NUMERICAL
ANALYSIS WEEK 6 2015
CURVE
FITTING
PROBLEM
1
Fit a linear line
by using
polynomial least
square curve
fitting
|
X |
2 |
3 |
4 |
7 |
8 |
9 |
5 |
5 |
|
Y |
9 |
6 |
5 |
10 |
9 |
11 |
2 |
3 |
PROBLEM
2
|
X |
y |
|
0 |
-2.20 |
|
0.1 |
-1.90 |
|
0.2 |
0 |
|
0.3 |
-0.1 |
|
0.4 |
-0.01 |
|
0.5 |
-0.3 |
|
0.6 |
-0.2 |
|
0.7 |
-0.09 |
|
0.8 |
0.4 |
|
0.9 |
-0.1 |
|
1.0 |
-1.1 |
|
1.1 |
-1.3 |
|
1.2 |
-2 |
|
1.3 |
-1.8 |
|
1.4 |
-0.02 |
|
1.5 |
1 |
Data
in the above
table is given
a)
Fit a polynomial least
square curve
b) Use interpolation polynomials
to fit the
data
PROBLEM3
|
X |
y |
|
0 |
0 |
|
1 |
1 |
|
2 |
4 |
|
3 |
8 |
Fit
cubic spline
to the
data.
PROBLEM 4
Liquid
flows through
a sphere. Friction
drag coefficient
CD is given as a function of Reynolds
number. Fit a cubic spline
interpolation polynomial to
the data.
Find CD values for
Re = 5, 50, 500 and 5000 .
Note: Logarithmic scale can be used.
|
Re |
0.2 |
2 |
20 |
200 |
2000 |
20 000 |
|
CD |
103 |
13.9 |
2.72 |
0.800 |
0.401 |
0.433 |
PROBLEM 5
Solve problem 4 by using 4 point
Lagrange and
Newton interpolation formula
PROBLEM
6
Distance required to
stop a car is function of velocty. This
relation is given
in the table
below
|
Velocity (km/h) |
24 |
32 |
40 |
48 |
|
Stopping distance (m) |
4.8 |
6.0 |
10 |
12 |
If the vehicle
travels with
a velocity of 38 km/h, calculate stopping
distance by
using Newton interpolation
formula.