Algebraic Methods for Analytical Methods for Engineers

This text covers various algebraic methods used in analytical methods for engineers. It includes solved examples for formula manipulations, calculations, polynomial division, quadratic equations, and geometric progressions. The text also mentions the subject, course code, course name, and university.

MODULE TITLE: ANALYTICAL METHODS FOR ENGINEERS
TOPIC TITLE: ALGEBRAIC METHODS
TUTOR MARKED ASSIGNMENT 1 (v2.1)

1. Formula manipulations and calculations
a) Making v the subject of the formula:
f = uv
u+v multply by ( u+v ) both sides fu+fv=uv rewrite fu=v ( u−f ) divide by ( u−f ) both sides
v= fu
u−f
b) Finding the value of x in:
3.42 x+3=8.5ln ( 3.42 x+3 ) =ln ( 8.5 ) ( 2 x+3 ) ln ( 3.4 ) =ln ( 8.5 )2 x+3= ln ( 8.5 )
ln ( 3.4 ) x=
ln ( 8.5 )
ln ( 3.4 ) −3
2
¿−0.6256295444
c) Finding the value of t, when θ= 58, V= 255, R= 0.1 and L= 0.5 in;
θ=V e
−Rt
L Make t the subject of the formulae
− Rt
L = θ
V
−R
L t=ln ( θ
V )t=
ln ( θ
V )
−R
I
=
ln ( 58
255 )
−0.1
0.5
∴ t=7.404102673
d) Finding L, if ω= –2.6, L0= 16 and h= 1.5, in:
ω= 1
h ln ( L
L0
−1 )make Lthe subject of the formula ln ( L
L0
−1 )=ωh L
L0
−1=eωh
L=L0 ( eωh+1 )
∴ L=16 ( e−2.6 × 1.5 +1 ) =16.32387058
2. Division of a polynomial.
a) Determining the quotient using long division
3 x3−5 x2 +10 x+4
3 x+1 ∴ 3 x3−5 x2+10 x +4
3 x +1 =x2−2 x+ 4
b) Proofing

x3 −3 x2 +12−5
2−2 = ( x2−x+10 ) + 15
x−2 ∴ x3−3 x2 +12−5
2−2 = ( x2−x +10 ) + 15
x−2
3. Given:
s=u0 t + 1
2 g t2u0 =72 km
h s=125 m, g=10 m
s2 u0 =72 km
h ×
1000 m
1 km
3600 s
1 hr
=20 m
s
a) A ball to drop to a fifth of the height of the building, this means s= 25m
Solving for t,
25=20 t+ 1
2 ×10 t2
−5 t2−20 t+25=0 Solve the quadratic equation
t=− ( −20 ) ± √ ( −20 ) 2− ( 4 ×−5× 25 )
2× 5 t=1∨t=−5t=1 s
Time cannot be negative, thus the time taken for the ball to fall a fifth of the building is 1
second.
b) For the ball to fall to the ground, we solve as part a) above taking into account that
s=125. Thus:
−5 t2−20 t+1 25=0 Solve thequadratic equationt=− (−20 ) ± √ (−20 )2− ( 4 ×−5× 125 )
2×5
t=−2+ √ 29∨t=−2− √ 29 t=−2+ √ 29=3.385164807 s
Time cannot be negative, thus the time taken for the ball to fall to the ground of the
building is 3.385164807 second.
4. Rate of filling the tank is 150 litters during the first hour, 350 litters in the second hour and
550 in third hour and so on:

Volume of tank , V =16× 9 ×9=1296 m3V =1296 m3 ×1000 l
m3 =1296 000 lS= ( n
2 )¿
1296000= ( n
2 ) ( 150+ ( 150+n−1 ) 200 ) ¿1296000=n (100 n+50 )
n2 +n−25920=0; Solving the quadratic equation gives:n=113.5922703∨n=−114.0922703
n=113.5923 hours
Since, time cannot be negative, it will take approximately 113.5923 hours to fill the tank
5. a). The question is a Geometric progression with a=110 employees, r=6%, therefore the total
number of employees on the 20th week is:
E ( 20 )=110× ( 1+0.06 )20¿ 352.784902
Thus, the number of employees on 20th week is approximately 353 employees.
b). the profit is 6000 which diminishes by 5% each successive year.
It is a geometrical progression with r=0.95, a=6000, n=5
Sn= a|1−rn
|
1−r S5= 6000 |1−0.955
|
1−0.95 =27146.2875
The total profit after 5 years is ₤ 27146.29