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

Answer :

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 Answer it Givon data Depth Stress - 18 M - 3.1 - 28.7 ATED)-14 1324 /-/m-39.2 (m-२४.२) (२.3 13- 4.1 an Y ५434 - -m 178.8- OIL -6.6 34-43-4 2.5 8.75 ४३, ५२-५, 2-। 20. 7-14.4 - 11 क Y 13 - ५ - %3D m-28.7 m - 28.7 4.1 - 3.1 - ।। अपर 43.1 -m 43-1-m - 14-५, - 03 6.6 -५ । ३.51 ४३ ५.-921 (m-२8.7) - 11 : 4.1 m-39-7 M - -1.8 2.3 14,3,2 : ५५३ -५3,2 -(43-1-m) - (m-28-1) 10-2m 8.15 ५ Me %3D 19431 - 94,3,2 - 93,21 84 04 66 - 3 | - 71-8-2m 8.75 - m-35.7 2.3 - 18 66 Teacher's Signature

3 Divided difference 512.515 - 13.35m. Yuzas 96.6

Notes Given that good divided difference (Juner) 73 So from equation 6 94321 - 0.73 = 5 12.515 - 13.35m 96.6 so ma 517.515 + (966*0.73) 13.35 m = 43.672 8 839 2 43.67 a 43.7 b for N = 4.5 meter Newton's divided difference formula f(u) = Sf(mg) + (01-02) Y22 + (1-2) (M-a) Y321 +(91-012) (01-02) (01-05) /433.1 where L fore) = Y,

Pasel Date: Notes נור putting value of m in, Your M-39.7 1.74 243.7 -39.7. 29.3 and 14,3,25 - m=43-7 so given = (-0.73) for putting valves in equation ② 1914 + (4.5-1.8) ()+ (4-5-1-8)(45–3:2) (174) +(4:5-1-8) (45-3-1) (4.5-42) ( -0-73) so dolving thes, we get fla) 14,4 + 29.7 + 6577? +(-1•10376) f(o) = 49.57344 a 49.6

UL Notes Date ن (ع) According to Second or der Lagrange interpolation frov = f(ona) + (21-ora) (lyn - yed h - na + (0-ra) (a-na) [ devine - 2 putting valve of m fo Y = m-28.7 = 43.7-28.7 = 15 13 Rutting in equation ③ for N= 4.5m where fing) = 4 flon = 14.4+ (4.5 - 1.8) [ 14.4 -28.7 1.8-3.1 + 1.8-4.1 (4.5-108) (4.55-3-1)11-15 on Solving this, we get f(n) = 14+4 + 29.7 6.57 f f(0) = 50°6 50.67 kPa Teacher's Signature

Notes Data (C) (u) According to Mast appropriate numerical differentiation Second derivate f(a+h)-f(n) + f (0-h) ha where the ho putting n = 6.6, f166) and h f"con) f(6:6th) - 2x( 43.1) + f (01-h) ho h² and f(6.6-h we - 11 + for Value of f(6.6th) Rut from eq f/6.6th) 14.4+ (6.6th-1.8 18) [ 16.6th-1.8) 1/6.6th-3.1) [1.74. so f1616th) = 14.4 + 59.8 +11 h +1.74 b + 14.442 h + 29.232 1.74 h' +25.442 h +96.432 74] 2 f166+1) 1

Notes Similarly from eq 6 f16-6-6) = 14.47 14.4 f (6.6-h-18) (11) 166-6-18) (6.6-n-31) [ 1.749 :14 4 t 52.8 - 11h + 1.74n² +29.232 -14.442h F166-4)= 1.746 - 25.443b + 96.432 eq" putting in f" (n) = 3,486° +138.864h-86-2 he 86.2 he : 3.48 + 192.864 cha On to king Limit sat f(n) = 3.48 .

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