Lecture Note of Quadratic Interpolation

University of Babylon
College of Engineering
Department of Environmental Engineering
Engineering Analysis I (ENAN 103)








Quadratic Interpolation
Undergraduate Leve, 3th Stage



Mr. Waleed Ali Tameemi
Engineer/ College of Engineering/ Babylon University
M.Sc. Civil Engineering/ the University of Kansas/ USA



2016-2017
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2.2 – Quadratic Interpolation
If three points, (x_0,f(x_0 )), (x_1,f(x_1 )),and (x_2,f(x_2 )) , are given and we asked to evaluate the value of f_2 (x) corresponding to any x that located between (x_0,f(x_0 )) and (x_2,f(x_2 )) and near (x_1,f(x_1 )) , as shown in Figure 2.

f_2 (x)=b_0+b_1 ?(x-x?_0)+b_2 ?(x-x?_0)?(x-x?_1)
Where:
b_0= f(x_0 )
b_1=(f(x_1 )- f(x_0 ))/?x_1-x?_0
b_2=((f(x_2 )- f(x_1 ))/?x_2-x?_1 -(f(x_1 )- f(x_0 ))/?x_1-x?_0 )/?x_2-x?_0
Ex1: The data in Table 1 was obtained by observation, estimate the value of y at x=2.
x y
1 0
2.718 1
5.92 2

Solution:
By quadratic interpolation, x=2
b_0= f(x_0 )=0
b_1=((f(x_1 )- f(x_0 ))/?x_1-x?_0 )=(1-0)/(2.718-1)=0.58
b_2=((f(x_2 )- f(x_1 ))/?x_2-x?_1 -(f(x_1 )- f(x_0 ))/?x_1-x?_0 )/?x_2-x?_0 =((2-1)/(5.92-2.718)-0.58)/(5.92-1)=-0.05
f_2 (2)=b_0+b_1 ?(x-x?_0)+b_2 ?(x-x?_0)?(x-x?_1)=0+0.58(2-1)+(-0.05)(2-1)(2-2.718)=0.62

Ex2: If the temperature (T) of a lake is given in the following table as a function of the lake depth (d), estimate the lake temperature at d = -7.5 m.
d (m) T (oC)
0 19.1
-1 19.1
-2 19.0
-3 18.8
-4 18.7
-5 18.3
-6 18.2
-7 17.6
-8 11.7
-9 9.9
-10 9.1

Solution:
By quadratic interpolation, d = -7.5m
Choose (-7,17.6), (-8,11.7), and (-9,9.9) because they are located around d = -7.5m
b_0= f(x_0 )= f(-7)=17.6
b_1=((f(x_1 )- f(x_0 ))/?x_1-x?_0 )=((11.7- 17.6)/((-8)-(-7)))=5.9
b_2=((f(x_2 )- f(x_1 ))/?x_2-x?_1 -(f(x_1 )- f(x_0 ))/?x_1-x?_0 )/?x_2-x?_0 =((9.9-11.7)/((-9)-(-8))-5.9)/((-9)-(-7))=2.05
f_2 (-7.5)=b_0+b_1 ?(x-x?_0)+b_2 ?(x-x?_0)?(x-x?_1)=17.6+5.9((-7.5)-(-7))+(2.05)((-7.5)-(-7))((-7.5)-(-8))=14.138 oC

Ex3: The velocity (v) of a rocket is given in the following table as a function of the time (t), estimate the rocket velocity when t =16 seconds.
t (s) v (m/s)
0 0
10 227.04
15 362.78
20 517.35
22.5 602.97
30 901.67

Solution:
By quadratic interpolation, t = 16 m/s
Choose (10 , 227.04), (15 , 362.78), and (20 , 517.35) because they are located around t = 16 m/s
b_0= f(x_0 )= f(10)=227.04
b_1=((f(x_1 )- f(x_0 ))/?x_1-x?_0 )=((362.78- 227.04)/(15-10))=27.148
b_2=((f(x_2 )- f(x_1 ))/?x_2-x?_1 -(f(x_1 )- f(x_0 ))/?x_1-x?_0 )/?x_2-x?_0 =((517.35-362.78)/(20-15)-27.148)/(20-10)=0.3766
f_2 (16)=b_0+b_1 ?(x-x?_0)+b_2 ?(x-x?_0)?(x-x?_1)=227.04+27.148(16-10)+0.3766(16-10)(16-15)=392.19m/s
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Homework 7
For the data set, shown in the following table, estimate f(3) using:
1- Quadratic polynomial (Second Degree Polynomial).
i x f(x)
0 -1 13
1 1 15
2 2 13
3 4 33
4 5 64



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