HNCB036 Applied Mathematics For Engineers

HNCB036 APPLIED MATHEMATICS FOR ENGINEERS
ASSIGNMENT
Student Name
[Pick the date]

Task 1
Question 1
(a) y=6 sin ( t −450 )
Amplitude = 6
Period ¿ 3600
1 =360 °
Sketch
(b) y=4 cos (2 θ+30 °)
Amplitude = 4
Period ¿ 3600
2 =180°
It can be seen that y=4 cos (2 θ+30 °) leads
y=4 cos 2 θ by 30 °
2 =15 °
Sketch
1

(c) Prove the identity
sin2 x ¿ ¿
LHS
¿ sin2 x ¿ ¿
¿
sin2 x ( 1
cosx + ( 1
sinx ) )
cos x sin x
cos x
¿ sinx( 1
cosx
+1
sinx )
¿ sin x (¿ sinx+ cos x
cos x sin x )¿
¿ sinx+ cos x
cos x
¿ tan x +1
¿ 1+ tan x
RHS
2

¿ 1+ tan x
Hence, the identity is provided LHS =RHS
(d) sin ( x+ π
3 )+sin ( x+ 2 π
3 )= √3 cos x
LHS ¿ {cos ( x )sin ( π
3 ) +cos ( π
3 ) sin ( x ) }+{cos ( x ) sin ( 2 π
3 ) + cos ( 2 π
3 ) sin ( x ) }
Here,
sin(¿ 2 π
3 )= √3
2 ¿
cos ( 2 π
3 ) =−1
2
sin ( π
3 )= √3
2
cos ( π
3 )= 1
2
¿ { √3
2 cos ( x )+ 1
2 sin ( x ) }+ { √3
2 cos ( x )− 1
2 sin ( x ) }
¿ √3
2 cos ( x ) + 1
2 sin ( x ) + √3
2 cos ( x )− 1
2 sin ( x )
¿ 2∗√ 3
2 cos ( x )
¿ √ 3 cos( x )
RHS ¿ √ 3 cos( x )
Hence, the identity is provided LHS =RHS
3

Task 2
Question 2
Gaussian elimination
5 T1 +5 T 2+5 T 3=7
T 1+2 T 2+ 4 T 3=2.4
4 T1 +2 T2 +0 T 3=4
Matrix form
4

Hence,
T 1=0.8
T 2=0.4
T 3=0.2
5

Question 3
x2+ 3 x −5=0
Let
f ( x )=x2 +3 x−5=0
1st iteration
f ( 1 ) =−1<0∧f ( 2 )=5> 0
Roots between 1 and 2
xo= 1+2
2 =1.5
f ( xo ) =f ( 1.5 ) =1.75>0
2nd iteration
f ( 1 ) =−1<0∧f ( 1.5 )=1.75> 0
Roots between 1 and 1.5
x1= 1+1.5
2 =1.25
f ( x1 ) =f ( 1.25 )=0.3125> 0
3rd iteration
f ( 1 ) =−1<0∧f ( 1.25 )=0.3125> 0
Roots between 1 and 1.25
6

x2= 1+1.25
2 =1.125
f ( x2 ) =f ( 1.125 )=−0.35938<0
4th iteration
f ( 1.125 )=−0.35938<0∧f ( 1.25 ) =0.3125>0
Roots between 1.125 and 1.25
x3= 1.125+1.25
2 =1.1875
f ( x3 ) =f ( 1.1875 ) =−0.02734<0
5th iteration
f ( 1.1875 ) =−0.02734< 0∧f ( 1.25 ) =0.3125> 0
Roots between 1.1875 and 1.25
x4 =1.1875+1.25
2 =1.21875
f ( x4 )=f ( 1.21875 ) =0.1416> 0
6th iteration
f ( 1.1875 ) =−0.02734< 0∧f ( 1.21875 ) =0.1416> 0
Roots between 1.1875 and 1.21875
x5= 1.1875+1.21875
2 =1.20312
f ( x5 ) =f ( 1.20312 ) =0.05688> 0
7th iteration
f ( 1.1875 )=−0.02734<0∧f ( 1.20312 )=0.05688>0
7

Roots between 1.1875 and 1.20312
x6= 1.1875+1.20312
2 =1.19531
f ( x6 ) =f ( 1.19531 )=0.01471> 0
8th iteration
f ( 1.1875 )=−0.02734<0∧f ( 1.19531 )=0.01471>0
Roots between 1.1875 and 1.19531
x7= 1.1875+1.19531
2 =1.19141
f ( x7 ) =f ( 1.19141 ) =−0.00633>0
9th iteration
f ( 1.19141 ) =−0.00633<0∧f ( 1.19141 ) =0.01471>0
Roots between 1.19141 and 1.19531
x8= 1.19141+1.19531
2 =1.19336
f ( x8 ) =f ( 1.19336 ) =0.00418>0
10th iteration
f ( 1.19141 ) =−0.00633<0∧f ( 1.19336 ) =0.00418>0
Roots between 1.19141 and 1.19336
x9= 1.19141+ 1.19336
2 =1.19238
f ( x9 ) =f ( 1.19238 ) =−0.00107>0
11th iteration
8

f ( 1.19238 )=−0.00107< 0∧f (1.19336 )=0.00418>0
Roots between 1.19238 and 1.19336
x10=1.19238+1.19336
2 =1.19287
f ( x10 )=f ( 1.19287 )=0.00155>0
12th iteration
f ( 1.19238 )=−0.00107< 0∧f (1.19287 )=0.00155>0
Roots between 1.19238 and 1.19287
x11=1.19238+1.19287
2 =1.19263
f ( x11 )=f ( 1.19263 ) =0.00024 >0
Summary Table
It can be said that the root of the equation x2+3 x −5 using bisection method is 1.19.
Question 4
Critical speed of oscillations = x
9

3 x3−10 x−14
Newton Rapson Method
Let
f ( x )=3 x3−10 x−14
f ' ( x )=9 x2−10
Let the initial guess is x1=1
xn+1=xn − f ( xn)
f ' ( xn )
 n = 1
x1+1=x1− f ( x1 )
f ' (x1 )
x2=1− 3 ( 1 )3− ( 10∗1 )−14
9 ( 1 )2−10 =−20
 n =2 23814
x2+1=x2− f ( x2)
f ' ( x2)
x3=−20−3 (−20 )3− ( 10∗(−20 ) ) −14
9 (−20 )2−10 =−13.34
In a same way, the iteration can be performed and the table is shown below:
10

11

Hence, the critical speed of oscillations x would be 2.313.
Part 2
Task 3
Question 5
Rectangular box shape
Minimum surface area =?
Volume = 250 m3
Figure
Let the volume of marque is V and surface area is S.
V =xyz =250
S= xy+ yz+ 2 zx
Now,
z= 250
xy
Let put the value of z in S
S= xy+ y (250
xy )+2 x ( 250
xy )
12

S= xy+ 250
x + 500
y
S= xy+250 x−1 +500 y−1
Let take the derivative of S with respect to x and y.
∂ S
∂ x = y− 250
x2 … … … … … ..(1)
∂ S
∂ y =x −500
y2 … … … … … ..(2)
At any stationary point
∂ S
∂ x =0 , ∂ S
∂ y =0
Hence,
y− 250
x2 =0
y= 250
x2
x2 y =250 …… … ….(3)
And,
x− 500
y2 =0
x= 500
y2
xy2=500 … … … … ..(4)
Let’s solve equation (3) and (4)
13

x2 y
xy2 =250
500
x
y =1
2
y=2 x
Put the value of y in equation (3)
x2 (2 x )=250
2 x3 =250
x=5 m
Hence,
y=2 x=2∗5=10 m
The volume V =xyz
250=5∗10∗z
z=5 m
Further,
Minimum surface area S= xy+ yz+ 2 xz
¿ ( 5∗10 ) + ( 10∗5 )+ ( 2∗5∗5 )
¿ 150 m2
Therefore, the minimum surface area of the canvas is 150 m2 .
Question 6
Base radius = 5 cm
14

Height = 12 cm
The radius is increasing = 5 mm/sec
The height is decreasing = 15 mm/sec
Rate of change of total surface area of right circular cone =?
Here,
Total surface area of cone = A
A=π rl+ πr2
l= √(r ¿¿ 2)+(h¿¿ 2)¿ ¿
A=¿
The rate of change of surface area
15

¿ ( 78.298∗0.5 ) + ( 14.50∗(−1.5 ) )
¿ 17.4 cm2 /sec
Rate of change of total surface area of right circular cone would be 17.4 cm^2/sec.
Question 7
Area between the two curves
y
x2 =1∧ y+ x2=8
Rotated about 3600 about x- axis
Volume produced =?
The equation of x- axis is y = 0
Hence
y
x2 =1
y=x2
Same way,
16

y + x2=8
y=8−x2
Now,
x2=8−x2
2 x2 =8
2 x2 −8=0
x2−4=0
( x−2 ) ( x+ 2 )=0
x=2 ,−2
Volume ¿∫
−2
2
π ¿ ¿
¿∫
−2
2
π ( x ¿¿ 4 ¿+64+ x4)dx−∫
−2
2
π ( x¿ ¿ 4¿) dx ¿ ¿¿ ¿
¿ π ∫
−2
2
64−16 x2 dx
17

V =170 ( 2
3 π )unit3
Therefore, the volume would be 170 ( 2
3 π )unit3 .
Question 8
Differential Equation
dy
dx =3− y
x
Initial conditions
x=1 , y=2
Range
x=1 ¿ 1.5
Interval
h=0.1∨ 1
10
18

Euler’s Method
Step 1
Step 2
Step 3
19

Step 4
Step 5
20

Summary table
Hence, the solution is 2.5714.
Graph
Task 4
Question 9
21

Probability of success p = 0.07
Probability of failure q = 1 – 0.07 = 0.93
Number of blocks n = 9
(a) Probability that three blocks will fail to meet the requisite specification
x=3
P(x=3)= n !
x ! ( n−x ) ! px ¿
¿ 9 !
3! ( 9−3 ) ! (0.07)3 ¿
¿ 0.01864
(b) Probability that less than four blocks will fail to meet the requisite specification
x <4
P ( x< 4 ) =P ( x =0 ) + P ( x=1 ) + P ( x=2 ) + P (x=3)
¿ { 9!
0 ! ( 9−0 ) ! ( 0.07 )0 (1−0.07 ¿¿¿ 9−0 }+{ 9!
1 ! ( 9−1 ) ! ( 0.07 )1 ¿
P ( x< 4 ) =0.99983
22

Question 10
Poisson distribution
Mean value γ = 0.7
Probability of 0, 1 2, 3, 4 and 5 failure in a week =?
P= e−γ γ x
x !
P ( x=0 ) =e− γ γ x
x ! = e−0.7 γ 0
0 ! =0.497
P ( x=1 )= e−γ γ x
x ! = e−0.7 γ1
1 ! =0.844
P ( x=2 )= e−γ γ x
x ! =e−0.7 γ2
2! =0.966
P ( x=3 )= e−γ γ x
x ! = e−0.7 γ3
3! =0.99
P ( x=4 )= e−γ γx
x ! =e−0.7 γ4
4 ! =0.999
P ( x=5 )= e−γ γ x
x ! = e−0.7 γ5
5! =1.000
23

0 1 2 3 4 5
0.000
0.200
0.400
0.600
0.800
1.000
1.200
0.497
0.844
0.966 0.994 0.999 1.000
Histogram
Number of Failures
Probability
Question 11
The independent variable = Time
The dependent variable = Force
Regression equation
y=a+bx
24

a= ( 107.2 )∗( 1.4056 )− ( 3.06 )∗(44.549)
{7∗(1.4056)2− ( 3.06 )2 } =30.194
b=7∗44.549− ( 3.06 )∗(107.2)
{7∗(1.4056)2− ( 3.06 ) 2 } =−34.039
Hence,
Regression equation
y=30.194 +(−34.039∗x )
y=30.194−34.039 x
Force ( N ) =30.194−(34.039∗Time)
25

26