HNEE2 Engineering Mathematics: Solved Problems and Concepts

Desklib provides past papers and solved assignments for students studying HNEE2 Engineering Mathematics.

HNEE 2 Engineering Mathematics

Contents
LO1.............................................................................................................................................................3
A..............................................................................................................................................................3
B..............................................................................................................................................................4
C..............................................................................................................................................................5
D..............................................................................................................................................................6
E..............................................................................................................................................................8
F..............................................................................................................................................................9
LO2...........................................................................................................................................................10
A............................................................................................................................................................10
B............................................................................................................................................................10
C............................................................................................................................................................11
D............................................................................................................................................................12
E............................................................................................................................................................13
LO3...........................................................................................................................................................13
A................................................................................................................................................................13
B............................................................................................................................................................15
D............................................................................................................................................................15
E............................................................................................................................................................16
F............................................................................................................................................................16
G............................................................................................................................................................17
LO4...........................................................................................................................................................18
A............................................................................................................................................................18
B............................................................................................................................................................18
C............................................................................................................................................................18
D............................................................................................................................................................19
E............................................................................................................................................................20
References.................................................................................................................................................22
List of figures
Figure 1 Question part D.............................................................................................................................7
Figure 2 radioactivity curve.........................................................................................................................9

Figure 3 Normal distribution curve..........................................................................................................12
Figure 4 P(t) curve....................................................................................................................................15
Figure 5 test values....................................................................................................................................18

LO1
A
The value of variables x,z and y needs to be found out using the equation provided,
In order to know the values, the unit dimensions in each side should be the same, therefore,
M represents mass, L represents the length and T represents time, so to calculate the powers of
M, L and T, the units of all quantities have been written on the right side and on the left side the
velocity dimension and power has been written.
The powers of each dimensions has to be equated to form the equations.
For L:
1 = (-x) + (-3y) + (3z) (equation 1)
For M:
0 = x + y (Equation 2)
(-1) = (-2x) (Equation 3)
On solving these equations, x is ½; y is -1/2.
Put x, y values in (1), and solving the equation further, the value of z comes as 0.
(1 = -1/2 + 3/2 +3z, 3z = 1-1, 3z = 0, z = 0)
To form the equation, x,y and z are put again into the equation, the resultant equation is
 = √C (P/ )

B
To find the equation in order to relate force with the quantities velocity, viscosity and radius,
Force is proportional to the multiplication of each quantity raise to a certain power, which gives
the equation:
F ∝ vx × ry × ηz
To remove the sign of proportionality, a constant C is place to make the equation correct.
F = C× vx × ry × ηz
In order to know the values, the unit dimensions in each side should be the same, therefore,
[M1L1T-2] = [[L1T-1]x [L1]y]*[[M1L-1T-1]z]
[M1L1T-2] = [(Mz) (Lx+y-z )(T-x-z)]
To find the values of x, y and z respectively, powers of each dimension is equated.
 z= 1
 -x -1= -2 gives x = 1
 1+y -1 = 1 gives y = 1
On putting the x, y and z with their respective values, the resultant equation becomes
F = C v1 r1 η1

C
Sn = ( n /2 (2a + (n-1) d) ) is the formula to find out the sum of first n numbers of an arithmetic
progression, where a stands for as first, n stands as the number of terms present in the arithmetic
progression, d stands for the difference present in each two consecutive term.
To calculate the sum for 8 terms, a = 3(given), n = 8,
S8 = (8/2 × ((2 × 3) + (8-1) × d))
S8 = (4 × (6 + 7d))
To calculate the sum for 5 terms, a = 3(given), n = 5,
S5 = (5/2 × ((2 × 3) + (5-1) × d))
S8 = (5/2 × (6 + 4d))
Comparing the above two to get the value of d,
4 × (6 + 7d) = 2 (5/2 × (6 + 4d)) (2 is multiplied because the sum of 8 terms is twice to that of 5
terms)
24 + 28d = 2 (15 +10d) gives 8d = 6, on further solving, d = 6/8 = 0.75
Therefore the difference present in the arithmetic progression is 0.75.
Sn = a × {(1- rn) / (1-r)) is the formula to find out the sum of first n numbers of a geometric
progression, where a stands for the first term, n is the number of terms present in the geometric
progression and r stands for the ratio between two consecutive term.
To calculate the sum for first 5 terms s, putting the values of a, r and n into the formula,
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
Figure 1 Question part D
The altitude of plane at point B makes an angle of 200 with the ground and the altitude at the
point C makes an angle of 600 with the ground.
To find out the value of h, the triangles BAB’ and the triangle CAC’ have to be considered.
Using the tan of angle 200 , tan 200 = h / (d + x)
To eliminate the value of d from the equation so that only 2 variables are left, the value of d is
calculated using distance formula.
[ Distance = speed × time ]
Speed = 600 miles per hour,
Time = 1/60 hours
Hence d becomes (600 × 1)/ 60 which gives the value of d as 10 miles.
Steps followed to convert x into terms of h,
In triangle BAB’, tan 200 = h / (10 + x)
x = (h – 3.639)/ 0.3639 (putting the value of tan 200 in the equation and solving it further)
In triangle CAC’, tan 600 = h/x
√3 = (h / ((h – 3.639)/ 0.3639))
√3 = ((h × 0.3639) / (h- 3.639))
Solving it futher,

(1.7320 h – 0.3693 h) = 5.8352 (putting the value of √3)
1.3627 h = 5.8352
h = 4.28 miles (final answer as the altitude of plane)

E
1)
Figure 2 radioactivity curve
2)
Formula that represents the amount of material decayed at a particular time,
N = N0 (1/2)t/t ½ where, N0 represents the amount at the initial point of time and the t1/2 represents
the time period when the amount will be reduced to half, also known as the half-life.
R = R0 e−λt, is the formula in order to find the rate at which the radioactive element is decaying at
a given point of time, where t represents the time, R0 represents the decay rate present at initial point
of time and λ represents the constant of radioactive decay.
3)
In order to find out the rate of decay present on completion of 3 weeks, t will be 3, t1/2 is 1 week
(given),
R = 20 × e−λ×3
To calculate , the formula is  = Ln 2 / t ½ which gives  = 0.693 / 1 = 0.693
R = 20 × e−0.693×3 (on putting the value of )
R = 2.50 (unit: counts/ second)

F
1)
Exponential equation given in the question is Y = c*ax, where y stands for the difference between
the given year and 2000 and x stands for the count of people with cell phones in that given year.
To calculate the value of constant a,
a = (375 / 275)1/2, value of a comes out to be 1.29.
To eliminate the variable c, its value needs to be calculated using a calculated above.
Y= c ×(1.29)x , to calculate y, putting y = 1725
And x comes out to be, x = 8
On solving the equation, c is 225.
Hence the resultant equation is, Y= (225) (1.29x) (Sosmathcom, 2019)
2)
In order to calculate the cell-phones bought in year 2025, x will be 2025- 2000 which is 25.
On putting the constant values and x value in the Exponential equation,
Y= (225) multiplied by (1.2925) gives 130895.6685 on solving as a result.
3)
In order to calculate the change occurred, value of x is 6 for year 2006 and 10 for 2010.
Y2= (225) × (1.296) gives 2971.307
Y1= (225) × (1.2910) gives 1036.86
To calculate the change inY with respect of X,
Y2 – Y1 divided by X2 – X1 = (2971.307-1036.86) divided by (10-6) = 483.61
4)
In order to calculate the instantaneous change of rate in the year of 2006, let us take two values
approximate to 6 but not exactly 6.
Hence x= 5.999 and x= 6.001 can be the two possible values.

To calculate the rate of change in the year 2006, putting the x and Y values respectively into the
equation gives,
Y2 – Y1 upon X2 – X1 = (1037.125636 – 1036.597579) upon (6.001- 5.999) = 264.0285
LO2
A
Formula to calculate the mean value is,
Where x stands for the Mean value, x stands for middle value and n stands for the number of
entries in sampled data
, gives 49.2 as the answer.
Therefore the mean value is 49.2
The formula to calculate the standard deviation is,
S= √ [ 1
1−n ]∗∑
i=1
n
( x−x)2
On Putting the mean value and value of n as mentioned in above equation gives, S = 17 as the
answer on solving. Therefore the standard deviation is 17.
B
Normal distribution makes the data easy to evaluate because the records are represented in
simpler ways.