Question

The shear stresses in (kPa) of four specimens taken at various depths in a clay stratum are given by Depth, meter 1.8 3.1 4.1 6.6 Stress ,kPa 14.4 28.7 m 43.1 (a) Find the value of m if the Newton's third divided difference is -0.73. (10 marks) (b) Hence, estimate the shear stress at a depth of 4.5 meter. (5 marks) (c) Using the value of m obtained in (a), (i) Approximate the shear stress at a depth of 4.5 meter using second-order Lagrange interpolation. (5 marks) (ii) Estimate the second-order derivative using most appropriate numerical differentiation at a depth of 6.6 meter. (5 marks)

Answer 1

3rd and let f(2) x 14-4 m-39- 16 108 11 39-7 mo 1.3 114-85-3.5m 8.75 2.3 12-3 I 6-8 28-7 2-3 m- 28.7 Ń -8 3.5 In Bol 12:41 2-5 . m 4301 -m 4301 2-5 114-85-3.5 m 8.75 17266 4301 -m -2.5m+71-75 119.85-3.5m 2-5 8.75 N (434 -m 2.5 -fon-25-7) 3.5 3.5 0-390 47619 m-39-7 203 --0-73 114-85-3-5m m-39-7 2-3 Given 1 4.8 8.75 m = 40.5980 9524

The interpolation formula in 14.4+ (2-1-8) 11 + (1 - 1-8) (~- 3-1) (+0.39047619) + Cor 118) (2-8-1) (3-4-) ( -0.73) Stren at 4.5 meter lohy (144) + (1182-7) + (2-7* 0-39047619) + (2.7 x 164 x 0.4% -0-73 = 14.4+ 29.7 +1•475999998 + (-1310376 = 44.47224 -) Second order Lagrange inter folating polynomial is giventy (2-2) (21-* flaz) P(2) (1-22) (R-8) (2,-2) (:21-3) f(n) + (?-2) (223 + (x-n ) (2-2) f(²3) Where (23-2,) (23-22) the conresponding table in in il fim:) 11 311 28-7 21 401 40.59809524 31 6-6 113.1 so P(4-5) 1 861 + 47.74336 +2.7584 125 = 43.61376

2 n, m Ncymerical differentiation wiring Devided Difference I'C) 28 [m 4 6%-2 (nota,+2) f (2012, 2, 3) 5'66)= 2(0.3904 7619) +(6x64) -2(10825144.) (-2) 0-78095238 + 52-744 53.52095238 1J

Lagrange's formula is an interpolation formula so the nodes has been chosen in such a way that 4.5 lies in the range of the given data points.

The differentiation formula has been obtained by twice differentiating the divided difference formula obtained in the 2nd question.