Direct Method of
Interpolation
Major: All Engineering Majors
Authors: Autar Kaw, Jai Paul
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
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1
Direct Method of
Interpolation
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What is Interpolation ?
Given (x0,y0), (x1,y1), …… (xn,yn), find the value of ‘y’ at a
value of ‘x’ that is not given.
Figure 1 Interpolation of discrete.
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Interpolants
Polynomials are the most common
choice of interpolants because they
are easy to:
Evaluate
Differentiate, and
Integrate
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Direct Method
Given ‘n+1’ data points (x0,y0), (x1,y1),………….. (xn,yn),
pass a polynomial of order ‘n’ through the data as given
below:
y  a0  a1 x  ....................  an x .
n
where a0, a1,………………. an are real constants.
 Set up ‘n+1’ equations to find ‘n+1’ constants.
 To find the value ‘y’ at a given value of ‘x’, simply
substitute the value of ‘x’ in the above polynomial.
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Example 1
The upward velocity of a rocket is given as a
function of time in Table 1.
Find the velocity at t=16 seconds using the
direct method for linear interpolation.
Table 1 Velocity as a function
of time.
6
t , s 
vt , m/s 
0
0
10
227.04
15
362.78
20
517.35
22.5
602.97
30
901.67
Figure 2 Velocity vs. time data for the
rocket example
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Linear Interpolation
vt   a0  a1t
y
v15  a0  a1 15  362.78
v20  a0  a1 20  517.35
Solving the above two equations gives,
a0  100.93 a1  30.914
x1 , y1 
x0 , y0 
f1 x 
x
Figure 3 Linear interpolation.
Hence
vt   100.93  30.914t, 15  t  20.
v16  100.93  30.91416  393.7 m/s
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Example 2
The upward velocity of a rocket is given as a
function of time in Table 2.
Find the velocity at t=16 seconds using the
direct method for quadratic interpolation.
Table 2 Velocity as a function
of time.
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t , s 
vt , m/s 
0
0
10
227.04
15
362.78
20
517.35
22.5
602.97
30
901.67
Figure 5 Velocity vs. time data for the
rocket example
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Quadratic Interpolation
y
vt   a0  a1t  a2t 2
x1 , y1 
v10  a0  a1 10  a2 10  227.04
x2 , y2 
2
v15  a0  a1 15  a2 15  362.78
2
v20  a0  a1 20  a2 20  517.35
2
f 2 x 
 x0 , y 0 
x
Figure 6 Quadratic interpolation.
Solving the above three equations gives
a0  12.05 a1  17.733 a2  0.3766
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Quadratic Interpolation (cont.)
517.35
vt   12.05  17.733t  0.3766t , 10  t  20
550
500
2
v16  12.05  17.73316  0.376616
2
450
ys
400
f ( range)

f x desired

 392.19 m/s
350
300
250
227.04
200
10
10
12
14
16
18
x s  range  x desired
The absolute relative approximate error a obtained between
the results from the first and second order polynomial is
392.19  393.70
100
392.19
 0.38410%
a 
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Example 3
The upward velocity of a rocket is given as a
function of time in Table 3.
Find the velocity at t=16 seconds using the
direct method for cubic interpolation.
Table 3 Velocity as a function
of time.
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t , s 
vt , m/s 
0
0
10
227.04
15
362.78
20
517.35
22.5
602.97
30
901.67
Figure 6 Velocity vs. time data for the
rocket example
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Cubic Interpolation
y
vt   a0  a1t  a2t  a3t
2
x3 , y3 
3
x1, y1 
v10  227.04  a0  a1 10  a2 10  a3 10
2
3
v15  362.78  a0  a1 15  a2 15  a3 15
2
3
x0 , y0 
v20  517.35  a0  a1 20  a2 20  a3 20
2
x2 , y2 
f 3 x 
x
3
Figure 7 Cubic interpolation.
v22.5  602.97  a0  a1 22.5  a2 22.5  a3 22.5
2
a0  4.2540
12
a1  21.266
3
a2  0.13204
a3  0.0054347
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Cubic Interpolation (contd)
vt   4.2540  21.266t  0.13204t 2  0.0054347t 3 , 10  t  22.5
v16   4.2540  21.26616  0.1320416  0.005434716 
 392.06 m/s
2
602.97
700
The absolute percentage relative
approximate error a between
second and third order polynomial is
600
ys
3
500
f ( range)

f x desired

400
300
227.04
200
10
10
13
392.06  392.19
100
392.06
 0.033269%
a 
12
14
16
18
x s  range  x desired
20
22
24
22.5
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Comparison Table
Table 4 Comparison of different orders of the polynomial.
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t(s)
v (m/s)
0
0
10
227.04
15
362.78
20
517.35
22.5
602.97
30
901.67
Order of
Polynomial
1
2
3
vt  16 m/s
393.7
392.19
392.06
Absolute Relative
Approximate Error
----------
0.38410 %
0.033269 %
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Distance from Velocity Profile
Find the distance covered by the rocket from t=11s to t=16s ?
vt   4.3810  21.289t  0.13064t 2  0.0054606t 3 , 10  t  22.5
16
s16  s11   vt dt
11
16


   4.2540  21.266t  0.13204t 2  0.0054347t 3 dt
11
16

t2
t3
t4 
  4.2540t  21.266  0.13204  0.0054347 
2
3
4 11

 1605 m
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Acceleration from Velocity Profile
Find the acceleration of the rocket at t=16s given that
 t   4.2540  21.266t  0.132042  0.0054347t 3 ,10  t  22.5
at  
d
vt 
dt
d

 4.2540  21.266t  0.13204t 2  0.0054347t 3
dt
 21.289  0.26130t  0.016382t 2 , 10  t  22.5


a16  21.266  0.2640816  0.01630416
2
 29.665 m/s 2
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Additional Resources
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice
tests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, please
visit
http://numericalmethods.eng.usf.edu/topics/direct_met
hod.html
THE END
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