Engineering Mathematics: Unit 2 Assignment Solutions

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Unit 2
Engineering Maths

Contents
LO1........................................................................................................................................................3
LO2........................................................................................................................................................8
LO3......................................................................................................................................................11
LO4......................................................................................................................................................16
References...........................................................................................................................................20
2

LO1
a.
First equate the both sides dimension:
Exponent’s comparison on both the sides we get:
-x-2y+2z =
x+y = 0
-x = -1
By (3) equation we get x=1
By putting the value of x in equation (2) we get y = -1
By putting the value of x and y in the equation (1) we get z = 0
Hence we get:
b.
Assume F ∝ ra
F ∝ vb
F ∝ncvbra
Or F ∝ nc
Or F ∝kncvbra [Equation 1]
k=constant value
by putting the dimensional formula we will get:
[M1L1T-2] = [McLa+b-cT-b-c]
Now equate the power we will get
-b-c = -2
c = 1
3

a+b-c=1
bysolving the above equations
b=1, a=1, c=1
Putting the value of a,b,c in equation 1 we will get:
F=krvn
c.
Here we have a=3, n=8 and n=5
2{5/2[2(3) + (5 – 1) d]} = 8/2[2(3) + (8 - 1) d]
30+20d = 24+28d
-8d = -6
d = 3
4
Sum of the given series where a=8, n=5
S5 = 8 (1 – (3/4)5)
(1 – 3/4)
S5 = 24.40
d.
First we have to find the value of d in order to find the value of h
d= 600 * (1/60)
d=10
Tangent of elevation angles:
4

tan (20o) = h and tan(60o) = h
(d+x) x
Eliminate the x:
d = h
[1 / tan (20o) – 1 / tan (60o)]
h = 4.6 miles
e.
1.
0 1 2 3 4 5 6 7 8 9 10 11
0
5
10
15
20
25 Radioactivity per week
Radioactivity
2.
X (sub t) = X (sub 0) *(1
2) time/ half life
Where,
X (sub t) = the remaining radioactivity amount at time t
X (sub 0) = the initial radioactivity amount
5

3.
Provided,
X (sub 0) = 20
Time (t) = 3 * 7 days = 21 days
Half-life = 7 days
X (sub t) = X (sub 0) * (1
2) 21/ 7
= 20 * (1
2) 3
= 20/ 8 = 2.5 counts/ sec
f.
(i) y =c.ax
Where x is no. of people who own the phones.
And y is years since 2000
a = (375/225)1/2 = 1.29
y = (1.29)8. c
c.(1.29)8 = 1725
1725 = 7.669c
7.669c 7.669c
c = 225
put the value of c and a we get
y = 225* (1.29)x
(ii) when the year = 2000 or 0.25
6

y = 225 * (1.29)25
y = 130895.67
(iii) x=10 & x=6 in y = 225 * (1.29)x
y = 225 * (1.29)10
= 2971.31
y = 225 * (1.29)6
= 1036.86
(y2-y1)= (2971.307 - 1036.86)
(x2-x1) (10 - 6)
Avg rate of change= 483.61
(iv) Let 2 nearest points are:
x = 5.999
x = 6.001
y = 225 * (1.29)6.001 = 1037.126
y = 225 * (1.29)5.999 = 1036.597
Avg rate of change = 1037.126 - 1036.597 = 264.03
6.001 – 5.999
7

LO2
a. Solution:
Mean=∑x / N
= 44+50+38+96+42+47+40+39+46+50
10
= 49.2
Standard Deviation =
(44 - 49.2)2 + (50 - 49.2)2 + (38 - 49.2)2+ (96 - 49.2)2 + (42 - 49.2)2+ (47- 49.2)2 + (40
- 49.2)2 + (39 - 49.2)2 + (46 - 49.2)2 + (50 - 49.2)2
10
= 16.12
b.
n = 20, p = .25
= 20 * (.25)
= 5
= 5(1 - .25)
= 3.75
8

= 1.94
np is more than or is equal to 5 and n(1 – p) is also greater than or is equal to 5
Normal distribution is consider as good approximation in case of binomial distribution.
Both Normal and binomial distribution are very close.
c.
Motor’s Avg. life = 10 years = μ
Standard deviation = 2 = σ
From the formula
Z = X−μ
σ
Therefore,
P (Z less than z) = 3/100
After calculating,
Z = - 188/ 100
Putting the values of σ, μ & Z above:
9

-188/ 100 = X−10
2
(-376/ 100) = X – 10
X = 10 – 3.76
= 6.24 Years
d.
1. Assume that the X is the number of piston which are rejected.
n = 10
p = 0.12
q = 0.88
Number of rejection (x) = 0
P(X) =
=
= 0.28
Number of rejection (x) = 1
P(X) =
=
= 0.38
Number of rejection (x) = 2
P(X) =
10

=
= 0.23
Probability of getting not more than 2 pistons are rejected = P(X<=2)
= 0.28+0.38+0.23
= 0.89
2. At least 2 pistons are rejected = 1 – P(X <= 1)
= 1 - (P(x0) + P(x1))
= 1 – (0.28 + 0.39)
= 0.34
3. The probabilities of the rejection from the above result will be 0.89, if this will be 0.89 then there
will be almost 100% benefit and the two piston which might get reject will be 0.341 and to this
manufacture will provide benefit.
e.
Formula using: z = (- μ+ z) / σ
1250 is (- 1000+ 1250) / 125 = 2
980 is (- 1000 + 980) / 125 = -0.16
1150 is (-1000 + 1150) / 125 = 1.2
LO3
a.
(i)
P (t) = a * cos[b(t-d)]
Pmax= 20 &Pmin= 4
11

12