Holistic Numerical Methods Institute

committed to bringing numerical methods to undergraduates

Multiple Choice Tests

Direct Method of Interpolation

 

 

Q1. Given n+1 data pairs, a unique polynomial of degree ____________ passes through the n+1 data points
n+1

n+1 or less
n
n or less


Q2. The following data of the velocity of a body is given as a function of time.

Time (s)

0

15

18

22

24

Velocity (m/s)

22

24

37

25

123

The velocity in m/s at 16 s using linear polynomial interpolation is most nearly

27.867
28.333
30.429
43.000


Q3. The following data of the velocity of a body is given as a function of time. 

Time (s)

0

15

18

22

24

Velocity (m/s)

22

24

37

25

123

The velocity in m/s at t=16 s using quadratic polynomial interpolation is most nearly

27.867
28.333
30.429
43.000


Q4. The following data of the velocity of a body is given as a function of time

Time (s)

0

15

18

22

24

Velocity (m/s)

22

24

37

25

123

Using quadratic interpolation, the interpolant

approximates the velocity of the body between 18 and 24 seconds.  From this information, one of the times at which the velocity of the body is 35 m/s during the above time interval of t=18 s to 24 s is

18.67 s
20.85 s

22.20 s
22.29 s


Q5. The following data of the velocity of a body is given as a function of time.

Time (s)

0

15

18

22

24

Velocity (m/s)

22

24

37

25

123

One of the interpolant approximations from the above data is given as

Using the above interpolant, the distance in meters covered by the body between t=19 s and t=22 s is most nearly

10.50

88.50

93.00

168.00


Q6. The following data of the velocity of a body is given as a function of time

Time (s)

0

15

18

22

24

Velocity (m/s)

22

24

37

25

123

If you were going to use quadratic interpolation to find the value of the velocity at t=14.9 seconds, the three data points of time that you would you choose for interpolation are

0, 15, 18
15,18,22
0, 15, 22
0, 18, 24


 

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