HNEE2 Engineering Mathematics: Solved Problems and Concepts

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HNEE 2 Engineering Mathematics

Contents
LO1.............................................................................................................................................................3
A..............................................................................................................................................................3
B..............................................................................................................................................................4
C..............................................................................................................................................................5
D..............................................................................................................................................................6
E..............................................................................................................................................................7
F..............................................................................................................................................................9
LO2...........................................................................................................................................................10
A............................................................................................................................................................10
B............................................................................................................................................................10
C............................................................................................................................................................11
D............................................................................................................................................................11
E............................................................................................................................................................12
LO3...........................................................................................................................................................13
A................................................................................................................................................................13
B............................................................................................................................................................14
D............................................................................................................................................................15
E............................................................................................................................................................15
F............................................................................................................................................................16
G............................................................................................................................................................16
LO4...........................................................................................................................................................17
A............................................................................................................................................................17
B............................................................................................................................................................17
C............................................................................................................................................................17
D............................................................................................................................................................18
E............................................................................................................................................................19
References.................................................................................................................................................20

LO1
A
in order to determine the values of x, y and z in the equation
The dimensions have to be equated,
The power of M, T, and L has been calculated according to the unit of pressure, density and
Volume.
Now the power of L will be equated both sides, therefore the equation comes out to be:
1 = (-x) + (-3y) + (3z) (1)
In the similar way, comparing the powers of M and T both the sides, two more equations come
out to be:
0 = x + y (2)
(-1) = (-2x) (3)
From equation (3), value of x can be determined which is ½,
From equation (2), x = (-y), therefore y = -1/2
Putting the values of x and y in equation (1),
 1 = -1/2 + 3/2 +3z
 3z = 1-1
 3z = 0
 Z = 0
Putting the values back into the main equation,  = C P1/2 -1/2 V0
which gives the equation,  = C (P/ )1/2

B
as stated in the question that the force depends upon velocity, radius and viscosity, let us assume
that the equation is
F ∝ vx × ry × ηz
removing the proportionality sign, and placing a proportionality constant C in this equation,
F = C vx × ry × ηz
To find the values of x, y and z, both sides of the equation should have the same dimensions of
M, L and T.
Hence to equate the equation,
[M1L1T-2] = {[L1T-1]x [L1]y}*{ [M1L-1T-1]z}
[M1L1T-2] = [Mz Lx+y-z T-x-z]
Directy comparing each side, we get z= 1, putting it into the equation –x-z = -2,
 -x -1= -2,
 X = 1,
Putting the value of x in x+y-z = 1,
 1+y -1 = 1,
 Y = 1
Putting the values of x,y and z = in the main equation, we get the final equation as the answer,
F = C v1 r1 η1

C
In an arithmetic progression, the sun of first n terms is Sn = { n/2 (2a + (n-1) d) }
Where a is the first term and n is the number of that term and d represents the difference between
two consecutive terms (Murray bourne, 2019).
Therefore, the sum of first 8 terms is,
 S8 = {8/2 × ((2 × 3) + (8-1) × d)}
 S8 = {4 × (6 + 7d)}
Calculating the sum of first 5 terms,
 S5 = {5/2 × ((2 × 3) + (5-1) × d)}
 S8 = {5/2 × (6 + 4d)}
Equating the two equations,
 4 × (6 + 7d) = 2 (5/2 × (6 + 4d)) (as the sum of 8 terms is twice of sum of 5 terms)
 24 + 28d = 2 (15 +10d)
 24 + 28d = 30 + 20d
 8d = 6
 D = 6/8 = 0.75
Sum of geometric series is, Sn = a × {(1- rn) / (1-r)}
Sum of 5 terms of geometric series, where a = 8, r = -4/8 = -1/2
 S5 = 8 × {(1-(-1/2)5)/ (1- (-1/2))}
 S5 = 8 × {(1+ 1/32) / (1+1/2)}
 S5 = 8 × {(33/32) / (3/2)}
 S5 = (8 × 33 × 2) / (32 × 3)
 S5 = 11/2 = 5.5

D
As in the diagram there are two triangles are formed, in the first triangle of 200,
tan 200 = h / (d + x)
in the question there is a need to find the h, therefore x needs to be converted in terms of h.
for this, calculating d first,
 As the airplane has travelled d distance in 1 minute (= 1/60 hours)
 Distance = speed × time
 d = 600 miles/ hour × 1/60 hours
 d = 10 miles
 tan 200 = h / (10 + x)
 0.3639 = h / (10 + x)
 X = (h – 3.639)/ 0.3639
tan 600 = h/x
 (3)1/2 = h / ((h – 3.639)/ 0.3639)
 (3)1/2 = (h × 0.3639) / (h- 3.639)
 1.7320 h – 0.3693 h = 5.8352
 1.3627 h = 5.8352
 h = 4.28 miles
therefore, the altitude is 4.28 miles.

E
1)
2)
The method to solve radioactive decay at any point of time is given as,
N = N0 (1/2)t/t-half
Where N0 is the amount present initially and the t-half is the half-life of the radioactive element.
And The method to solve radioactivity rate at any particular point of time is R = R0 e−λt, where t is
the time, R0 is the rate initially and λ is the radioactivity constant.
3)
After the completion of one week the amount left is the half of the amount present previously, let
us assume that the amount present initially was R, after completion of 1 week the amount gets
reduced to R/2. After 2nd week the amount will be R/4 and after the third week the amount left
will be R/8. Therefore, the amount left will be R-R/8 = 7R/8.
And the radioactive decay rate after 3 weeks will be R = 20 × e−λ×3
 = Ln 2 / t half
hence  = 0.693 / 1 = 0.693

R = 20 × e−0.693×3
R = 2.50 counts/ second (Opentextbcca, 2019)

F
1)
To form the exponential equation Y = c*ax , y represents the number of years after 2000.
And x represents the number of people who have bought the cell phones
A can be calculated by dividing the cell phone numbers with power raised to it,
Hence a = (375 / 275)1/2
Which gives a = 1.29.
To calculate the value of c, putting the value of a in the equation,
Y= c * (1.29)x
outing the value of y as 1725 and x as 8, the value of c comes out to be 225.
Therefore, the equation can be written as Y= (225) ×(1.29x)
2)
To find the value of cell phones in 2025, x is 25 and y needs to be calculated.
Putting these values in equation,
Y= (225) ×(1.2925)
Y = 130895.6685
3)
To find the rate of change, x will be 6 and 10.
Y= (225) × (1.296) = 2971.307
Y= (225) × (1.2910) = 1036.86
Rate of change of Y will be,
 Y2 – Y1 / X2 – X1
 (2971.307-1036.86) / 10-6
 483.61
4)
To find the instantaneous rate change in 2006, taking the values of x very close to the value 6.

Let us have 5.999 and 6.001 as the two values.
Putting these values in the equation one by one, and calculating the rate of change,
 Y2 – Y1 / X2 – X1 = (1037.125636 – 1036.597579)/ 6.001- 5.999
 264.0285
LO2
A
The sample has been collected by the restaurant owner and to calculate the mean and standard
deviation, the following are the formulas,
x=(1/n)∑
i=1
n
x
Where x represents the Mean, x represents middle value, n is the total sampled data,
Putting the values in the equation, x=(1/10)∑
i=1
n
x so value of x comes out to be 49.2
To find standard deviation in the question,
S= √ [ 1
1−n ]∗∑
i=1
n
( x−x)2
Putting the value of x and n in the equation and solving it further gives,
S = 17
B
To easily analyse and evaluate the data, distribution when done normally can be done in the way
which the records and data is made simpler to analyse it.

The value of n is 20, p is 25/100
Formula for calculation is μ = np, putting the values of n and p into the equation,
 μ = (20) x (0.25)
 μ = 5
 σ2 = np (1-p)
 putting the value of n and p, in the equation, σ2 = 3.75
 So, σ = 1.94
C
Average life has been given which is, 10 years and the standard deviation is σ = 2,
Z = X−μ
σ
To find the value of x, put the values of Z, μ & σ
 -1.88*2 = X −10
 X comes out to be 6.24
D
1)

Total number of pistons (n) are 10, the success rate (q) is 0.88 and the average rejection (p) is
0.12.
The value of x will be the numbers of piston rejected.
To find the probability, when x is 0,
 10C0 (0.12)0 (0.88)10-0 = 0.2785
When x is 1,
 10C1 (0.12)1 (0.88)10-1 = 0.3797
To find the probability, when x is 2,
 10C2 (0.12)2 (0.88)10-2
When x is greater than 2, P will be,
 P (x=0) + P (x=1) + P (x=2)
 0.8912
2)
When the rejections are at least 2,
 P = 1 – P (x<= 1)
 P = 1 – (P(x=0) + P(x=1))
 P = 0.341
3)
The value of rejected pistons tells the profits and loss of the organization which has been
beneficial because the value of rejected piston is 0. .8912
E
The method to solve conversion is, (Z) = X−μ
σ
putting the values of the variables in the equation,
when 1250 hours is the lasting time ; 1250−1000
125 , Z = 2