NAME: EMMANUEL YILLIA
ID:21446
DEPARTMENT: BUILDING AND CIVIL ENGINEERING
FACULTY: ENGINEERING
LEVEL: YEAR TWO(2)
DATE OF SUBMISSION: 03/03/2025
QUESTION ONE(1)A
SOLUTION
2
1
Given that OA=a, OB=b, OC:CB=2:1[Meaning OC=3] OB and OC=(3)OB,
[i] To find OC
Since OC, CB=2:1
2
We have OC=(3)OB
2
OC=(3)b
[ii] To Find OM
OM=OC+CM=OC+½CB
2�
1 1
+ 2 ( 3b)
3
2� 1
+b
3 6
5
OM=6b
[iii]
To find ON
1
ON=2 [OA + OC]
2
½ [a + 3 b]
1
1
a+3b
2
[iv] To find OX
OX=OA+AX
=a + t [AM]
= a + t [AB+AC]
�
1
2
= a + 2[b-a+2(a+3b)]
�
1
1
= a+ 2 [b- 2 a + 3b}
� 4�
1
2�
�
�
2�
= a +2[ 3 – 2a]= a + 3 b +4a = [1-4]a + 3 b
�
2�
OX=(1-4] a + 3 b
[v] To find OY
BY=BN, BN=ON-OB
1
1
1
2
1
2
BN= (2)a + (3)b-b = (2)a - -3b
�
2�
BY=S[(2)a-(3)b] =(2)a-( 3 )b
OY=OB+BY
2
2�
b+(3)a -( 3 )b=
2
2�
=( 3)a +(1- 3 )b
[ Iv] To Find OP
At the intersection point p, OX=OY, We equates the coefficient of a and b
OX and OY
�
(1-t)=3…………..eq1
5�
2�
( 6 )=1- 3 ………..eq2
From 1
5 =2(1-t) we sub into two
5�
2(2(1−�)
+=1- 3
=
6
5t=-2+8t
5
=
6
4
4�
1-3+- 3
3t=2
2
T=3, sub t into 1
2 5
1-3= 3
2
So s=3
2
2
Sub t=3 into OX or s=3 into OY
2
5
2
OP=( 1-3)a + 63b
1
5
OP= 3 a + 9 b
[VII] Collinearity of C,P and Q
1
1
Q is the midpoint of AB so OQ=2(OA+OB)=2(a+b)
1
5
2
1
1
2
CP= OP-OC= 3a + 3b - 3b
1
1
=3a - 9 b
CQ=OQ-OC=2a + 2b - 3b
1
1
CQ =2a - 6b
We need to see if CP is a scalar multiple of CQ
Let cheek
2
2 1 1
CP=(3)[2a- 6b]
3
1
1
=3a - 9b =CP
2
Since CP=3CP, CP and CQ are parallel, and since they share the point c
They are collinear
[QUESTION ONE (1)B
SOLUTION
�−4
Let line =L1 = 2 =
Let line L2 =
�−3 �−7
= 2
1
�+1 � �+1
=2= 6
3
Find the direction vector of the two line
The direction vectors of the first line is V1=(21 )
2
The direction vectors of the second line is V2=(32 )
Determine if the lines intersect
6
Let parameterize the lines
Line 1: x=2t-4, y=t+s, z=2t+t
Line 2: x=3s-1, y=3s-1, y=2s, z=6s-1
For the lines to intersect we need to first find the value of t and s that satisfy the equation
simultaneously. Let equate the x, y, z co ordinate
2t+4=3s-1
2t+4=3s-1……..eq1
t+3=2s…..eq2
2t + 7=6s-1…..eq3
From the second equation
t=2s-3
Substitute into equation 1
2[2s-3]+4=3s-1
4s-6+4=3s-1
4s-2=3s-1
4s-3s=-1+2
S=1
Now substitute S=1 into t=2s-3
t=2s-3
t=2[1]-3
t=2-3
t=-1
Let check if these value satisfy the third equation
2t+t=5
2[-1]+7=5
-2+7=5
5=5
The third equation is satisfied therefore the lines intersect
To find the point of intersection
Substitute t=-1 into the parametric equation for line1[ or s-1] into line two
X=2[-1]+4=2
Y=-1+3=2
Z=2[-1]+7=5
The point of intersection is [2,2,5]
To find the angle between the lines, let Ɵ be the angle between the directions vectors V1 and V2 is given
�1−�2
by Cos Ɵ=|�1||�2|
V1×V2= (21 )× (32 )
2
6
=[2×3] + [1×2] +[2×6]
=6+2+12=20
|V1|=
�2 + �2 + �2 =
|V2|=
�2 + �2 + �2 =
20
20
Cos Ɵ= 3×7 = 21
22 + 12 + 22 = 9= 3
32 + 22 + 62 = 49 = 7
20
Cos Ɵ = 21
Cos Ɵ= 0.9524
Ɵ=���−1 [0.9514]
Ɵ=17.7°
The angle between them is 17.7°
QUESTION TWO(2)A
SOLUTION
Standardized math series
[I] 90th percentile
Using the z-table
We find the z-score for the percentile is approximately 1.28
Use the z-score formula
[x−Ɲ]
Z= σ
Z=1.28, x=1, Ϭ=500
�−500
1.28= 100
1.28x100=x-500
128=x-500
X=128=500
X=628
Approximately 628 is the 90th percentile score
[ii] Third quartile
From the z-table
The third quartile corresponds to 75th percentile using z-table
From the z-score formula for the 75th percentile score is approximately 0.67
Using the z-score formula again
[x−Ɲ]
Z= σ
Z=0.67, x=1, Ɲ=500, Ϭ=100
0.67×
�−500
100
0.67×100=x-500
67=x-500
X=67+500
X=567
Approximately 567 is the 75th percentile score
QUESTION TWO(2) B
SOLUTION
The average arrival rate is 15 cars/hour
=[15 cars/hours] × [1hours/60 minutes] ×5minutes
=1.25 cars per 5 minutes
This is a poison distribution
P[x=15] =
�� ×�−�
k!
K=1car in this case
λ=1.25cars per 5 minutes
e=Euler numbers=2.7128
[i] exactly one car in five minutes
P[x=1]=
=
1.25× �−1.25
1!
1.25×0.2865
1
P[x=1] =0.3561
[ii] Fewer than two arrivals in 15minutes
Convert this averages arrivals rate to cars per 15 minutes
[15cars/hours] × [1hours/60 minutes] × 5minutes
=3.75cars per 15 minutes
P[x<2]= p[x=0]+ p[x=1]
Using the poison distribution formula
P[x=0] =
P[x=1] =
3.75◦ × �−3.75
=0.0235
0!
3.75◦ × �−3.75
=0.0881
1!
P[x=2]=p[x=0] + p[x=1]
=0.0235+0.0881
P[x=2] =0.1116
The probability of fewer than two arrivals in 15 minutes is = 0.116
QUESTION THREE(3)A
SOLUTION
For a probability density function the integral over is entire range must equal to 1
[a] Solving the integrals
�1
1
[2K x62 ] + 2� [3-X] + X=1
Substitute the limits of integration
1
1
3
1
3
1
[2 k[1]2 - +2K(0)2 + (2 K(3)-4k[3)2 -2k[1] - 4k[1)2 ]=1
Simplify and solve for k
2k+18k+9k−6k+k
=1
4
6�
=1
4
6� 4
=
6 6
4 2
K=6 =3
2
K=3
[b] Expectation E[x]
The expectation is calculated as
E(x)=∫x.f[x]dx
1
3
3
1 2
E[x]= 0 � × 2 × dx + 1 �( 2)(3)(3-x)dx
1
1
2 �3
1 3�2
2
3
E[x]=3 0 �2 dx+3 1 �(3 − �)��
�3
E[x]=3[ 3 ]10 +3[ 2 - 3 ]31
Sub the limits
2 1
1 27 27
2
1
E[x]=[3][3-0]+[3][ 2 - 3 ]-[3-− 3]
2 20
E[x]=3+18
4 20
E[x]=18+18
24
E[x]=18
4
E[x]=3
[c] Expectation [EY]
Use the linearity of expectation
1
E[y]=E[18�2 - 12x - 2]
1
E[y] = 18E[�2 ] - 12E[x]- 2
Calculate E[x^2]
2
1
E[�2 )=[10 �2 (3 �)dx + [31 �2 ( 3 (3 − �)]
1
1
1
=[6 �4 ]10 + [3 �3 − 12 �4 ]31
1
27
1
1
=6 +(9- 4 ) − ( 3 − 12 )
1
9
1
14
7
+ − = =
6
4
4
6
3
Substitute into the equation for E{y]
7
13
1
E[y] = 18[3] - 12[18]- 2
252−52−3
6
E[y]=
197
E[y]= 6
[d] standard deviation
We need to calculate the variance
Var[x]=E[�2 ]-[E(x)]2
7
13
7
169
Var [x]=3 − ( 18 )2
Var [x]=3- - 324
756−169
324
Var [x]=
587
Var[x]=324
1
(e) Pr(|x-1|<2)
Absolute vale the inequalities
|x-a|<b means a-b<x<a+b
1
Solve for PR(|x-1|<2)
1
1
1
3
|x-1|<2 means 1-2<x<1+2<x<2
Calculate the probability
3
1
3
1
Pr(½<x<2)= ½ ⅔��� + 12 3 (3 − �)��
Evaluate the integral
�2
1
�2
2
⅔( 2 )11 +3(3X- 2 )31
2
1 1
+( ) =0.583
4 5
QUESTION THREE(3)B
SOLUTION
�
P[x=x]= { 3|x|}
-3≤x≤3
The values are -3, -2, -1,0,1,2,3,
Ep (x-x)=1 for X=-3, -2, -1, 0, 1, 2, 3
�
Let plug in the values of x 3 (| − 3| + | − 2| + | − 1| + |0| + |1| + | + |2| + |3|) = 1
�
(3+2+1+0+1+2+3)=1
3
�
(12)=1
3
12�
=1
3
1
Calculate the variance(σ2)
σ2 =E(�2 ) - (E(�)2
σ =E(�2 ) - Ɲ2
K= 4
First let find E(�2 )
3
�2 ×P(X=X)
�=3
E(�2 ) =
=
=
1
3
�2 × |x|
�=3
12
(�2 ) = 6
Therefore the variance is
σ=6
Calculate the standard deviation
The standard deviation is the square root of the variance
σ= �2
σ= 6
σ=2.449
Mean Ɲ=0 and standard deviation
(Q4)
To find the fourier series
� −2<�<−1
F(t)=｛ �� −1<�<1
1<�<2
Determine the period
The function defined on the interval (-2,2)
L = 2-(-2)=4
2�
2� 2� �
T = L = 4, Ɯ= � = 4 = 4 =2
Cheek for symmetry
F(t)=f(t)
Bn=0
2
�
2
4
(0,2)�(�) dt
Calculate �0 =(�) (0, 2) f(t)dt
�0 =
1
�0 = 2[ (0,1)��� + (1,2)���
1
�0 = 2[kt(0,1)+0)
1
2
�0 = [k(1)-K(0]
�0 =
�
2
Let calculate again for ��
2
�
�� =( � ) (0, 2)ft cos (n(1)t)
2
���
1
���
�� +
2
��=( ) (0,2)�(�) ���( )��
4
2
��=2( (0,1)� ���
�
���
�� =(2 ) (0,1) cos( 2 )dt
�
2
���
(1,0)���( 2 )dt]
���
�� =(2 )[ (��[ sin( 2 )-(0,1) ]
��=(
�
��
)[ ( sin( 2 )-sin (0)]
��
�
��
��=(�� ) sin( 2 )
�
� � = 2� +
�
4
� � = +
∞
�=1
∞
�� ��� ���
�
��
�=1
��
���
sin 2 ) cos( 2
QUESTION FIVE(5) A
SOLUTION
Let prove the vector identity
Define the vector and it’s magnitude
R=xi+yj+zk
Magnitude |r|= �2 + �2 +�2
Express �� in the terms of x,y and z
�
��=(�2 +�2 +�2 )2
Apply the operations ∇ to ��
�
�
�
∇=��I +��j +��k
�
�
�
�
=(��I +��j +��k)((�2 +�2 +�2 )2
Now I calculate the partial derivative
(�� ) � 2 2 2 �
= (� +� +� )2
�� 2
11
nx(�2 +�2 +�2 )( 2 )−1 =n�11−2
(�)�
(�)�
J �� =ny��−2 and J �� =nz��−2
Now I combine partial derivatives to find the gradient
∇(rn)=nx��−2 I +ny��−2 j +nz��−2 k
When I factor out and identify the positive vector,
I notices that (xi +yj+2k) is simply the position vector r
So my final result is
∇(rn)=n��−2 r
Our answer have prove the vector identity
∇(rn)=n��−2 r=∇rn=n��−2 r is prove
QUESTION FIVE(5)B
SOLUTION
When we are proving irrotational and solenoidal properties
Our part one says Q∇4 is irrotational if curl is zero(∇xf = 0)
If we say let calculate for the curl of 4∇4
Curl of scaler times vector=
∇x(4A)=∇4 ×A+4(∇ ×A)
A=∇4
∇ ×(4∇4)=∇4 × ∇4+Q(∇ × 4)
∇4 × ∇4=0
Our curl of a gradient is also zero
∇x∇4=0
∇x(4∇4)=0
But since Our curl of 4∇4 is zero it is irrotational
Part 2
∇4x∇4 to solenoidal
Vector field is solenoidal if it is diver-gem is zero (∇f = 0)
We calculate the diver-gem of ∇4 × ∇4
So ∇ × (∇4 × ∇4)=∇4 ×( ∇ × ∇4)-∇4 ×(∇ × ∇4)
The curl of a gradient is always zero
( ∇ × ∇4 =0 and ∇ × ∇4=0)
So since our divergem of ∇4 × ∇4 is zero its solenoidal
There fore 4∆4 is irrotational because ∇ × (4∇4)=0 and (∇4 × ∇4) is solenoidal because of
∇ × (∇4 × ∇4)=0