Engineering Maths

This document provides solutions for various topics in Engineering Maths including mean and standard deviation, normal distribution approximation, and more. It also includes calculations and explanations for each topic. The subject is Engineering Maths and the course code and name are not mentioned. The document type is a solved assignment.

ENGINEERING MATHS

TABLE OF CONTENTS
LO2..................................................................................................................................................3
A. Mean and standard deviation..................................................................................................3
B. Normal distribution approximation for binomial distribution................................................3
C...................................................................................................................................................4
D..................................................................................................................................................4
E.).................................................................................................................................................6
LO 3.................................................................................................................................................7
A..................................................................................................................................................7
D................................................................................................................................................10
E.................................................................................................................................................10
G................................................................................................................................................11
LO 4...............................................................................................................................................13
A................................................................................................................................................13
B.)..............................................................................................................................................13
C.)..............................................................................................................................................13
E.)...............................................................................................................................................15

LO2
A. Mean and standard deviation
Solution
Mean (X') = Sum of observations / Total number of observations
= (44+50+38+96+42+47+40+39+46+50) / 10 = 49.2
X (X-X') (X-X')^2
44 -5.2 27.04
50 0.8 0.64
38 -11.2 125.44
96 46.8 2190.24
42 -7.2 51.84
47 -2.2 4.84
40 -9.2 84.64
39 -10.2 104.04
46 -3.2 10.24
50 0.8 0.64
∑ = 2599.6
Variance = ∑(X-X')^2 / (n-1)
Variance = 2599.6 / 9 = 288.844
Standard deviation = √288.844 = 16.99
Mean = 49.2
Standard deviation = 16.99
B. Normal distribution approximation for binomial distribution
Solution

N = 20
P = 0.25
Mean = np = 20 * (0.25) = 5
Variance = np*(1-p) = 5*(1-0.25) = 3.75
Standard deviation = √3.75 = 1.94
Thus normal approximation: mean = 5 and standard deviation = 1.94
B (20, 0.25) and N (5, 1.94)
C.
Average life of motor = X' = 10
Standard deviation =σ = 2
Solution
Let x denotes the life of motors
Z value = (x - X') / σ
If life time is distributed normally then
3% area = Z – value of -1.88
-1.88 = (x-10) / 2
On solving we have x =6.24 years
Thus motors should be guaranteed for 6.24 years for normal
distribution.

D.
Probability of rejection or failure = q = 12% or 0.12
n = 10
p = 0.88
Solution
Let x denotes number of rejected pistons
1.) Probability of not more than 2 pistons are rejected = P (x<=2)
= P (x=0) + P (x=1) + P (x=2)
P (x) = nCx*(pn) * [q(n-x)]
P (0) = 0.2785
P (1) = 0.3797
P (2) = 0.2330
P (x<=2) = 0.891
2.) Probability of at least 2 pistons are rejected = P (x>=2)
P (x>=2) = 1 - P (x<=1)
= 1 – [ P (x=0) + P (x=1) ]
P (0) = 0.2785
P (1) = 0.3797
P (x>=2) = 0.341

3.) Results in terms of profitability and revenue losses
On the basis of above results it can be concluded that as the number of pistons increases there is
less probability of getting them rejected. Thus with increasing number of pistons the profitability
will increase and revenue losses due to rejected pistons will also decrease.
E.)
Mean = 1000 = X'
Standard deviation = 125 = σ
Life time of bulbs = 1250, 980, 1150
Solution
Standardise normal value Z = (x - X') / σ
x = 1250 Z = 2
x = 980 Z = -0.16
x = 1150 Z = 1.2

LO 3
A.
P max = 20 kwh / day
P (t) = a cos [b (t-d)] +c
t = 0 denotes 1st January
Period of P (t) = 365 days [ February has 28 days]
P min = 4 kwh / day
Solution
1.)
C = (P max + P min) / 2
=( 20 +4 )/2 = 12
C = 12
Similarly |a| = (P max - P min) / 2
|a| = ( 20 -4 )/2 = 8
|a| = 8
Time over which p(t ) is maximum
Month Number of days
January 31
February 28
March 31
April 30
May 31

June 21
Total days 172
Total number of days = 172
Time period T = 365 = 2π / b
So b = 2π / 365
For cos function without shift maximum value occurs at t = 0
P (t) max is at t = 172
Thus d = 172
Hence a = 8 b= 2π / 365
c=12 d= 172
P (t) = a cos [b (t-d)] +c = 8 cos [2π / 365 (t-172)] +12
2.)
3.)

P (t) = 8 cos [2π / 365 (t-172)] +12
P min =4
Thus
4 = 8 cos [2π / 365 (t-172)] +12
8 cos [2π / 365 (t-172)] = -8
cos [2π / 365 (t-172)] = -1
t = 354.5 days = first 11 months
Thus minimum power is produced by solar system on 354.5 th day which is on 21 December
4.)
Power produced > = 16 kwh/ day
P (t) = 8 cos [2π / 365 (t-172)] +12 = 16
8 cos [2π / 365 (t-172)] = 4
cos [2π / 365 (t-172)] = 0.5
t = 232.8
as t > 172 which is t for maximum power
t1 = 172 – (232.8 - 172) = 111.2
t2 – t1 = 232.8 – 111.2
t2 – t1 = 121.6 days
Thus the power produced by system is sufficient for 121.6 days.

D.
Vector P = 4i +0j + 7k
Vector q = -2i + j + 3k
Solution
Dot product between two vectors is given by the formula:
p.q = |p| |q|. Cos ϴ
p.q = (4i +0j + 7k). (-2i + j + 3k) = 13
|p| = √65
|q| = √14
13 = √65 * √14 Cos ϴ
Cos ϴ = 13
ϴ = 64.47 degree = 1.12 radian
E.
Solution
Sin (a + b) = sin a cos b + cos a sin b
Sin (x + π) = sin x cos π + cos x sin π
Sin π = 0 and cos π = -1
Sin (x + π) = - sin x
Similarly
Sin (a - b) = sin a cos b - cos a sin b
Sin (x - π) = sin x cos π - cos x sin π

Sin (x - π) = - sin x
Thus Sin (x - π) = Sin (x + π)
G.
Sin a = 3/5
Sin b = 5/13
Solution
Sin (a + b) = sin a cos b + cos a sin b
Sin2 ϴ + cos2 ϴ = 1
cos2 a = 1 - Sin2 a
cos2 a = 1 – (3/5) 2 = 0.64
cos a = 0.8
Similarly
cos2 b = 1 – (5/13) 2 = 0.92
cos b = 0.92
sin (a +b) = (3/5) * 0.92 + (0.64*(5/13))
= 0.798
Sin (a + b) = 0.798

LO 4
A.
(1/R) = (1/R1) + (1/R2)
Solution
R is constant so d (1/R2) / dt = 0
(-1) * (R-2) * dR /dt = (-1) * (R-1) * dR1 /dt
(1/R) = (1/R1) + (1/R2)
= R1 R2 / (R1 + R2 )
On substituting value of R above we get:
dR/ dt = [ R1 R2 / R *(R1 + R2 ) ] 2 * dR1 /dt
dR/ dt =[ R2 /(R1 + R2 ) ] 2 * dR1 /dt
B.)
Velocity = v(t)= 1 +4t + 3t2
Solution
Distance (d) = dv / dt
d = 4 + 6t
At t = 3 minute = 180 seconds
Distance = 1084 meter
C.)

Radius = 0.5 mm
Length = 3 m
Young’s modulus Y = 2.1 * 1011 N/m3
Force F = 40 N
Solution
1. Longitudinal strain
Area of wire = πr2
= π (0.5 * 10-3)2
Area A = 7.85 * 10-7
Stress = Force / Area
Stress = 40 / (7.85 * 10-7)
Stress = 5.09 * 107 N/m
2.) Longitudinal strain
Y = Stress / Strain
Thus, strain = Stress / Y
Strain = 5.09 * 107 / 2.1 * 1011
Strain = 2.4 * 10-4
3.) Elongation
Stain = Elongation / Length

Elongation I = Strain * length
= 2.4 * 10-4 *3 = 7.2 * 10-4 meter = 0.72 mm
Elongation = 0.72 mm
E.)
y= x3 – 6x2 +9x
Solution
y= x3 – 6x2 +9x
dy/dx = 3x2 – 12x +9
dy/dx = 0
3x2 – 12x +9 = 0
3x2 – (9 + 3)x +9 = 0
3x2 – 9x – 3x +9 = 0
x= 3 and x = 1
At x = 3, y = 0
At x = 1, y = 4
Thus stationary points are (3, 0) and (1, 4)
On double differentiating y we have: y’’ = 6x – 12
x= 3 y’’ = 6 > 0 so it is minima point
x= 1 y’’ = -6 < 0 so it is maxima point