Mathematical Methods in Engineering: A Practical Approach

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LO1: Identify the relevance of mathematical methods to a variety of conceptualised
engineering examples
A) The dimension has to be equated of equation which is given.
This has to be done to equate the equation and the comparison which be performed on both
side.
-3y –x +3z = 1
x + y=0
(-2)x = -1
By evaluating the equation we will have,
0= z. -1/2 = y, x = ½
Now,
[L] / [T] = ([M] / [T] 2 [L]) x and ( p) = ([M] / [L] 3) y [L3] z
s= [L] / [T]
Therefore we get,
s =(([M]/[T] 2 [L]) x )
s = (([M]/[L]3)y[L3]z)
Ans: C √s /s= V
B) Let us assume that F prepotional ra,
F prepotionalvb
F prepotional v b vbranc
F prepotional v b nc
Or,
F = kvbranc........................................ (1)
In equation 1 there is k value which will be constant
Substituting the dimensional equation (1) in formula
([M1L1T-2] = [McLa+b-cT-b-c])
Now power has to be equated.
– c -b = -2
– c + a + b = 1
c = 1
Now solve the a, b,c to get the values:-
(b = 1, c = 1, a = 1)
Substituting the values eq……(1)

Ans: F:k multiple r multiple v multiple n
C) To solve the d to the given equation thee formula which will be used is the ¿divide by
2(2a+d(n–1))) the given equation (S of8 = 2 (S of5)) and also a=3,
Now put the values in the formula:
2(5/2 (3) + d (–1+ 5))) = 4(3 * 2) + d ( - 1+8))
28d +24 = 20d +30
- 28d + 20d = – 30 + 24
-8d = -6
¾= d
series calculated with the formula (Sn = (1- rn) a1/(1- r))
(1 minus (3 divide 4)5) 8 / (1–0.75) = S5
(1minus (243/1024) 8 / (4 minus 3) divide 4)= S5
24.40= S5
D) To find the value of h there is need to solve d with the time and the speed after that there
can be conversion of the 1hour =60 minutes
d = (1/60) *600
d = 10
Now, the elevation which is given it consist of the tangent,
h /(x+d) = tan (20o)……….(1)
h*1/ x = tan (60o)…………(2)
For x can be eliminated from the equation (1) and (2)
h = 1/[– 1/tan (60o)+1/tan(20o)]*d
h= 4.60 miles answer
E)
E.1) below the graph given:

0 1 2 3 4 5 6 7 8 9 10 11
0
5
10
15
20
25 Radioactivity per week
Radioactivity
E.2) Formula:
S * (0.5) time/1* half life (sub 0) = S (sub t) here,
S (sub t) radioactive at t where the S (sub 0) = initial amount of radioactive.
E.3) 20 = S (sub 0)
Time (t) = 7 x 3 days = 21 in days
T/2=7 in days
S(sub 0)x(0.5)21/ 7=S (sub t)
20/ 8=20 *(0.5)3
= 2.5counts/ sec
F)
F.1) The equation needed the exponential that is: y = c.bx
let’s say
x = no. of people have cell phones
y = years lets have 2000
1.29 = b
y = c * (1.29)8
c = let (8,1725)

sub x = 08,y=1725 in y=c *(1.29) x
c(1.29)8 = 1725
225 = c
Substitution of the values in a and c b = 1.29 and c = 225 equations:
225 (1.29) x = y
Equation we get: y=(225 *(1.29) x))
F.2) Year = 1/4 year
Substitute y in x = 25
y = 130895.668
Therefore, 130895.668 thousand people will have cellphones in 2025 in Canada.
F.3) Here, Substitute x = 10 and x = 6 by the value of x in y
y =225(1.29)10 = 2971.31 (10, 2971.307)
y =225(1.29)6 = 1036.9 (6, 1036.86)
= (-y1 + y2) / (-x1 + x2)
= (- 1036.86+ 2971.307) / (- 6+ 10)
= 483.61
The average is 483.61 will change between 2006 to 2010
F.4) Taking the closet point x=5.999 & x=6.001
Placing the values of x=05.999,x=06.001 in equation
y =225(1.29)6.001 = 1037.125636 (6.001, 1037.125636)
y =225(1.29)5.999 = 1036.597579 (5.999, 1036.597579)
= (-y1 + y2) / (-x1 + x2)
= (- 1036.597579+ 1037.125636) / (– 5.999 + 6.001)
= 264.0285
Changed rate is2006 is 264.0285 per year(people).
LO2: Investigate applications of statistical techniques to interpret, organise and present
data by using appropriate computer software packages
Taking out the Mean: (040+050+038+096+ 042+047+040+039+046+050)/010=0488 / 010=
048.8
To find
44.00 48.80 – 44.00 4.80
50.00 50.00 – 48.80 1.20

38.00 48.80 – 38.0 10.80
96.00 96.0 – 48.80 47.20
42.00 48.80 – 42.0 6.80
47.00 48.80– 47.0 1.80
40.00 48.80 – 40.0 8.80
39.00 48.80 – 39.0 9.80
46.00 48.80 – 46.0 2.80
50.00 50.0 – 48.80 1.20
((-04.80)2 + (01.20) + (-010.80)2 + (047.20)2 + (06.80)2 + (-01.80)2 + (-08.80)2
+ (-09.80)2 + (- 02.80)2 + (01.20)2)/10=0260.120
b)
np= μ = 0.25 * 20
= 5σ2 = (1-p) np
= 0.18* 20
= 03.75(σ)
= 1.94
n(1-p)=15greater 0r equal to 5 and np =greater or equal to 5, it concludes that
B (20, 0.25) ~ N (5, 1.94)
c) Replacing manufacturing with 3% motors it takes about 6.24yrs for
manufacturing.
Let us assume, the Motor’s Average life μ=10yrs
Deviation or σ = 2
To solve this formula used is, Z = X−avg life
deviation
P (Z < z)=3 divide100
Calculate Z=- 1.88 , putting the value in
We get, 6.24yrs
d) Rejected piston x will be success as rejected.
According to question: p=00.12,q=0.88 and n=10s
(i) When no rejections at 0 x,
P(x) = (Cnxpxqn-x)* s
P(x) = 0.28
There is one rejection when x=1
P(x) = C110 (00.120)1*(.88)9
P(x) = 0.37977
There is rejection of two when x is2,
P(x) = C210 (.120)2 (.88)8
P(x) = .23304
Probability = P(x less than or equal to 2)

=.37977+.23304+.2785= .89131
(ii)Rejection of atleast 2 pistons
= 1- (0.28 + 0.38)
= 0.34
(iii) No rejection possibilities will be 0.89 out of ten the poses manufacturing
is utmost 100% o benefit. The two piston rejected is .341
e) Using the formula: z = (- μ+ x) / σ for solution:
980 is -0.16
1150 is 1.2
1250 is 2
LO3: Use analytical and computational methods for solving problems by relating
sinusoidal wave and vector functions to their respective engineering applications.
a) (1) Lets take P(t) = c +a cos( b (t-d)) given: Pmax = 20 and Pmin = 4.
So, c = ( Pmax + Pmin) / 2 = ( 4+ 40) by2 = 12 and |a| = (– Pmax+ Pmin ) by 2 = ( – 4+ 20) by 2 = 8
The no. of days after the 1st January in which is P(t) is max from January the 5th month and
adding twenty one days of the month June.
t = 30 + 31 + 21+ 31 + 28 + 31 = 172
Period : 365 = 2 π by b
b = 2 π by 365
P(t) is maximum at t = 172 and t=0, the shift of cos(x) is 172.
P(172) is equal to 8
12+ Cos of [(0)] = 20.
2)

3) The difference of the half period at max value at t equals 127 minimum after the maximum
therefore P(t) at maximum at: t = 0.5 *365 + 172 = 354.5, the days in 11 months 0354.50 at
21st of December.
4) Interaction point t1& t2 of P(t) & y = 16
16 = P(t)
8 cos((2 π / 365)*(t – 172)) + 8 = 16
t = 365 arc cos(0.5)+ 172 / 2π
t = 232.8
172 is larger than corresponding to the maximum that why t2 in the graph
– t1+ t2 =– 111.2+ 232.8
= 121.6
Approx 121 days for the power systems.
B: OA & OB component will give cosine
T1 for horizontal component will be: - √ 2 = Cos (1350)
T2 for Horizontal components will be: √3
2 = Cos (30º)
OA & OB component:
T1 for Vertical components will be: Sin (135º) = / √ 2
T2 for Vertical component will be ss: Sin (30º) = / √ 2
c) Given in question:
F1equals 20lb, F2 equals 30lb
F1 force:
= (1 /√2). 20
= 14.1442
F1y
= (1 /√2). 20
= 14.1442

F2 force:
F2x
= ( √3 / 2).30 = 25.98
F2y
= 1 / 2). 30 = 15
Fx net = -11.83l0b
Fy net = 29.14410b
D) P equals 4i + 0j + 7k
Q equals -2i + j + 3k
using Cos α=[(v. u)/(|u|.|v|)]
P x Q = (4 x (- 2 ) ) + (1 x 0) + (3 x 7) = 13
Take P mod and Take Q mod
|P| = 651/2
|Q| = ((-2)2 + 12 + 32)1/2
|Q| = 141/2
Put the values at equation
Cos x=0.430
x = 64.5 º
e) Prove to be done:
sin (x - π) = sin (x + π) = cos (x + π/2)
Since
= -sin x
sin(x+π) =(cos x)(sin π)+(sin x)(cos π)
cos(x+π/2)=(cos x)(cos π/2)–(sin x) (sin π/2)
Answer: angles are same.
f) Sin (x + π) = Sin (x - π)
Sin (x + π) = Cos (x + π
x )

Let’s see recall the trigonometric table from that
The values are true to prove the expression.
g)
Since,
sin(a)=3/5
y2 + x2 = r2
Substituting the values:
32 +x2 = 52
x2 = 25 – 9
x = 4
sin b = 5/13
y2 +x2=r2
Placing values:
52 + x2 = 132
x2=169–25
x=12
cos b( x/r) = (-4/5s)
=- 0.86.
LO4: Examine how differential and integral calculus can be used to solve engineering
problems.
LO4
A) H= dR/dt
R=( R2 * R1)/( R2 pluse R1) , in this R2 is kept const.
H = [(R2 + R1) (dR1/dt) R2 – (R2 * R1) (dR1/dt)]/(R2 + R1)2
H =dR1/dt[(R2 * R2)/(R2 + R1)2]
B) Given values: V = 1 + 3t2 +4t,

While, integrating V wrt t:
¿ ¿∫¿ ¿ + 3t2 +4t) dt)
X = y + t + 3t3
3 + 4 t2
2
At, x =0 and t =0,
0 = 0 +y,y =0
If t =3,
Put 3=t:
X = 4
2 *(32)+3+ 3
333=18+3+27=48
48 units is the total unit travelled.
C) Given:
R =5 * 10-4
L =3mt
Y=(2.1)1011 N/m3
F =40N
. Stress =(40/25 * 3.14 * 10-8)= 0.5095 * 10-8
While,
Strain * Y=Stress put the value of Y
Strain=I/L
So, I=strain* L
I =3*0.2426*10-19 =0.7278*10-19
So, Longitudinal stress is=0.5095*10* 10-7
Longitudinal strain is=0.2426*10 * 10-18
Elongation is = 0.7278*10-19
d: IL=5 A
¿) = 20 * ( dt
L )(current law)
Integrate the expression:

- Log I=(Log L )+20 ¿) --------------------------- (1)
We get:
VL = (L dI
dt ¿=100e(-20t/)L ------------------------(2)
1. Current= 5 e(-20t/)L A and VL=(100)e(-20t/L) V
2. I:5A
V:100V
3. t=0 the current is 25 A/sec.
(e) Given in the question: y =– 6x2 + 9x + x3
On differentiation
dy
dx =– 12x + 9+3x2s
putting ¿ = 0),
(0=(x – 1) (3x – 9))
Solving value x are:
X values are:
x = 3 and x = 1
for (x = 3, y = 0) and (x = 1, y = 4)
The points which are stationary ((1,4),(3 0))
Finding max. and min. to differentiate with respect to x:
d 2 y
dx 2 = - 12+ 6x……….(1)
d 2 y
dx 2 (1, 4) = - 6………(2)
d 2 y
dx 2 (3, 0) = 6………….(3)
d 2 y
dx 2 (1, 4) this is the negative value, so it will have local maxima point.
d 2 y
dx 2 (3, 0) this is the positive value, so it will have local minim point.