Unit-2 Engineering Maths-Level 4 Complete Solution

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Unit-2 Engineering Maths-Level 4
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Contents
LO1............................................................................................................................................3
A)............................................................................................................................................3
B)............................................................................................................................................3
C)............................................................................................................................................4
D)............................................................................................................................................5
E)............................................................................................................................................5
LO2............................................................................................................................................8
A)............................................................................................................................................8
B)............................................................................................................................................8
C)............................................................................................................................................9
D)............................................................................................................................................9
E)..........................................................................................................................................10
LO3..........................................................................................................................................11
A)..........................................................................................................................................11
B)..........................................................................................................................................12
C)..........................................................................................................................................12
D)..........................................................................................................................................13
E)..........................................................................................................................................13
F)...........................................................................................................................................14
G)..........................................................................................................................................14
LO4..........................................................................................................................................16
A)..........................................................................................................................................16
B)..........................................................................................................................................16
C)..........................................................................................................................................16
D)..........................................................................................................................................17
E)..........................................................................................................................................18
Reference..................................................................................................................................20

LO1
A)
Given,
Speed of sound=v
Pressure =Ƥ
Density=ρ
Volume=V
Formula, v=C Ƥ x ρ y V z, where C is dimensionless constant.
Equation, [ L ]
[T ] = ( [ M ]
[T ]2 [ L ] ) x ( [ M ]
[L]3 ) y [L3] z
Equating the dimension on both side of equation,
1 = -x -3y+3z
0 = x + y
-1 = -2x
On solving the above equations, the values of x, y and z are:
x = 1
2, y = - 1
2, z = 0,
Putting this in formula,
Ƥ= ( [ M ]
[T ]2 [ L ] ) x and ρ= ([ M ]
[L]3 ) y [L3] z
Solution:
V=C√ Ƥ
ρ .
B)
Given, Force of Viscosity= F
Velocity= v, Radius= r, coefficient of viscosity= ƞ.
To find F=?

Consider, F directly proportional to r a v b ƞ c,
F=k r a v b ƞ c, k is constant …………………………………….. (1)
Substituting formula in equation (1)
[M1L1T-2]= [M0L1T0] a [M0L1T-1] b [M1L-1T-1] c
[M1L1T]= [M c L a +b-c T - (b +c)]
On equating both sides:
M1=M c
L1=L a +b-c
T-2=T- (b +c)
By equating powers, we get the following equations,
c=1, a + b + c=1, -b-c = -2,
Solving the above equation, we get the value of a, b and c as 1.
Substituting the value in equation (1),
Solution: F= k r v ƞ.
C)
Given, first term of AP, a = 3,
Sum of first 8 terms= 2(sum of first 5 terms)
For solving formula used are, S n =
n
2(2a + (n-1) d)
8
2 (2)(3) + (8-1) d= 2{¿ ) ((2) (3) + (5-1) d)}
4(6 + 7d) = 5 (6+4d)
24 +28 d = 30 +20 d
(28-20) d = 30-24
8 d= 6
d= 6
8 = 3
4
Using formula to find the sum of series, 8-4+2-1+…. S n =
a 1(1−r n)
1−r ,

Substituting the value,
8(1−( 3
4 )
5
)
1−( 3
4 )
]
=
8(1−( 243
1024 )
5
)
4−3
4
=24.40
D)
Given, angle of elevation at 10:00 am= 20
Angle of elevation at 10:01 am= 60
Speed= 600 miles/hour
To find altitude h=?
First, calculate distance by time speed formula
(1 minute= 1/60 hr)
d= 600(1/60)
=10
tan (20) = h/ (d + x), tan (60) = h/x
By eliminating x from above equation:
h=4.6 miles
E)
(i) Graph of the amount of radioactivity against time in weeks is shown below:

(ii) Given, Half-life of radioactive substance= 1 week
Initial level of radioactivity =20 counts per second.
Formula used: X t = X0 ( 1
2) time/half-life ……….. (1)
Where, X t denotes remaining radioactivity at time t.
X0 denotes the initial radioactivity amount.
(iii) To find radioactivity after 3 weeks=?
Given, Initial level of radioactivity =20 counts per second, i.e. X0=20
Half-life of radioactive substance = 1 week = 7 days.
To calculate for 3 week = 3 *7 days = 21 days.
Substituting the value in equation (1)
X t = X0( 1
2) 21/7.
= 20( 1
2) 3.
= 2.5 counts per second.
F)
(i) Exponential equation used to solve the question is:

Y= C * AX……………… (1)
Assume, number of people having cell phones = X
A = √375/225
= 1.29
Y= C *(1.29)8
C *(1.29)8 = 1725 and Y= C * (1.29) X.
1725= C *(1.29)8
1725/7.669C= 7.669C/7.669C
C= 225
Substituting the value of C and A in equation 1.
Y=225 *(1.29) X which is required exponential equation.
(ii) To find number of people who will own cell phone in 2025=?
Initially in the year 2000, number of people who owns cell phone = 225
X= 2025-2000 = 25
Substituting values in equation 1, Y=225 *(1.29)25.
Y= 130895.6685, these many people will have cell phone in Canada during 2025.
(iii) X1=2006-2000=6, X 2 2010-2000=10
Substituting these value in equation 1,
Y1= 225 *(1.29) 6 = 1036.86
Y2= 225 *(1.29) 10 = 2971.307
Average = (Y2 -Y1) / (X 2 –X1)
Average = 483.61
(iv)Determine instantaneous rate of change in 2006 =?
Let’s take value of X1 =5.999, X2= 6.001
Y1 = 225*(1.29) 5.999 = 1036.5975
Y2 = 225*(1.29) 6.001 = 1037.1256
Average rate of change = (Y2 -Y1) / (X 2 –X1)

= 264.0285 thousand people per year.

LO2
A)
Given data 44, 50, 38, 96, 42, 47, 40, 39, 46, 50
Mean= (Sum of all values)/total number of values.
Total number of values= 10.
Mean= (44 + 50 + 38 + 96 + 42 + 47 + 40 + 39 + 46 + 50) / 10
= 488/10
= 48.8
Calculation of standard deviation
Value Difference Difference value
44 44 - 48.8 -4.8
50 50- 48.8 1.2
38 38- 48.8 -10.8
96 96- 48.8 47.2
42 42- 48.8 -6.8
47 47- 48.8 -1.8
40 40- 48.8 -8.8
39 39- 48.8 -9.8
46 46- 48.8 -2.8
50 50- 48.8 1.2
Standard deviation: {(-4.8)2 + (1.2) 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)
The normal binomial distribution approximation is defined as:
If X~B (n, p) and if n is larger and p is near to half, then X ~ N (n p, n p q), (where, q =1-p)
Given n = 20, p = 0.25
μ = n p = 20 (0.25)
= 5σ2 = n p (1- p)
= 20 (0.25) (0.75)
= 3.75 σ = √3.75 = 1.94

n p =15 ≥ 5 and n x (1-p) = 15 ≥ 5
Then according to definition: B (20, 0.25) ~ N (5, 1.94)
C)
Given, Average life of motor = 10 year
Standard Deviation, σ =2 year
Manufactures can replace =3 %
Let, μ= 10 years = motor’s Average life
Formula Z= (X- μ)/ Standard deviation ……… (1)
P (Z < z) = 3 / 100
On solving, Z= -188/100
Putting the values of σ, μ and Z in the equation (1)
-188/100 = (X-10) /2
X-10 = -376/100
X = 10 - 3.76
= 6.24 years.
Hence, 6.24 years guarantee should be provided when manufacture take 3% failed item.
D)
Average piston rejected p = 12 % =0.12
q = 0.88
Number of piston = 10
Let x be the number of rejected piston
(i) When x = 0, means no piston is rejected.
Applying permutation and Combination
p (0) = C n x p x q n-x
= C10 0(0.12)0 (0.88)10
= 0.2785
When x=1, one piston is rejected.
p (1) = C n x p x q n-x
= C10 1 (0.12)1 (0.88)9
=0.37977

When x=2, two piston is rejected.
p (2) = C n x p x q n-x
= C10 2(0.12)2 (0.88)8
=0.23304
Probability of getting not more than two rejections p (0) + p (1) + p (2)
=0.2785 + 0.37977 + 0.23304
=0.89131
(ii) At least two piston gets reject
Probability of at least two piston implies two or more than two pistons can be
rejected.
Therefore, Probability of at least two rejected is equal to 1- (probability of zero
piston + probability of one piston getting rejected)
p (0) = 0.2785, p (1) = 0.37977
P (x ≤2) = 1- (0.2785 + 0.37977)
Answer = 0.34173
(iii) According to the above calculated results
Probability of zero rejections = 0.89, out of 10 probabilities.
This will give almost 100 % profit to the manufacture.
P (2) that is probability of getting 2 pistons rejected = 0.341. then also
manufacture will gain benefit.
E)
Given, normal Distribution with mean lifetime= 1000 hours
Standard deviation= 125 hours.
Three bulb have respectively life time of 1250, 980, 1150 hours.
Using the formula to convert it into standardized normal score z = x−μ
σ
First bulb having 1250 hrs is converted into (1250 -1000) = 250 divided by standard
deviation 125 = 2
Second bulb having 980 hrs is converted into (980 -1000) = - 20 divided by standard
deviation 125 = - 0.16
Third bulb having 1150 hrs is converted into (1150 -1000) = 150 divided by standard
deviation 125 = 1.2

LO3
A)
Given, maximum Power, P1= 20 kwh/day
Minimum Power, P2= 4 kwh/day
Sinusoidal function given P(t) = a cos [b (t - d)] + c, t=0…………… (1)
(i) To find a, b, c and d.
(P1+P2)/2= (20+4)/2=12
Modulus od a = (P2) - ( P1)/2= 20 - 4 divided by 2= 8
Total number of days starting from first January to May + 21 days of June
= 31 + 28 + 31 + 30 + 31 + 21= 172
Given period =365= 2( π/b)
b= two πdivided by 365
cos function without shifting has a maximum when t=0
P(t) is maximum at t=172
P(t) =8*cos [(2π/b) *(t-172)] + 12
Substitute the value in equation 1
P (172 ) = 8
Cos [0] +12= 20
(ii) The graph will look like this
(iii)
P(t) is minimum at t= 172 +half (365) = 354.5