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

We have given data and asked to solve through three different ways,

1.Newton's divided difference law

2. Second order lagrange interpolation

3. Appropriate numerical differentiation

All three laws have different ways to solve for the function so i have written notes for step by step solution.

Kindly go through them :
Notes Page Date dewer i Givon data Depth Stress 18 AARDP)-14 - 3.4 -28.7 321 (m-39.7 D m-२४.२) (२.3 13 24.1 an ५43 | 43-1-m 178.8- 14-6.6 8.75 -43.1 मा ५२-५. 20. 7-14.५ - 11 2-गा ४३, 13 -५ - %3D m-28.7 m - 28.7 4.1 - 3.1 g-11 431 -m 43:1-m D प -14-५ 4-03 6.6 -५.। २.5 १३॥ ५.-981 (m-28-7) - 11 : 4.1 -1.8 m - 39.7 -ग. -(43-1-m) - (m-28-1) 14,3,2 : ५५२ - 93,2 Me 10-2m 8.5 ५ 2.5 6-6-3.1 71-8-2m 8.75 |M-35.7 84 M 2.3 Teacher's Signatural

rd 3 Divided difference Y432 517.515 - 13-35 m 96.6

Notes Given that good divided difference (Juner) -0. 73 So from equation ③ - 0.73 = 5 12.515 - 13.35m Y4321 96.6 so 519.515 + (96.6*0.73) 13.35 m = 43.672 8839 7 43.67 / 43.7 By (b for Ni 4.5 meter Newton's divided difference formula f(x) = f(ng) + (1-2) Yat +60-012) (01-02) (07-08) 44381 where (0-24) (01-02) Jazz L f(ne) = Y,

Pages Data: Notes נור GA putting value of m in, Your - M-39.7 243.7 -39.7. 1.74 and Y413725 m=43-7 + given = (-0.73) for putting values in equation ④ 1914 + (4.5-1.8) (11)+ (4-5-1-8)(45–3:2) (174) +(9:5-18) (4.5-3-1) (4.5-42) (-0-73 ) solving this, we get So f(a) = 14.4 + 99.7 + 6:5778 +(-1•103 76) 49.57344 a 49.6 kPa

Notes - ن ) According to Second order Lagrange interpolation fro= f(ona) + (21-ou) (lyn - Ye da ne + (-03) (0-10) [ Helime - 32 ] putting value of n fo Y = m-28.7 = 43.7-28.7= 15 13 Rutting in equation ③ for N = 4.5m where fine) = 4 flan) = 14.4+ 14.5 – 1.8) [ 14.4-90.77 1.8-3.1 (415-108) (4.5-3.1) [11-15 on solving this, f(n) = 14.4 + 29.7 6.57 f(0) = 5006 КРg + 1.8-4.1 we get + Teacher's Signature

Notes PA Date (C) () According to Most appropriate numerical differentiation Second derivate f/m) - film+h)-f(n) + f(n) he where the ho putting n= 6:6, f166) and h f"come f(6.6+h) - 28(43.1) + f(m-h) ho h² and f(6.6-h) 11 + for Value of f(6.6th) put from eq 176-6th) = 24.4+ 14. 4 + (6.6th-1.8 18) [ (6.6th-1.8) x/6.6th-3.1) [1.74 şo f 16.6+h) 19.4 + 59.8 +11 h +1.7 b +_14.442h + 29.232 1.74 h' +25.448 h + 96.432 2 f66-645) 1

Notes Similarly from ea 6 $16.6-h) = 24.47 14.4 f (6.6-h-18) (11) 166-h-18) (6.6-h-31) [1.74] : 14.4 t 52.8 -1ch + 1.746² +29.232 -14.442h f166-4)= 1.746 - 25.443b + 96.432 w putting in eq f" (n) = 3,486° +13,864h-86-2 hr 86.2 ha : 3.48 + 192.864 b? On ta king Limit f'(o) 3.48 .