multiple choice questions - 6te.netmathskthm.6te.net/numerical analysis 2.pdf · multiple choice...

NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE

L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3

Multiple Choice Questions



24. The following n data points, (x1, y1), (x2, y2),...,(xn, yn) are given. For conducting quadratic spline

interpolation the x-data needs to be

equally spaced

in ascending or descending order

integers

positive

25. In cubic spline interpolation,

the first derivatives of the splines are continuous at the interior data points

the second derivatives of the splines are continuous at the interior data points

the first and the second derivatives of the splines are continuous at the interior data points

the third derivatives of the splines are continuous at the interior data points



26. The following incomplete y vs. x data is given

x 1 2 4 6 7

y 5 11 ???? ???? 32

The data is fit by quadratic spline interpolants given by

,

where a, b, c, and d, are constants. The value of c is most nearly

-303.00

-144.50

-0.0000

14.000

27. The following incomplete y vs. x data is given.

x 1 2 4 6 7

y 5 11 ???? ???? 32


,

where a, b, c, d, e, f, and g are constants. The value of df/dx at x=2.6 is most nearly

-144.50

-4.0000

3.6000

12.200

1 axxf 21 x

42 ,91422

xxxxf

64 ,2

xdcxbxxf

76 ,928303252

xxxxf

1 axxf 21 x

42 ,91422

xxxxf

64 ,2

xdcxbxxf

76 ,2

xgfxexxf



28. The following incomplete y vs. x data is given

x 1 2 4 6 7

y 5 11 ???? ???? 32


,

where a, b, c, d are constants. What is the value of ?

23.50

25.67

25.75

28.00

29. A robot needs to follow a path that passes through six points as shown in the figure. To find the shortest

path that is also smooth you would recommend which of the following?

Pass a fifth order polynomial through the data.

Pass linear splines through the data

Pass quadratic splines through the data

Regress the data to a second order polynomial

21 ,1 xaxxf

42 ,91422

xxxxf

64 ,2

xdcxbxxf

76 ,928303252

xxxxf

5.3

5.1

dxxf

Path of a Robot

0

2

4

6

8

0 5 10 15

x

y



30. The two-segment trapezoidal rule of integration is exact for integrating at most ______ order polynomials.

first

second

third

fourth

31. The value of by using the one-segment trapezoidal rule is most nearly

11.672

11.807

20.099

24.119

32. The value of by using the three-segment trapezoidal rule is most nearly

11.672

11.807

12.811

14.633

33. The velocity of a body is given by

where t is given in seconds, and v is given in m/s. Use the two-segment Trapezoidal Rule to find the distance

covered by the body from t=2 to t=9 seconds.

935.0 m

1039.7 m

1260.9 m

5048.9 m

34. The shaded area shows a plot of land available for sale. The numbers are given in meters measured from

the origin. Your best estimate of the area of the land in square meters is most nearly

2.2

2.0

dxxex

2.2

2.0

dxxex

51,2)( tttv

145,352

tt



2500

4775

5250

6000

35. The following data of the velocity of a body as a function of time is given as follows.

Time (s) 0 15 18 22 24

Velocity (m/s) 22 24 37 25 123

The distance in meters covered by the body from t=12 s to t=18 s calculated using using Trapezoidal Rule with

unequal segments most nearly is

162.9

166.0

181.7

436.5

36. The highest order of polynomial integrand for which Simpson’s 1/3 rule of integration is exact is

first

second

third

fourth

37. The value of by using two-segment Simpson's 1/3 rule is most nearly

7.8306

7.8423

2.2

2.0

dxex

60

75

45

25

100 60



8.4433

10.246

38. The value of by using four-segment Simpson's 1/3 rule is most nearly

7.8036

7.8062

7.8423

7.9655

39. The velocity of a body is given by

where t is given in seconds, and v is given in m/s. Using two-segment Simpson's 1/3 rule, the distance covered

in meters by the body from t=2 to t=9 seconds most nearly is

949.33

1039.7

1200.5

1442.0

40. The value of by using two-segment Simpson’s 1/3 rule is estimated as 702.039. The estimate

of the same integral using four-segment Simpson’s 1/3 rule most nearly is

702.39 + 8/3 [2f(7)-f(11)+2f(15)]

702.39/2 + 8/3 [2f(7)-f(11)+2f(15)]

702.39 + 8/3 [2f(7)+2f(15)]

702.39/2 + 8/3 [2f(7)+2f(15)]

41. The following data of the velocity of a body is given as a function of time.

Time (s) 4 7 10 15

Velocity (m/s) 22 24 37 46

2.2

2.0

dxex

51,2)( tttv

145,352

tt

19

3

)( dxxf



The best estimate of the distance in meters covered by the body from t=4 to t=15 using combined Simpson’s

1/3rd

rule and the trapezoidal rule would be

354.70

362.50

368.00

378.80

42. If is the value of integral using n-segment Trapezoidal rule, a better estimate of the integral can

be found using Richardson’s extrapolation as

43.The estimate of an integral of is given as 1860.9 using 1-segment Trapezoidal rule.

Given f(7)=20.27, f(11)=45.125, and f(14)=82.23, the value of the integral using 2-segment Trapezoidal rule

would most nearly be

787.32

1072.0

1144.9

1291.5

44. The value of an integral given using 1, 2, and 4 segments Trapezoidal rule is given as 5.3460,

2.7708, and 1.7536, respectively. The best estimate of the integral you can find using Romberg integration is

most nearly

1.3355

1.3813

15

2

2

nn

n

III

3

2

2

nn

n

III

nI 2

n

nn

nI

III

2

2

2

nI b

a

dxxf

19

3

dxxf

b

a

dxxf



1.4145

1.9124

45.Without using the formula for one-segment Trapezoidal rule for estimating the true error, can

be found directly as well as exactly by using the formula

, for

46. For , the true error, in one-segment Trapezoidal rule is given by

, .

The value of for the integral is most nearly

2.7998

4.8500

4.9601

5.0327

47. Given the velocity vs. time data for a body

t(s) 2 4 6 8 10 25

0.166 0.55115 1.8299 6.0755 20.172 8137.5

The best estimate for distance covered between 2s and 10s by using Romberg rule based on Trapezoidal rule

results would be

33.456 m

36.877 m

37.251 m

81.350 m

12

3ab

Et

"f ba

xexf

xxxf 33

352 xxf

xexxf

25

12

3ab

Et

"f ba

dxe

x

2.7

5.2

2.03

sm /

b

a

dxxftE

b

a

dxxftE



48. is exactly

49. For a definite integral of any third order polynomial, the two-point Gauss quadrature rule will give the same

results as the

1-segment trapezoidal rule



Simpson's 1/3 rule

50. The value of by using the two-point Gauss quadrature rule is most nearly

11.672

11.807

12.811

14.633

51. A scientist uses the one-point Gauss quadrature rule based on getting exact results of integration for

functions f(x)=1 and x. The one-point Gauss quadrature rule approximation for is

52. A scientist develops an approximate formula for integration as

1

1

)5.75.2( dxxf

1

1

)5.75.2(5.2 dxxf

1

1

)55(5 dxxf

1

1

)()5.75.2(5 dxxfx

)()(2

bfafab

2)(

bafab

23

1

223

1

22

ababf

ababf

ab

)()( afab

b

a

xfcdxxf ),()( 11

10

5

)( dxxf

2.2

2.0

dxxex

b

a

dxxf )(



where

The values of c1 and x1 are found by assuming that the formula is exact for the functions of the form a0x + a1x2 polynomial. Then the resulting formula would therefore be exact for integrating

53. You are asked to estimate the water flow rate in a pipe of radius 2m at a remote area location with a harsh

environment. You already know that velocity varies along the radial location, but you do not know how it

varies. The flow rate Q is given by

To save money, you are allowed to put only two velocity probes (these probes send the data to the central office in New York, NY via satellite) in the pipe. Radial location, r is measured from the center of the pipe, that is r=0 is the center of the pipe and r=2m is the pipe radius. The radial locations you would suggest for the two velocity probes for the most accurate calculation of the flow rate are

0,2

1,2

0,1

0.42,1.58 54. To solve the ordinary differential equation , by Euler’s method, you

need to rewrite the equation as

55. Given

and using a step size of h=0.3, the value of y(0.9) using Euler’s method is most nearly

-35.318

-36.458

.1 bxa

2)( xf2

532)( xxxf 2

5)( xxf

xxf 32)(

2

0

2 rVdrQ

50,5sin2

yyxdx

dy

50,5sin3

1 2 yyx

dx

dy

50,3

5cos

3

13

y

yx

dx

dy

50,sin3

1 yx

dx

dy

53.0,sin532

yxydx

dy

50,sin532

yxydx

dy



-658.91

-669.05

56. Given

and using a step size of h=0.3, the best estimate of dy/dx(0.9) using Euler’s method is most nearly is

-0.37319

-0.36288

-0.35381

-0.34341

57. The velocity (m/s) of a body is given as a function of time (seconds) by v(t)=200 ln(1+t) -t, t≥0

Using Euler’s method with a step size of 5 seconds, the distance in meters traveled by the body from

t=2 to t=12 seconds is most nearly

3133.1

3939.7

5638.0

39397

58. Euler’s method can be derived by using the first two terms of the Taylor series of writing the value

of , that is the value of at , in terms of and all the derivatives of at . If

, the explicit expression for if the first three terms of the Taylor series are chosen

for the ordinary differential equation

would be

59. A homicide victim is found at 6:00PM in an office building that is maintained at 72˚F. When the

victim was found, his body temperature was at 85 ˚F. Three hours later at 9:00PM, his body

temperature was recorded at 78˚F. Assume the temperature of the body at the time of death is the

normal human body temperature of 98.6˚F.

53.0,31.0

yeydx

dy x

1iy y 1ix iy y ix

ii xxh 1 1iy

70,325

yey

dx

dy x

hyeyy ix

iii 3

2

1 51

22

53

2

1 255

1

hehyeyy ii x

i

x

ii

24

9

4

133

2

1 255

1

hyehyeyy i

x

i

x

iiii

2

332

12

5

1

hyhyeyy ii

x

iii



The governing equation for the temperature θ of the body is

where

= temperature of the body, ˚F

θa = ambient temperature, ˚F

t = time, hours

k = constant based on thermal properties of the body and air.

The estimated time of death most nearly is

2:11 PM

3:13 PM

4:34 PM

5:12 PM

60. To solve the ordinary differential equation

by the Runge-Kutta 2nd order method, you need to rewrite the equation as

61. Given

and using a step size of h=0.3, the value of y(0.9) using the Runge-Kutta 2nd order Heun's method is

most nearly

-4297.4

-4936.7

-0.21336

-0.24489

62. Given

,

)( akdt

d

50,sin2

yxyxdx

dy

50,sin3

1 2 yxyx

dx

dy

50,3

cos3

1 3

y

xyx

dx

dy

50,sin3

1 yx

dx

dy

53.0,sin532

yxydx

dy

1410

1410

53.0,531.0

yeydx

dy x

50,sin32

yxxydx

dy



and using a step size of h=0.3, the best estimate of dy/dx(0.9) using the Runge-Kutta 2nd order

midpoint-method most nearly is

-2.2473

-2.2543

-2.6188

-3.2045

63. The velocity (m/s) of a body is given as a function of time (seconds) by

Using the Runge-Kutta 2nd order Ralston method with a step size of 5 seconds, the distance in meters

traveled by the body from t=2 to t=12 seconds is estimated most nearly is

3904.9

3939.7

6556.3

39397

64. The Runge-Kutta 2nd order method can be derived by using the first three terms of the Taylor

series of writing the value of yi+1 (that is the value of y at xi+1 ) in terms of yi (that is the value of y at xi)

and all the derivatives of y at xi . If h=xi+1-xi, the explicit expression for yi+1 if the first three terms of

the Taylor series are chosen for solving the ordinary differential equation

would be

65. A spherical ball is taken out of a furnace at 1200K and is allowed to cool in air. Given the following,

radius of ball = 2 cm

specific heat of ball = 420 J/(kg-K)

density of ball = 7800 kg/m^3

convection coefficient = 350 J/s-m^2-K

The ordinary differential equation is given for the temperature, of the ball

0,1ln200 tttt

70,352

yey

dx

dy x

2

5532

3

1

hhyeyy i

x

iii

2

2521532

22

1

hyehyeyy i

x

i

x

iiii

2

6532

22

1

hehyeyy ii x

i

x

ii

2

56532

22

1

hehyeyy ii x

i

x

ii

841310811020673.2

dt

d



if only radiation is accounted for. The ordinary differential equation if convection is accounted for in

addition to radiation is

66. To solve the ordinary differential equation ,

by Runge-Kutta 4th order method, you need to rewrite the equation as

67. Given and using a step size of , the value of

using Runge-Kutta 4th order method is most nearly

- 0.25011

-4297.4

-1261.5

0.88498

68. Given , and using a step size of , the best estimate of

Runge-Kutta 4th order method is most nearly

-1.6604

-1.1785

300106026.110811020673.228413

dt

d

300103982.410811020673.228413

dt

d

300106026.12

dt

d

300103982.42

dt

d

50,sin2

yxyxdx

dy

50,sin3

1 2 yxyx

dx

dy

50,3

cos3

1 3

y

xyx

dx

dy

50,sin3

1 yx

dx

dy

4010

50,sin32

yxxydx

dy

53.0,sin532

yxydx

dy3.0h

9.0y

53.0,32

yeydx

dy x 3.0h 9.0dx

dy



-0.45831

2.7270

69. The velocity (m/s) of a parachutist is given as a function of time (seconds) by

Using Runge-Kutta 4th order method with a step size of 5 seconds, the distance traveled by the body

from to seconds is estimated most nearly as

341.43 m

428.97 m

429.05 m

703.50 m

70. Runge-Kutta method can be derived from using first three terms of Taylor series of writing the

value of , that is the value of at , in terms of and all the derivatives of at . If

, the explicit expression for if the first five terms of the Taylor series are chosen for the ordinary differential equation

, would be

0),17.0tanh(8.55 ttt

2t 12t

1iy y 1ix iy y ix

ii xxh 1 1iy

2

553

22

1

hhyeyy i

x

iii

24

3906253009096

625483

2252153

42

32

222

1

hye

hye

hyehyeyy

i

x

i

x

i

x

i

x

ii

ii

ii

24

24

612

2653

42

32

222

1

he

he

hehyeyy

i

iii

x

xx

i

x

ii

24

24

612

25653

42

32

222

1

he

he

hehyeyy

i

iii

x

xx

i

x

ii

70,352

yey

dx

dy x



71. A hot solid cylinder is immersed in an cool oil bath as part of a quenching process. This process

makes the temperature of the cylinder, , and the bath, , change with time. If the initial temperature of the bar and the oil bath is given as 600° C and 27°C, respectively, and

Length of cylinder = 30 cm

Radius of cylinder = 3 cm Density of cylinder = 2700 kg/m^3 Specific heat of cylinder = 895 J/kg-K Convection heat transfer coefficient = 100 W/(m^2-K) Specific heat of oil = 1910 J/(kg-K) Mass of oil = 2 kg The coupled ordinary differential equations governing the heat transfer are given by

The following equations are used to answer questions#2, 3, and 4

c b

bc

c

dt

d

4.362

cb

b

dt

d

5.675

bc

c

dt

d

4.362

cb

b

dt

d

5.675

bc

c

dt

d

5.675

cb

b

dt

d

4.362

bc

c

dt

d

5.675

cb

b

dt

d

4.362

hkkkkyy ii 43211 226

1

ii yxfk ,1

Oil

Cylinder



72. The differential equation is

linear

nonlinear

linear with fixed constants

undeterminable to be linear or nonlinear

73. A differential equation is considered to be ordinary if it has

one dependent variable

more than one dependent variable

one independent variable

more than one independent variable

74. Given

, y(2) most nearly is

0.17643

0.29872

0.32046

0.58024

75. The form of the exact solution to is

76. The following nonlinear differential equation can be solved exactly by separation of variables.

The value of θ(100) most nearly is

-99.99

hkyhxfk ii 12

2

1,

2

1

hkyhxfk ii 23

2

1,

2

1

hkyhxfk ii 34 ,

60,2sin32 yxydx

dy

xxBeAe

5.1

xx BxeAe 5.1

xx BeAe 5.1

xxBxeAe

5.1

10000,811026

dt

d

50,32 yey

dx

dy x



909.10

1000.32

1111.10

77. A spherical solid ball taken out of a furnace at 1200K is allowed to cool in air. Given the following radius of ball=2 cm density of the ball=7800 kg/m^3 specific heat of the ball=420 J/kg-K emmittance=0.85 Stefan-Boltzman constant=5.67E-8 J/s-m^2-K^4 ambient temperature=300K convection coefficient to air=350 J/s-m^2-K.

The differential equation governing the temperature, of the ball as a function of time, t is given by

841210811020673.2

dt

d

3001060256.12

dt

d

300106026.11081102067.2128412

dt

d

300106026.11081102067.228412

dt

d

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