Lagrange Interpolation and Extrapolation
- My sister Julie planted a tomato plant and she measured and kept track of its growth every other day. Julie is a curious person, and she would like to estimate how tall her plant was on the fourth day, do so by using LaGrange method for Linear Interpolation.
Her table of observations looked like this:
[pic 1]
For this first order polynomial we chose the height given by:
[pic 2]
Expanding equation within limits we get;
(d) H() +(d) H()[pic 3][pic 4][pic 5][pic 6]
;[pic 7]
[pic 8]
[pic 9]
Subbing function into original equation we et:[pic 10][pic 11]
[pic 12]
= 3 , H() = 2.1 = 5, H() = 4.7[pic 13][pic 14][pic 15][pic 16]
[pic 17]
[pic 18]
= 3.4 mm
- A straight galvanize pipe was purchased to be used on a water dredge ,the said pipe have to be curved to be installed through three points, the pipe must reach a temperature of 775K before it changes shape but the specific capacity () of the material is unknown at that temperature , find the specific capacity at 775K using the Quadratic Interpolation method? [pic 19][pic 20][pic 21][pic 22]
Table of specific as a function of temperature
(T) K | 650 | 700 | 750 | 800 | 850 | 900 |
(Cp) kJ/kg.K | 0.697 | 0.759 | 0.802 | 0.829 | 0.849 | 0.873 |
[pic 23]
(T) = (T) () + (T) () + (T) ()[pic 24][pic 25][pic 26][pic 27][pic 28][pic 29][pic 30][pic 31][pic 32][pic 33]
[pic 34]
[pic 35]
[pic 36]
(T)= ()+ ()+ ()[pic 37][pic 38][pic 39][pic 40][pic 41][pic 42][pic 43][pic 44][pic 45][pic 46][pic 47][pic 48][pic 49]
(775K)= 0.829[pic 50][pic 51]
(775K) = (-0.094875) + (0.6015) + (0.310875) [pic 52]
(775K) = 0.8175 kj / kg. k[pic 53]
- The acceleration of a Bugatti Veyron is being tested on how fast it goes from zero to sixty seconds. The test shows in our table in respect of time their speed. Using cubic LaGrange interpolation determine how fast it will be in 25 seconds.
Time(s) | mph |
0 | 0 |
10 | 20 |
20 | 60 |
30 | 105 |
40 | 128 |
50 | 150 |
60 | 180 |
Values:
[pic 54]
[pic 55]
[pic 56]
[pic 57]
[pic 58]
[pic 59]
[pic 60]
[pic 61]
[pic 62]
[pic 63]
[pic 64]
[pic 65]
[pic 66]
[pic 67]
[pic 68]
[pic 69]
[pic 70]
[pic 71]
[pic 72]
[pic 73]
[pic 74]
mph=83.5625[pic 75]
- Given: x = [5 10 15 20]; and y = ln (x). By means of linear extrapolation compute the value of y when x = 25.
Compare this value to the actual value of ln (25) and find the Absolute Relative Approximate Error.
x | y = ln (x) |
5 | 1.6094 |
10 | 2.3026 |
15 | 2.7081 |
20 | 2.9957 |
25 | ??? |
[pic 76]
[pic 77]
[pic 78]
[pic 79]
[pic 80]
[pic 81]
[pic 82]
[pic 83]
[pic 84]
[pic 85]
[pic 86]
[pic 87]
[pic 88]
Actual value of ln(25): 3.2189
Absolute Relative Approximate Error:
= x 100[pic 89][pic 90]
= x 100[pic 91][pic 92]