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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,


=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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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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