29)8
1725 / 7.669c = 7.669c / 7.669c
c = 225
Substitute c = 225 and a = 1.29 in the equation:
y = 225 (1.29) x
Therefore the equation will be: y = 225 (1.29) x
(ii) Year = 2000 which is the 0.25 year
Now, substitute x = 25 in y = 225 (1.29)25
Therefore, y = 130895.6685
Hence, 130895.6685 thousand people own cell phone in Canada during 2025.
(iii) Substitute x = 6 and x = 10 into y = 225(1.29) x
y =225(1.29)6 = 1036.86 (6, 1036.86)
y =225(1.29)10 = 2971.307 (10, 2971.307)
A.R.O.C. = (y2 – y1) / (x2 – x1)
= (2971.307 - 1036.86) / (10 - 6)
= 483.61
Hence, the average rate of change between year of 2006 and 2010 is nearly,
483.61thousand people per year.
(iv) Now use two nearest point x = 5.999 and x = 6.001
Substitute x = 5.999, x = 6.001 in y = 6.001in the equation of y.
y =225(1.29)5.999 = 1036.597579 (5.999, 1036.597579)

y =225(1.29)6.001 = 1037.125636 (6.001, 1037.125636)
A.R.O.C. = (y2 – y1) / (x2 – x1)
= (1037.125636 - 1036.597579) / (6.001 – 5.999)
= 264.0285
Therefore, the rate of change in 2006 is 264.0285 thousand people per year.
LO2
a) Mean of the given data is: (40 + 50 + 38 + 96 + 42 + 47 + 40 + 39 + 46 + 50) / 10
= 488 / 10 = 48.8
Standard deviation is:
44 48.8 – 44 4.8
50 50 – 48.8 1.2
38 48.8 – 38 10.8
96 96 – 48.8 47.2
42 48.8 – 42 6.8
47 48.8 – 47 1.8
40 48.8 – 40 8.8
39 48.8 – 39 9.8
46 48.8 – 46 2.8
50 50 – 48.8 1.2
((-4.8)2 + (1.2) + (-10.8)2 + (47.2)2 + (6.8)2 + (-1.8)2 + (-8.8)2 + (-9.8)2 + (-2.8)2 +
(1.2)2) / 10 = 260.12
b) First define some parameters,
μ= np = 20 * 0.25
= 5σ2 = np (1-p)
= 20 * 0.25 * 0.75
= 3.75 σ = √ 3.75 = 1.94
As, np = 15 ≥ 5 and n (1-p) = 15 ≥ 5
Now, according to the concept of the corollary 1 it can be concluded that B (20,
0.25) ~ N (5, 1.94)
c) If the manufacturer will replace the 3% of the motors that fail then it will take
6.24 years long guarantee should be offered by the manufacturer. The solution is
given here:

Let say, μ= 10 years = Motor’s Avgerage life
Standard deviation = σ = 2
Formula used, Z = X−μ
σ
P (Z is less than z) = 3/100
On calculating, Z = -188 / 100, put the values of σ, μ & Z in the given question:
-188/ 100 = X−10
2 = (-376/ 100) = X – 10 = X = 10 – 3.76 = 6.24 Years
d) Consider x as the rejected pistons and for this question success is considered as
rejection.
Given that: n=10, p=0.12 and q=0.88
(i) No rejections when x=0,
P(x) = Cnxpxqn-x = C100 (0.12)0 (0.88)10 = 0.2785
One rejection is when x = 1,
P(x) = C110 (0.12)1 (0.88)9 = 0.37977
Two rejection when x = 2,
P(x) = C210 (0.12)2 (0.88)8 = 0.23304
Therefore, the probabilities of getting no more than two rejects are:
Probability = P(x ≤ 2)
=0.2785 + 0.37977 + 0.23304 = 0.89131
(ii) At least two pistons are rejected
Probability of two rejects is equal to 1 – P(x ≤ 1)
= 1 – (P (x0) + P(x1)) = 1- (0.2785 + 0.37977) = 0.34173
(iii) According to the above results if out of 10 the probabilities of no rejections
are 0.89 then the manufacturer will poses almost 100% of benefit and probability
of two pistons getting rejected is 0.341 then also the manufacturer will provide
benefit.
e) By Use formula: z = (x - μ) / σ in this question:
1250 is converted to (1250 - 1000) / 125 = 2
980 is converted to (980 - 1000) / 125 = -0.16

1150 is converted to (1150 - 1000) / 125 = 1.2
LO3
a) (i) Let say, P(t) = a cos [b (t-d)] + c and the given quantities are: Pmin = 4 and Pmax = 20.
So, c = (Pmin + Pmax) / 2 = (20 + 4) / 2 = 12 and |a| = (Pmin – Pmax) / 2 = (20 – 4) / 2 = 8
Now find the number of days after 1st January on which the P(t) is maximum by counting
the days from Jan to May and add 21 days of june:
t = 31 + 28 + 31 + 30 + 31 + 21 = 172
Now use period for finding b (b>0):
Period = 365 = 2 π / b, therefore, b = 2 π / 365
Cosine function without shifting has a maximum on t=0 and P(t) is max at t = 172 so P(t)
can be model by cos(x) function is shifted to 172 to right :
P(t) = 8 cos[(2 π / 365) (t – 172)] + 12
Now, check P(t) is max at t=172 by substituting t = 172 in the equation:
P(172) = 8 cos[(2 π / 365) (172- 172)] + 12 = 8,
Cos [(0)] + 12 = 20
(ii) The graph will look like:
(iii) The difference among half period between the max value at t=172 and min after the
max so, P(t) is min at: t = 172 + 0.5(365) = 354.5
The first 11 months have 334 days and therefore, 354.5 is 21st of dec, the day on P(t) is
min.
(iv) There is a need to find the point of intersection among t1 and t2 of P(t) and y = 16
shown in graph with the help of the equation:
P(t) = 16
8 cos[(2 π / 365) (t – 172)] + 12 = 16
cos[(2 π / 365) (t – 172)] = ½

t = 172 + 365 arc cos(1/2) / 2π = 232.8
The result is greater than 172 correspond to maximum therefore the result corresponds to
t2 in graph.
Answer is:
t1 = 172 – (232.8 – 172) = 111.2
t2 – t1 = 232.8 – 111.2 = 121.6
Therefore, system will produce power to approximately 121 days.
b)
The components of the OA and OB will be the cosines of the given angles:
Horizontal components for T1 will be: T1 Cos (1350) = - T1 √ 2
Horizontal components for T2 will be: T2 Cos (30º) = √ 3
2 T2
Now, the components of the OA and OB will be the cosines of the given angles:
Vertical components for T1 will be: T1 Sin (135º) = T1 / √ 2
Vertical component for T2 will be: T2 Sin (30º) = T2 / √ 2
c) Given:
F1= 20lb
F2 = 30lb
On dividing F1 force we’ll get:
F1x = F1 cos45o
= 20 x (1 /√2) = 14.1442
F1y = F1 sin45o
= 20 x (1 /√2) = 14.1442
Now divide F2 force:
F2x = F2 cos30o
= 30 x (√3 / 2) = 25.98
F2y = F2 sin30o
= 30 x (1 / 2) = 15
Therefore, net force will be:
Fx net = 14.1442 – 25.98 = -11.83lb
Fy net = 14.1442 +15 = 29.144 lb

d)
P = 4i + 0j + 7k , Q = -2i + j + 3k are given:
On applying the formula: Cos α = [(u. v) / (|u| . |v|) ]
P x Q = (4 x (- 2 ) ) + (0 x 1) + (7 x 3) = 13
Now, take mod of both P and Q:
We’ll get: |P| = (42 + 02 + 72)1/2 = 651/2 and |Q| = ((-2)2 + 12 + 32)1/2 = 141/2
Now, on putting the values we’ll get: Cos x = [13/ (651/2 * 141/2 )] = Cos x = 0.430
x = 64.5 º
e) To prove: sin (x - π) = sin (x + π) = cos (x + π/2)
Proof:
sin (x - π) = (sin x) (cos π) – (cos x) (sin π)
= (sin x) (-1) – (cos x) (0) = -sin x
sin (x + π) = (sin x) (cos π) + (cos x) (sin π)
= (sin x) (-1) + (cos x) (0) = -sin x
cos (x + π/2) = (cos x) (cos π/2) – (sin x) (sin π/2)
= (cos x) (0) – (sin x) (1) = -sin x
Now, by solving it can be seen that all the given angles are similar.
f) Sin (x + π) = Sin (x - π) = Cos (x + π
x ) is the given expression:
On analyzing, the following table will form:

Angle (x) π /4 π / 2 3 π/4 Π
Sin ( x – π ) - 1
√2
- 1 - 1
√2
0
Sin ( x + π ) - 1
√2
- 1 - 1
√2
0
Cos ( x + π
x ) - 1
√2
- 1 - 1
√2
0
The values present in the above table prove that the expression which is given is true.
g) Solve sin (a + b) when sin a = 3/5 and sin b = 5/13
sin a = 3/5 = y/r
x2 + y2 = r2
Substitute values:
x2 + 32 = 52
x2 = 25 – 9
x = -4
sin b = 5/13 = y/r
x2 + y2 = r2
Substitute values:
x2 + 52 = 132
x2 = 169 –25
x = -12
On solving and sketching graph we’ll get:
Cos a = x/r = (-4/5) and cos b = x/r = -(12/13)
Now solve,