Numerical DifferentiationII

8/3/2019 Numerical DifferentiationII

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10www.mathematicsondemand.com Prof OP Dogra ( 9876486598)

i.e 010.0 4 dp

dyei

dp

dyor

0

2

120

20

y

py

02

1266 0

2 yp

or 0492

1266

p

pp 18491324912132 i.e. 8469.198

181p

05.8469.160. xifrom

app692.092345.60. , .692 is close to .60

So we useNewtons forward difference formula from (2)

02

49.8469..8469.1668469.16221104 y 426.6137

y = .6137 (app) .6137 .692the Min value of y at x

OR we can use Backward formula 75.692. tonearx

Derivative using Newtons backward interpolation formula

..............3

!3

212

!2

1

ny

pppny

ppnypnyy

hpnxxwhere

....................3

3

12

2

11nynyny

hnxx

dx

dy

2

2

51 112 3 4 5

2 12 6

d yy y y y

dx n n n n hx xn

.......................4

2

333

1

3

3

nynyh

nxxdx

yd

Q1. Find the tablefollowingthefrom4.0xatdx

dy

4.3.2.1.;x

49182.134986.122140.110517.1:y

Sol: 4.x lies at the end we use Newtons Backward formula

4.0&4.0,1. xnxh


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hpnxx 01

4.4.

h

nxxp

Table:

x y510 y510 y2510 y3510

.1

.2

.3

.4

110517

122140

134986

149182

11623

12846

14196

1223

1350

127

By Newtons Backward formula

1491331273

11350

2

114196

1.

1510

nxxdx

dy

49133.14.

xdx

dy

Q2. The distance (s) covered as a function of time (t) by an athlete during his/ her run for

the 50 metres race is given in the following table:

6543210sec sTime

505.365.245.155.85.20tan ceDis

Determine the speed of the athlete at t = 5 secs.

Sol: t = 5 near the end we use backward formula


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04,0.13,0.32,12 nsnsnsnsHere

By Newtons backward formula

nsnsnsnsns

hntt

dx

dy 5

5

14

4

13

3

12

2

11

5.35

1013

132

1121

1

= 13.13

or sec/13.135

mdt

ds

t

Q3. The following data given the corresponding values of heat generated by the carnot

engine (Q) in calories and time (t) in seconds.

6543210:t

t s s s2 s3 s4 s5 s6

0

1

2

3

4

5

6

0

2.5

8.5

15.5

24.5

36.5

50

2.5

6.0

7.0

9.0

12.0

13.5

3.5

1.0

2.0

3.0

1.5

-2.5

1.0

1.0

-1.5

3.5

0.0

-2.5

-3.5

-2.5

1


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505.365.245.155.85.20:Q

Find the instantaneous rate of heat flows in calories per second at t = 5 secs.

Q4. A rod is rotating in a plane about one of its ends. If the following table gives the

angle radians through which the rod has turned for different values of time t

seconds. Find its angular velocity and angular acceleration at t = 0.7 secs.

0.18.06.04.02.00secint

20.30.210.048.012.00)( radiansin

Sol: t = .7 sec lies near the end of table we use Newtons backward formula

Table:

7.,2.,0.1 thnt

nthpt ort t

nt t p h phn

5.12.

3.

2.

0.17.

p

By Newtons backward formula 0p we use the formula

t

2 3 4 5

0

.2

.4

.6

.8

10

0

.12

.48

1.10

2.0

3.20

.12

.36

.62

.90

1.20

.24

.26

.28

.30

.02

.02

.02

0

0

0


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..............................3

6

26232

2

121n

ppn

pn

hdt

d

2

2 1.5 1 3 1.5 6 1.5 211.20 .30 .02

.2 2 6

sec/496.4 rad ,

.............................312

2

1

2

2

npnhdt

d

02.15.130.

2.

12

= 7.25 radian / sec.

Q5. From the table values of x and y, obtain first & 2nd

derivatives for the function

xey at x = 2.0 & x = 2.2

2.228.16.14.12.10.1x

0250.93891.70496.69530.40552.43201.37183.2y

Sol: As the derivative are required at x = 2 & x = 2.2 which are near the end of table.

we use Newtons backward formula

x y410 y410 y2410 y3410 y4410 y5410 y6410

1.0

1.2

1.4

1.6

1.8

27183

33201

40552

49530

60496

6018

7351

8978

10966

13395

1333

1627

1988

2429

294

361

441

535

67

80

94

13

14

1


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2

2.2

73891

90250

16359

2964

At x = 2

1.2.

2.

2.

2.22,2. 0

h

xxphHere

By Newtons formula

nyny

nynynyny

hnxx

dx

dy

6

6

15

5

1

4413

312

21

1

&

......

5

6

54

24

11322

1

2

2

nynynynyh

nxxdx

yd

2n

xat

13

5

180

4

1441

3

12429

2

113395

2.

110

0.2

4

xdx

dy

73896

2.0

7.3896x

dy

dx

&

2

4

22

2.0

1 11 510 2429 441 80 1324 6.2x

d ydx

58.7293783.1067.36441242925

2938.7

0.22

2

xdx

yd

(ii) Take 2.2n

x

1

6

114

5

194

4

1525

3

12964

2

116359

2.

110

20.2

4

dx

dy

90228

0228.92.2

dx

dy


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and

1

180

13714

6

594

24

115352964

2.

1

2.2

1022

24

xdx

yd

75.8861276.67.14308535296425

8613.8

2.22

2

xdx

yd

Derivatives using Striling formula:

.2!3

1

22

23

1322

12

210

0

yyppypyy

pyy

..........

2!5

21

!4

1 35

252222

24

222

yypppy

pp

1

20

0

h

xx

phpxx , hdx

dp 1

from (2) p = 0 at 0xx

....

2120

4155

24

24

26

13

21

35

2524

24

3

23

132

1210

0 yyppy

pp

yypyp

yy

hdx

dy

xx

If thenxxatp 00

...................

230

1

26

1

21

35

25

23

13

10

0 yy

yyyy

hdx

dy

xx

................212

32

1216

22

1&

35

253

2423

13

12

22

2

yypp

ypypy

hdx

yd

0pputting

...............................

90

1

12

113

62

41

222

2

yyyhdx

yd


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Q1. Finddx

dyat x = .6 of the function xfy where

8.7.6.5.4.:x

6510818.23275054.20442376.27974426.15836494.1:y

6.x (is in middle of table)

we use Striling formula for differentiation

Table:

Here 1.,8.,7.,4.,5.,6. 21210 hxxxxx

34710,38358,2467950,2832678 23

13

10 yyyy

..................330

1

26

1

21

0 35

25

23

13

10

yy

yyyy

hxxdx

dy

p x y710 y710 y2710 y3710 y4710

-2

-1

0

1

2

.4

.5

.6

.7

.8

15836494

17874426

20442376

2327054

26510818

2137932

2467950

2832678

3235764

330018

364728

403084

34710

38358

3648


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....

1

3471038358

12

1

2

24679502832678

2

1

1.

1

6.107

xdx

dy

thereisyoftermNo 2526442250

.644225.26.

Ans

xdx

dy

Q2. A rod is moving in a plane the following table given the angle in radian through

which the rod has turned for various values of t seconds.

2.118.6.4.2.0:t

67.420.302.212.149.12.0:

Calculate the angular velocity & angular acc. Of the rod

When x = .6

Sol: Angular velocitydt

d & Angular acc.

2

2

dt

d

Since .6 lies in the middle. We use striling formula

H t p 2 3 4 5 6

0

.2

.4

.6

.8

1.0

1.2

-3

-2

-1

0

1

2

3

0

.12

.49

1.12

2.02

3.20

4.67

.12

.37

.63

.90

1.18

1.47

.25

.26

.27

.28

.29

.01

.01

.01

.01

0

0

0

0

0

0


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here 2.,6.0 ht

27.,63.,90. 12

10

01.,01. 23

13

Now

....

230

1

26

1

2

1 35

25

23

13

10

6.

hdt

d

t

01

01.01.121

263.90.

2.1

815.30017.765.2.

11202.

256.1

2.1

..........12

112

41

2

26.

2

2

hdt

d

t

.75.627.250

12

127.

2.

1

2Ans

Q3. Find 93'f from the following table:

120105907560:x

7.379.402.432.382.28:xf

Hint: Here5

1

15

90930

h

xxp

...................

2

11

210 ypyy

hdx

dyuse

Derivatives using Bessels formula

Consider the Bessels form

13

02

12

00!3

12

1

2

1

!2

1

y

ppp

yypp

ypyy

1!4

2111

42

4

yypppp

hpxx 0

132

02

12

0!3

2

133

2

1.

!2

12

y

ppp

ydp

dyyy

......2

1.

!4

22641

42

23

yyppp

If p = 0 , at 0xx


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Then

..................240

1

120

1

24

1

12

1

4

1

1

26

36

25

14

24

13

02

12

0

0

yy

yyy

yyyy

hdx

dy

xx

&

.........2!4

21212

!3

36

21

14

242

10

21

2

22

2

yypp

ypyy

hdx

yd

at 0xx when p = 0

.......

24

1

2

1

2

11

0

14

24

13

02

12

22

2

yyyyyh

xxdx

yd

Q1. Find the value of ' 4f from the following table using Bessels formula

7654321:x

181285310:xf

Sol:

x p y y y2 y3 y4 y5 y6

1

2

3

4

5

6

-3

-2

-1

0

1

2

0

1

3

5

8

12

1

2

2

3

4

1

0

1

1

2

-1

1

0

1

2

-1

1

-3

2

5


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7 3 18

6

2

63

62

5

14

24

13

02

12

0

4

240

1

120

1

[24

1

12

1

4

1

1

yyy

yyyyyy

hdx

dy

x

Here h = 1 , 0,1,1,3 13

12

02

0 yyyy

0,5,2,11 26

36

25

14

24 yyyyy

Putting in above formula we get

1 13 1 1 1

1 4 12

1 1 114 1 1 2 5 0

24 120 240

dy

dxx

4625.2240

591

240

54120720

48

1

60

100

2

13

Derivative using Newtons divided difference formula

210101000 ,,, xxxfxxxxxxfxxxfxf

...................,,, 3210210 xxxxfxxxxxx

2101010' ,,, xxxfxxxxxxfxf

.....................,,, 3210212010 xxxxfxxxxxxxxxxxx

Q1. Given the following pairs of values of x & y.

108421:x

2721510:xfy


13/14


Determine first derivative at x = .4

Sol: x values are not equally spaced we use Newtons divided difference formula

x xf xf xf2 xf3 xf4

1

2

4

8

10

0

1

5

21

27

112

01

224

15

448

521

3810

2127

333.14

12

333.2824

1667.410

43

0

.1667 .333

10 2

0.0625

-.0069

4,8,4,2,1 3210 xxxxx

02

10000 xfxxxxxfxxxfxf

30 1 2 0x x x x x x f x

40 1 2 3 0 1x x x x x x x x f x

02

100' xfxxxxxfxf

30 1 0 2 1 2 0x x x x x x x x x x x x f x


14/14


0 1 2 0 1 3 40

0 2 3 1 2 3

x x x x x x x x x x x x f x

x x x x x x x x x x x x

0.44142414333.241414' f

0069.

844424844414

842414442414

= 1+ 1.665+0-24(-.0069)

= 2.8321

Q2. Using the following data, find 5'f

974320:x

93246611258264:xf

(Try) Ans. 98

Q3. Find 10'

f for the following data:34271153:x

35606173158992313: xf

(Try) Ans. 232.8688

Numerical DifferentiationII

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