Analytical Methods for Engineers: Algebraic Methods TMA 1 (v2.1)

This document presents a solved assignment for the 'Analytical Methods for Engineers' module, focusing on algebraic methods. It includes detailed solutions to five questions covering topics such as transposing formulas, solving equations using logarithms, finding values in given equations, polynomial long division, and applying quadratic formulas to solve problems related to motion and geometric progressions. The assignment provides step-by-step explanations and calculations, making it a valuable resource for students studying similar concepts. This solved assignment is available on Desklib, a platform offering a wide range of study tools and resources for students.

University
HNC/HND ELECTRICAL AND ELECTRONICS ENGINEERING
By
Your Name
Date
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© <Your Name> 2018

ANSWERS TO QUESTIONS
QUESTION 1
Part A
Problem statement
Given the formula:
f = uv
u+v
We are required to transpose it and change its subject to v.
Solution
Step1
Multiply the right hand side and the left hand side with the denominator on the right hand side to
eliminate the denominator (Watts, 2012).
f (u+ v)= uv
u+ v (u+ v )
f (u+ v)=uv
Step2
Open the brackets of the above equation through expansion.
fu+ fv=uv
Step 3
Collect the like terms.
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fu=uv−fv
Step 4
On the right hand side of the equation, we factor out v out of the brackets.
fu=v (u−f )
Step 4
Divide both sides of the equation by u-f.
v= fu
u−f
Part B
Problem Statement
Solving of the equation:
( 3.4 ) 2 x+3=8.5
Solution
Step 1
Take the logarithms of both the left hand side and the second hand side and multiply the
superscript on the left hand side with the log of the value on the left (Chapra and Canale, 2015).
2 x+3 ( log3.4 ) =log8.5
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Step 2
Divide both the logs as below:
2 x+3= log 8.5
log 3.4
2 x+3=1.7487
Step 3
Solve to find the value of x
2 x=1.7487−3
2 x=−1.2513
x=−0.6256
Part C
Problem Statement
To find the value of t in the equation below given θ=58, V =255, R=0.1 and L=0.5.
θ=V e
−Rt
L
Solution
Step 1
Make t the subject of the formula as below:
θ
V =e
−Rt
L
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−Rt
L =ln ( θ
V )
t=−L
R X ln ( θ
V )
Step 2
Substitute the given into the equation above
t=−0.5
0.1 X ln ( 58
255 )
Solving the above yield t=7.404
Part D
To find L in the equation below given ω=-2.6, Lo =16 and h=1.5.
ω= 1
h ln ( L
Lo
−1
)
Solution
Step 1
Make L the subject of the formula as below:
ωh=ln ( L
Lo
−1 )
ewh=( L
Lo
−1 )
( ewh . Lo =L−Lo )
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( L=ewh . Lo+Lo )
Step 2
Substitute the values given above into the equation where L is the subject
L=e
(−2.6 x1.5 ) . 16+16
L=16 ( 0.0202 ) +16
Solving the above yields L=16.3239.
QUESTION 2
PART A
Problem Statement
To use polynomial long division to find the quotient when
3 x3−5 x2 +10 x +4 is divided by 3 x +1
Solution
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Hence the quotient is x2−2 x+ 4
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Part B
Problem Statement
We show by polynomial long division that
x3 −3 x2 +12 x−5
x−2 = ( x2−x +10 ) + 15
x−2
Solution
The quotient is x2−x +10while the remainder is 15. Therefore, the complete answer as a result of
division is ( x2−x +10 ) + 15
x−2
QUESTION 3
Given the equation below, the initial speed as 72km/h, the gravitational acceleration as 10m/s2
and the height of the building as 125m, we are required to find the time when the ball travels
specific distances.
s=Uo t+ 1
2 g t2
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Part A.
The time for the ball to drop to a fifth of the height of the building.
Solution
A fifth of the height of the building is equivalent to:
h1
5
=125−(1
5 x 125)=100 m.
Convert the speed in km/h to m/s as below:
s=72 x 1000
3600 =20 metres per sec
Substitute the values into the equation.
100=20t + 1
2 x 10 t2
100=20t +5 t2
Rewriting the equation in quadratic form, we get
5 t2 +20 t −100=0
Apply the quadratic formula to solve for t. The quadratic formula is given by:
t=−b ± √b2−4 ac
2 a
Therefore,
t=−20 ± √400−−20 00
10
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t=−20 ± √24 00
10
t=−20 ± 48.99
10
t1=−20+48.99
10 =2.8 s
t2=−20−48.99
10 =−6.9 s
Since t cannot be negative we take the positive t as the answer hence time taken by the ball to
reach a fifth of the building t=2.8s.
Part B
The time for the ball to reach the ground.
Solution
125=20t + 1
2 x 10 t2
125=20t +5 t2
Rewriting the equation in quadratic form, we get
5 t2 +20 t −125=0
Apply the quadratic formula to solve for t. The quadratic formula is given by:
t=−b ± √b2−4 ac
2 a
Therefore,
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t=−20 ± √400−−2500
10
t=−20 ± √2900
10
t=−20 ±53.85
10
t1=−20+53.85
10 =3.3855
t2=−20−53.85
10 =−7.385
Since t cannot be negative we take the positive t as the answer hence time taken by the ball to
reach the ground t=3.3855s.
QUESTION 4
Problem Statement
To find the number of hours that a tank of 16m by 9m by 9m dimension will take to fill when it
fills at 150 litres in the first hour, 350 litres in the second hour and 550 litres in the third hour.
Solution
We let L be the terms, a be the first term in the series, n the nth term in the series, d the common
difference, and s the sum (Bird and May, 2014). The sum is found by finding the volume of the
tank and converting it into litres as below
s= LXWXHX 1000
S=16 X 9 X 9 X 1000=1296000 L
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The common difference is given by:
d=a2−a1∨a3−a2
d=350−150=200
Now,
Sn= n
2 [ 2 a+ ( n−1 ) d ]
1296000= n
2 [ 2(150)+ ( n−1 ) 200 ]
1296000= [ 150 n+n ( n−1 ) 100 ]
1296000= [150 n+100 n2−100 n ]
1296000= [50 n+100 n2 ]
Rewriting into quadratic form we have
100 n2 +50 n−1296000
Apply the quadratic formula to solve for t. The quadratic formula as below:
n=−b ± √b2 −4 ac
2 a
Therefore,
n=−50 ± √2500−−518400000
200
n=−50 ± √ 518402500
200
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n=−50 ± 22768.45
200
n1=−50+22768.45
200 =113.59 hrs
n2 =−50−22768.45
200 =−114.09 hrs
Since time cannot be negative, we take the positive time. Therefore, it will take 113.59hrs for the
tank to fill.
QUESTION 5
PART A
Problem Statement
We need to find the number of persons that will be employed by the firm in the 20th week if the
current rate of expansion continues at 6% per week.
Solution
The initial rate assuming no increase is 100%. With increase in rate at 6%, the new rate becomes
106% (1.06 as fraction). The number of employees employed by the 20th week is given by:
N20 th=a rn
N20 th=110 x 1.0620=352.7 ≅ 353 employees
Hence the firm will hire 353 by the 20th week.
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PART B
Problem Statement
To show that the annual profit the contractor gets form a geometric progression and find the total
of all the profit for the first 5 years given that in the first hiring year the profit was £6000 and
diminishes by 5% in successive years.
Solution
In geometric progression, the proceeding term is determined by multiplying the preceding term
by a common ratio r (Alldis and Kelly, 2012). This is as in the formula below:
an=a1 . rn−1
Now, using the first year profit and using the depreciation rate of 0.05, we get:
6000=6000 x 0.951−1
£ 6000=£ 6000
Hence, the profit follows a geometric progression.
The sum of profit for the first five years is given by:
Sn= a1 (1−rn )
1−r
Sn= 6000(1−0.955 )
1−0.95
Sn=£ 27146.28
References
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Alldis, B. K., and Kelly, V. A. (2012). Mathematics for technicians. 7th ed. North Ryde, N.S.W:
McGraw-Hill Education.
Bird, J. O., and May, A. J. (2014). Mathematics for electrical technicians. 2nd ed. London:
Routledge.
Chapra, S. C., and Canale, R. P. (2015). Numerical methods for engineers. 7th ed. New York,
NY: McGraw-Hill Education.
Watts, R. G. (2012). Essentials of applied mathematics for engineers and scientists. 2nd ed. San
Rafael, Calif.: Morgan & Claypool Publishers.
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