Engineering Maths Level 4 Unit 2 Complete Homework

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Unit 2: Engineering Maths- level 4

Contents
LO1............................................................................................................................................3
LO2..........................................................................................................................................10
LO3..........................................................................................................................................13
LO4..........................................................................................................................................21

LO1
A) Given,
Speed of sound inside a gas chamber = v
Pressure of the gas= p
Density of the gas= ρ
Volume of the gas = V
To equate the exponents of z and Iy for the formula v = cx x py x vz here c is a constant that is
dimensionless.
Pressure unit = kg per m per sec2
Solution:
Given Equation: v = C x px x ρy x Vz
On equating the dimensions that are provided in the given equation, the result obtained is:
[L] by [T] = ([M] by [T]2 x [L])x ([M] by [L]3)y [L]z
Bothe the right-hand side and the left-hand side will be compared for obtained the following
expression:
 (-x) + (-3y) +3z = 1
Again, x + y is equal to zero
 (-2x) = (-1)
Putting the values in the equation and solving for the equation, following are the values of x,
y and z that are obtained,
X = 1
2 , y = (−1
2 ) and z = 0
So, p = ([M] by [T]2 x [L])x
And, (p) is equal to ([M] by [L]3)y [L]z

Therefore, the required solution for the exponential value of the exponents z and Iy is v = c x
√ p
( p)
B) Given,
Viscosity force = F
Velocity = v
Radius of the spherical body = r
Viscosity coefficient = η
To determine an expression for F
Solution:
Let us assume that F is directly proportional to rn
Or, F is directly proportional to vb
Or, F is directly proportional to nc
From the above values,
 F is directly proportional to rn x vb x nc
 F = K x (rn x vb x nc) ………. (i)
Where, K = Proportionality constant for the equation (i)
The formula is then substituted to the equation (i)
Which gives, [M1 x L1 x T-2] = [M0 x L1 x T0]a x [M0 x L1 x T-1]b x [M1 x L-1 x T-1]c
 [M1 x L1 x T-2] = [Mc x La+b+c x T-b]
Equating the powers of M, L and T on both left-hand side and right-hand side,
 The value of c is 1
 (-c) +b = 1
 (-b) + (-c) = -2
Solving the equations,
 a = b = c = 1

On substituting the values of a, b and c in equation (i) we get,
F = k x r x v x n
Therefore, the requires equation for F is given as F is equals to k x r x v x n.
C) Given,
The first term of an arithmetic progression is 3
The sum of the 1st 8 terms = 2 x (Sum of 1st 5 term)
To determine the common difference and the sum total of the series 8 − 4 + 2 − 1 + . . ..
Solution:
The formula for the equation is,
Sn = ( n
2 )x {(2 x a) + (n - 1) x d} for calculating the value of d and the equation that is given
in the question is
 S8 = 2 x (S5), and first term (a) = 3
So, putting the values
= ( 8
2 ) x {(2 x 3) + (8 - 1) x d}
= 2 x { 5
2 x (2 x3) + 5 +(-1) x d}
= 4 x (6.0 + 7d)
= 5 x {6 + (4 x 3)}
 {(28 x d) + 24} = d multiplied by 20 plus 30
 24 + (-30) = (d multiplied by 20) – (d multiplied by 28)
 (-6) = (-8d)
 d = 3
4
Therefore, the required value of the common difference is d is 3
4 .
Now, the sum total of the series 8 − 4 + 2 − 1 + . . .. is given by,

Sn = a1 x (1−rn) / 1−r
 Sn = 8 x (1 – ( 3
4 ¿ 5 ) / (1- ( 3
4 ))
 Sn = 24. 400
Therefore, the required value of sum total of the series 8 − 4 + 2 − 1 + . . .. is 24.400.
D) Given,
Altitude at which the aeroplane was flying = h
Aeroplane is flying at a constant speed = 600 miles/ hour
To determine the altitude h of the aeroplane
Solution:
The altitude h is calculated with the use of speed as well as time.
Let us assume that d = 600 multiplied by ( 1
60)
1 minute is equal to 1
60 hours
Explaining the tangent that is related to the given elevation angle we get,
tan (20º) is equal to h
( x +d )
As well as tan 60º = h
( x )
On eliminating the value of x from the equations that are given above we get,
h = d
¿ ¿
 h = 4.60 miles
Therefore, the required value of the altitude h of the aeroplane at which it was flying is 4.6
miles.
E) Given,
The half life of the radioactive substance = one week

Initial radioactivity of the element is = 20 counts / second
To determine:
i) Radioactivity graph against the time in a week
ii) Formula of radioactivity considering the time
iii) The amount of radioactivity of the element left by the end of the third week
Solution:
i) Radioactivity graph against the time in a week is gives as:
ii) Formula of radioactivity considering the time is given as,
X. (sub t) equal to. {x (sub 0) x ( 1
2)time divided by half-life}
Where, X. (sub t) = remaining amount of radioactivity in the element during the
time t
X. (sub 0) = The amount of radioactivity in the element at the initial stage.
iii) It is known that, X. (sub 0) is equal to 20
Time t is equal to 7 days multiplied by 3 that is equal to 21 days
So, the half-life of the radioactive element would be 7 days
Then,
 X x (sub t) = X x (sub 0) x (1 by 2)21 by 7
 20 x (1 by 2)3 is equal to 20 by 8
This gives 2.5 counts per seconds.

So, the required amount of radioactivity of the element left by the end of the third week is
2.5 counts per seconds.
F) To determine:
i) Exponential Equation for the given data
ii) Number of people owning an apple phone by the end of 2025
iii) Average of the change rate between 2006 & 2010
iv) Instantaneous change rate in the year 2006
Solution:
i) The required exponential equation in form, y = c multiplied by ax
Let us consider, x is the total number of populations that is owning phones (that is in
thousands)
 a = ( 375
225 ) 1 divided by 2
 a = 1.290
y = c x 1.298
c x 1.298 = 1725.0 also, y = c x 1.298
1725 = c multiplied by 1.298
 7.6690c by 1725 is equal to 7.6690c by 7.6690c
 C = 225.0
Substituting the value of C = 225.0 and a = 1.290 in the above equation we get,
Y = 225 x 1.29x
Therefore, exponential equation for the given data is Y = 225 x 1.29x
ii) Year is equal to 2000, that is 0.250 years
Substituting the value of x equal to 25 in the equation that is obtained that is, Y =
225 x 1.29x
 Y = 130895. 66850
Therefore, the total number of populations owning a phone by the end of 2025 is 130895.
66850.

iii) Substituting the value of x equal to 6 and x equal to 10 in the equation, Y = 225 x
1.29x
 Y = 225 multiplied by 1.296
 Y = 1036.860………… (6, 1036.860)
 Y = 225 multiplied by 1.2910
 Y = 2971.3070 …………. (10, 2971.3070)
A. R. O. C is equal to y2 – y1 divided by x2 – x1
 A.R. O. C = (2971.3070 minus 1036.860) divided by (10 minus 6)
 A.R. O. C = 483.610
Therefore, average of the change rate between 2006 & 2010 is 483.610.
iv) By the use of the two nearest points that is 483.610 and 483.610
By substituting the value of 483.610 and 483.610 and y = 6.0010 in Y = 225 x 1.29x
We get,
 Y = 225 multiplied by 1.29509990 = ……….1036.5975790 (5.9990, 1036.5975790)
 Y = 225 multiplied by 1.296.0010 = ……….1037.1256360 (6.0010, 1037.1256360)
Or, A.R. O. C is equal to y2 – y1 divided by x2 – x1
 A.R. O. C = (1036.5975790 minus 1037.1256360) divided by (6.0010 minus 5.9990)
 A.R. O. C = 264. 02850
Therefore, instantaneous change rate in the year 2006 is given by 264. 02850 thousands /
people.

LO2
A) Given,
Data for the 10 selected people visiting the restaurant in a four parties
44, 50, 38, 96, 42, 47, 40, 39, 46, 50
To determine the standard deviation and mean
Solution:
To determine the mean of the data (44, 50, 38, 96, 42, 47, 40, 39, 46, 50) the values of the
data are added and then divided by 10.
That is 44+ 50+ 38+ 96+ 42+ 47+ 40+ 39+ 46+ 50 = 488
Or, 488 divided by 10 = 48.80
Therefore, the required mean of the data is 48.80
Standard deviation
44 48.80 - 44 4.80
50 50 – 48.80 1.20
38 408.80 - 38 10.80
96 96 – 48.80 47.20
42 48.80- 42 6.80
47 48.80-47 1.80
40 48.80- 40 8.80
39 48.80- 39 9.80
46 48.80- 46 2.80
50 48.80- 50 1.20
(-4.802 + 1.202 + (-10.802 )+ 47.202 + 6.802 + (- 1.802 )+ (- 8.802) +( - 9.802) + (- 2.802 )+
1.202 ) divided by 10 = 1260. 120
Therefore, the required standard deviation for the data is given as 1260. 120.
B) Given,
n = 20

p = 0. 250
To determine the normal approximation distribution
Solution:
First defining some of the parameters,
Μ is equal to n multiplied by p = 20 x 0.250
 5 x σ2 = n x p x (1 minus p)
= 20 x 0.250 x 0.750
= 3.750 x σ
= √ 3.750
= 1.940
Now, n x p = 15 which is grater than or equal to 5 and n multiplied by (1 minus p) is equal to
15 greater than or equal to 5.
So, as per the corollary concept 1 it can be assumed or concluded that B (20, 0.250) ˷ N (5,
1.940).
Therefore, the required normal approximation distribution is B (20, 0.250) ˷ N (5, 1.940).
C) Given,
Average motor lifetime = 10 years
Standard deviation = 2 years
Only 3% replacement is possible for any kind of fault
The life of the motor is represented by a normal distribution
To determine the guarantee period of the motor
Solution:
If the manufacturer of the motor will be replacing 3% of the motor on the failure o the motor
then it will be taking 6.240 years guarantee period that should be provided by the
manufacturer.
Let us consider that,
μ is equal to 10 years that is equal to the average life of the motor
the standard deviation of the motor life that is σ is equal to 2
Formula required,
 Z = {x minus (average life of motor)} divided by (standard deviation)

 Z = (x – 10) by (2)
P (Z < z) equal to 3 divided by 100.
On calculating the value of z, we get
 Z = 100 divides minus 188, Putting the numerical values of μ, z and σ in the equation of
z we get that is provided in the question we get,
Minus 188 ÷ 100 = (x minus 10) by σ, here σ = 2
 Minus 188 ÷ 100 = (x minus 10) by 2
 Minus 376.0 divided by 100.0 = x minus 10
 X = 10 – 3.760
 X = 6.240 years
Therefore the required guarantee period of the motor is 6.240 years.