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), (1) Approximate the shear stress at a depth of 4.5 meter using second-order Lagrange interpolation. (5 marks) Estimate the second-order derivative using most appropriate numerical differentiation at a depth of 6.6 meter. (5 marks) (ii)

answer C plz

make sure the answer is correct

Answer 1

solution For answer (C) part, first solve part (a). we will st rd Depth stress 1st Order divided 2 Order divided to order divided (meter) (in kPa) differences differences differences f(x) Af(x) $²f(x) AF%) 1.8 14.4 28. 7-14.4 3.1-1.8 = 11 fm-28.7)-11_m-39-7 401-108 2.3 3.1 28.7 Im-28 I=m-287 -0.73 4.1-3.1 14.3.l-M) (03-26-17 ) 4.1 m 6.6-3.1 143-1-m 43.tm 16-6-4,1 2.5 6.6 43.1 (413.. 143.1-ml 2.5") – (m–28.7) "7 m-39-7 2.3 3.5 → =-0.73 6.6 - 1.8 7 43.) - m -2.5m + 71.75 2.5 X 3.5 m-39.7 -60.73)x(418) [264.155 - 8.05 m] - [ 8.75m - 347-375 2.3 > =-3.504 8.75 x 2.3 => 611.53-16.8m = -70.518

= 16.8m 682.048 = m = 40.6 Answer Now, we will solve part (e) م (1) 70 %=1-8 ,-3.1 2₂=4.13=6.6 f(x) F(x) = 14.4 f(x) = 28.7|f(2)=40.6 f (23) = 43-1 Second order lagrange interpolation formula f(x) = (x-x,) (X=X3?f() + (76-4,) (20-22) (2,-Xo) (1, - 2) (4-x.) (4__3) xf (%) + (12H2) (22=21) at n=4.5 f (x=4.5) = (4.5-3.1) (4.5-4.1) A 14.4 (4.5+1.8)(4-5-3.1). 2x40.6 (3.1-1.8) (3.1-4.1) (4.1-1.8)(4.1-3.1) (4.5-1.8) (4.5-4.1) X26.7+ + → f(4.5) = 2.7 - 23,8 + 66-7 => f (4.5) = 45.6 Answer

from part (9) Depth stress (meter) (inkpa) 4f(x) 기 fla) 1.8 14.4 A f(x) 43 f(x) 11 0.4 3.1 28.7 -0.73 4.1 40.6 -3.1 1 16.6 43.1 Newton divided difference formula f(x) = f(40) +(x-xo) 4f(xo)+(x=) (>=60) 4F (20) +(x-xo) (1-x)(x-X») &F(40) +- - differentiation with respect to a f'(x) = = 0+ 4f () + (24-40-41)42f (la) + [332-24(40+4, + x2)+ xgx + x184 xo xz] 4fce) it otot - Again differentiation with respect to a F"(x) = 0 + 2. A fixa)+[64--2/20+%, tra]]476) at n=6.6 f"(6:6) = 2X (0.4) + [6x6-6 - 2 (1.8 +3:1 +4.160-73) f" (6.6) = -15 Answer