Using Newton’s forward formula compute f(142) from the following table: x 140 150 160 170 180 f(x) 3.685 4.854 6.302 8.076 10.225

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech I Year MATHEMATICAL METHODS Previous Question Paper (October/November - 2020 - R15)

B.Tech | I Year | I-I | R15 | October/November - 2020

Answer PaperPaper Code: 121ALMaximum Marks: 75Time: 2 hours

Answer any five questions. All questions carry equal marks.

Main

Using Newton’s forward formula compute f(142) from the following table:

x	140	150	160	170	180
f(x)	3.685	4.854	6.302	8.076	10.225

15 marks

Answer

To compute f(142) using Newton's forward interpolation formula, we first construct the forward difference table with h = 10 and x₀ = 140.

x	f(x)	Δf(x)	Δ²f(x)	Δ³f(x)	Δ⁴f(x)
140	3.685	1.169	0.279	0.017	0.007
150	4.854	1.448	0.296	0.024	-
160	6.302	1.774	0.320	-	-
170	8.076	2.149	-	-	-
180	10.225	-	-	-	-

The formula is given by f(x₀ + ph) = f(x₀) + pΔf(x₀) + [p(p-1)/2!]Δ²f(x₀) + [p(p-1)(p-2)/3!]Δ³f(x₀) + [p(p-1)(p-2)(p-3)/4!]Δ⁴f(x₀), where p = (x - x₀) / h.

For x = 142, p = (142 - 140) / 10 = 0.2.

Substituting the values from the table:

f(142) = 3.685 + (0.2)(1.169) + [(0.2)(0.2-1)/2](0.279) + [(0.2)(0.2-1)(0.2-2)/6](0.017) + [(0.2)(0.2-1)(0.2-2)(0.2-3)/24](0.007)

f(142) = 3.685 + 0.2338 - 0.02232 + 0.000816 + 0.0001075

f(142) = 3.8974035

Rounding to three decimal places, the value is 3.897.

By the method of least squares, find the straight line that best fits in the following data:

x	1	2	3	4	5
y	14	27	40	55	68

15 marks

Answer

To find the best-fit straight line of the form y = ax + b using the method of least squares, we solve the following normal equations:

Σy = aΣx + nb

Σxy = aΣx² + bΣx

Given the data points (1, 14), (2, 27), (3, 40), (4, 55), and (5, 68), we first calculate the necessary summations for n = 5:

x	y	x²	xy
1	14	1	14
2	27	4	54
3	40	9	120
4	55	16	220
5	68	25	340
Total	204	55	748

Substituting these values into the normal equations:

204 = 15a + 5b

748 = 55a + 15b

Multiplying the first equation by 3 gives 612 = 45a + 15b. Subtracting this from the second equation:

748 - 612 = (55 - 45)a

136 = 10a

a = 13.6

Substituting a = 13.6 into the first equation:

204 = 15(13.6) + 5b

204 = 204 + 5b

b = 0

Thus, the straight line that best fits the data is y = 13.6x.

315 marks

Determine a root of

xe^x - 1 = 0

by method of false position.

7 marks

Answer

Determine

dy/dx

and

d^2y/dx^2

x = 0

from the following data:

x	0	1	2	3	4	5
y	4	8	15	7	6	2

8 marks

Answer

415 marks

Given that

dy/dx = x^2 + y^2, y(1) = 1

. Find

y(2)

in steps of 0.2 using the Euler’s method.

8 marks

Answer

Evaluate

∫_{0}^{0.6} e^{-x^2} dx

by using Simpson’s 1/3rd rule taking seven ordinates.

7 marks

Answer

Find the half – range sine and cosine series expansions of the function:

f(x) = { kx/L if 0 < x < L/2; k(L-x)/L if L/2 < x < L }

15 marks

Answer

Find the Fourier transform of

f(x) = { 1-x^2 for |x| ≤ 1; 0 for |x| > 1 }

hence evaluate

∫_{0}^{∞} ((x cos x - sin x)/x^3) cos(x/2) dx

15 marks

Answer

715 marks

Derive the partial differential equation by eliminating the constants a and b from the equation

z = ax + by + a^2 + b^2

7 marks

Answer

Solve the partial differential equation

(y^2 + yz)p + (x^2 + xz)q = z(x + y)

8 marks

Answer

State Gauss divergence theorem and Verify Gauss divergence theorem for the function

F = 4xz i - y^2 j + yz k

over the cylindrical region bounded by

x^2 + y^2 = 4, z = 0 and z = 3

15 marks

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech I Year MATHEMATICAL METHODS Previous Question Paper (October/November - 2020 - R15)

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