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Analytical MethodsDilshad Mahmoud
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PROBLEM 1
Polynomial Division
(i)
82
8672
234
++
+++
xx
xxxx
867234 +++ xxxx
The Top row is the dividend.
822 ++ xx
The Bottom row is
the divisor.
i) X4+X3+7X2-6X+8X2+2X+8
X2-X+1X2+2X+8X4+X3+7X2-6X+8
X4+2X3+8X2
-X3-X2-6X+8
-X3-2X2-8X
X2+14X+8
X2+2X +8
12X
Simplified Result is
=X2-X+1+12XX2+2X+8
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Dividing the first term of the dividend by the first of the divisor, gives us2x
,
which is put above the first term of the dividend as show. Then the divisor is
multiplied by2x
, which is placed under the dividend as show, subtracting gives
us23 xx , the process is then repeated until the remainder, on subtraction is
zero, which completes the process.
So as we can see above:
828672234 +++++ xxxxxx
is equal to
12 + xx
We can check to see if our answer is correct by multiplying the divisor by the
answer:
( 822 ++ xx
( 12 + xxThis is equal to
867234 +++ xxxx
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(ii)
1
333107234
++
x
xxxx
333107234 ++ xxxx
Is the dividend and
1xis the divisor therefore:
337
3331071
23
234
+
++
xx
xxxxx
34
77 xx
023
33 xx +
23
33 xx +
0
33 x
33 x
0
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The same process as the question (i) is repeated until we end up:
333107234 ++ xxxx
1x=
33723 + xx
The function can again be check by multiplying the divisor with the answer.
Thus,
( )1x ( )337 24 + xx=
333107234 ++ xxxx
Hence,
Simplified Result is
=7X3-3X2+3+X-1
PROBLEM 2
Partial Fractions
Resolve the following into Partial Fraction:
I.
)3+(
3+7+2
2
SOLUTION: Integrate X2+7X+3X2X+3
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Notice that the linear factor (x) is repeated 3 times and the factor (x
1) is repeated 2 times. So, we set up the following partial fractions:
X2+7X+3X2X+3AX+BX2+CX+3
Now if I multiply throughout by the denominator of the left hand sideI will get:
X2+7X+3AXX+3+BX+3+CX2
There are two ways of finding A, B and C. either multiply out theRight hand side and arrange terms in X2, x and a constant term. ThenEquate the coefficients ofX2on each side, equate coefficients of xOn each side and finally equate the constant term on both sides. ThisGives three equations from which you can find A, B and C.
A second method is to give x any convenient value, and this will giveAn equation connecting both sides. Give x another value and get aSecond equation, and then, if necessary, a third value to get a thirdEquation. Often a mixture of the two methods is the quickest way toFind A, B and C.
If we use x = 0, then the left-hand side equals 3 and the right-handSide equals 3B, so B = 1.
Put in x = -3, then the left hand side equals -9 and the right hand sideEquals 9C, so C = -1.
Finally, the coefficient of x^2 on the left is 1, and on the right
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A+C, so we can put:
A+C = 1A = 1-CA = 1 - (-1)A = 2
So the fraction can be written 2X+1X2-1X+3
II.
)6_+(
8+9+2
2
We can decomposer in to simpler parts:
In this problem the nominator has a same degree with the denominator.
1
X2+X-6X2+9X+8
X2+ X-6 -
8X+14
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Simplified Result is
1+8X+14X2+X-6
III.
)2_x)(7+(
13__2
2
X2-X-13X2+7X-2 AX+BX2+7+CX-2
AX+BX-2+CX2+7X2+2X+1
Equation Numerator is X2-X-13= AX+BX-2+CX2+7
Let X = 2:
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Now we can find the value of C with this Equation.
22-2-13= C22+7
-11=11C C=-1
Now we can simplify the Equation Numerator is to find the value of A &B
X2-X-13= AX+BX-2+CX2+7
X2-X-13= AX2-2AX+BX-2B+CX2+7C
In term ofX2 A+C=1
A+-1=1
A=2
In term of X A+B=1
2+B =1
B=3
* X2-X-13X2+7X-2 2X+3X2+7+-1X-2
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PROBLEM 3
i) The current i amperes flowing in a time t seconds is given by
=
CR
t
et 10.8
,
Where the circuit resistance R is3
1025ohms and capacitance C is
61016
farads.
Determine: (a) the current i after 0.5 seconds and (b) the time, to the
nearest millisecond, for the current to reach 6.0A and (c) using Excel
produce a graph of current against time.
ii) The rate at which a body cools is given byte 05.0250 =where the
excess temperature of a body above its surrounding at time t minutes
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isC0
. Plot a graph showing the nature decay curve for the first hour
of cooling. Hence determine (a) the temperature after 25 minutes and
(b) the time when the temperatures is 195C.
Question (i)
a) Determination of current i.
i = 8.0
CR
t
e1
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i= 8.0
)1025)(1016(
5.036
1e
=
( )25.110.8 e
( )2865047969.010.8 ( )17134955203.00.8
Hence, i = 5.71 amperes
b) Determination of time to the nearest millisecond.
c) We know that:
=
CR
t
ei 10.8
We need to work the equation above by transposition to determine t, so wewill have,
CR
t
ei
=10.8
And
0.81
ie CR
t
=
So
= ieCR
t
0.8
0.8
= iCRt
0.8
0.8
ln
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Then finally we will have:
=
i
CRt
0.8
0.8ln
So when i = 6.0A we have,
( )( )
=
0.60.8
0.8ln10251016
36t
( )0.4ln4.0=t
( )386294361.14.0=t
Hence
st 5545.0=
And to the nearest millisecond this is equal to
ms554
d) Graph of current against time.
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Question (ii)
a) The temperatureafter25 minutes.
We have,
te 05.0250 =
( ) ( )2505.0250
= e
Hence
C071=
b) The time when the temperature is 195 C.
We know that:
te 05.0250 =and we also know
C0195=
So we will have:
te 05.0250195 =
te 05.0
250195 =
t05.0195
250ln =
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t=
05.0195
250ln
Hence
min5min969.4 =t
PROBLEM 4
Solve the following Hyperbolic Function:
a) A chain hangs so that its shape is of the form
=56
56x
chy
. Determine,
correct to 4 significant figures, (a) the value of y when x is 35, (b) the
value of x when y = 62.35.
b) Using Excel plot a graph of the above function and use it to verify your
answers.
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c) What is meant by the term `Catenary`? Also give brief examples of where
this is used in engineering.
Question a
a) The value of y when x is 35.
We have
= 5656
x
chy
, and we know that
35=x
So we will have:
=56
3556chy
8
556chy =
In our study we know that
2
xx eechx
+=
, that mean we will have now,
+=
256
85
85
eey
29.67=y
b) The value of x when y = 62.35.
We know
=56
56x
chy
, and
35.62=y
Thus,
=56
5635.62x
ch
56
35.62
56=
x
ch
and
1133.156
=
x
ch
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In our study
chx
can be written also as follow
+
2
xx ee
,
Thus,
1133.12
5656
=+
xx
ee
2266.25656 =+xx
ee
In our study we know that we can rearrange the equation above into the form
0=++ rqepe xxWhere
p
,
q
, and
r
are constants.
Then we will multiply each term byxe
which produces an equation of the form
( ) 02 =++ qreep xxThis can be solving as a quadratic equation.
So
02266.25656 =+xx
ee
02266.2 5656565656 =
+
xxxxxeeeee
012266.2 56
2
56 =+
xxee
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Our equation is on quadratic form
02 =++ cbxax
which gives us the following
form when solving
a
acbbx
2
42
=
then we will have now:
( ) ( ) ( )( )[ ]
( )12
114226.22266.22
2
9786.02266.2 =
Hence,56
x
e
= 1.6026 or 0.624
56
x = ln1.6026 or ln0.624
4716.04716.0
56= or
x
Therefore
( ) ( ) ( )( )4716.0564716.056 = orx
Hence
4096.264096.26 = orx
Thus x = 26.40 correct to four significant figures.
Question b
Graph plotting:
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Question c
The curve hanging flexible wire or a chain assumes when supported at its ends
and acted upon by a uniform gravitational force. Therefore cartenary which
derived from Latin word is a curve of a chain hanging under gravity.
The cartenary is given by the Cartesian equation: y = acosh(x/a)
Examples: an electric cable between two pots.
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PROBLEM 5
(a) An oil company bores a hole 80m deep. Estimate the cost of boring is the
cost is 30 for drilling the first metre with an increase in cost of 2 per
metre for each succeeding metre.
In our study this Arithmetic progressions can be solve by the equation as
follow:
( )[ ]dnan
sn 122
+=
With: n = 80; a = 30; d = 2;
Thus
( ) ( )[ ]21803022
80+=sn
( )185602
80+=sn
Hence
8720=sn
The increase in cost is 8720.
(b) Find the number of terms of the series 5, 8, 11 ... of which the sum is
1025.
We know that:
( )[ ]dnan
sn 122
+=
, with: a = 5; d = 3; and d =?
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Thus,
( ) ( )[ ]31522
1025 += nn
( )33102
1025 += nn
2
3
2
3
2
101025
2 nnn+=
23
271025
2nn+=
Now we can put our equation in quadratics form:
02 =++ cbxax
Thus,
010252
72
3
2
=++nn
and let multiply this equation by 2
Thus,
( )2010252
72
3
2
=++nn
02050732
=++ nn
Hence,
( ) ( )
( )32
205034772
=n
6
24600497 =n
95.24=n
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Hence, the number of term of series is:
25=n
(c) Find the sum to infinity of the series: 3, 1, 1/3 ...
We know that 3, 1, 1/3, is a geometric progressions which have a common
ration of 1/3, therefore in our study we know:
r
asn
=1
asn
Thus,
r
as
=1
Hence,
=
3
11
3s
3/2
3
2
33
5.42
9=
Hence the sum to infinity of the series 3, 1, 1/3 is: 4.5
(d) A drilling machine is to have 8 speeds ranging from 100rev/min to
1000rev/min. If the speeds from geometric progressive determine their
values, each correct to the nearest whole number.
Let the geometric progressions of the n terms is given by
,,,, 32 ararara
When the common ration is r we will have1nar
Our first term is a = 100rev/min.
Thus the 8th
terms is given by18ar
which is 1000rev/min.
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Thus,
100010018 =r
10001007 =r
100
10007 =r
710=r
3895.1=r
Thus, the each correct to the nearest whole number of the 8 speeds are as
follow.
The first term is a 100 100rev/min
The second terms is ar ( ) ( )3895.1100 139rev/min
The third terms is 2ar ( )( ) 23895.1100 193rev/min
The fourth terms is 3ar ( ) ( ) 33895.1100 268rev/min
The fifth terms is 4ar ( )( ) 43895.1100 373rev/min
The sixth terms is 5ar ( ) ( ) 53895.1100 518rev/min
The seventh terms is 6ar ( )( ) 63895.1100 720rev/min
The eighth terms is 7ar ( )( ) 73895.1100 1000rev/min
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