JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech I Year MATHEMATICAL METHODS Previous Question Paper (June - 2022 - R13)

B.Tech | I Year | I | R13 | June - 2022

Answer PaperPaper Code: 111ALMaximum Marks: 75Time: 3 Hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a

From the data find number of students who obtained marks between 40 and 45.

Marks30-4040-5050-6060-7070-80
Number of students3142513531
8 marks

Answer

To find the number of students who obtained marks between 40 and 45, we use Newton's forward interpolation formula. Let x be the marks and y be the number of students. The class interval h = 10. We set the origin at x0 = 40, so y0 = 42.

u = (x - x0) / h = (45 - 40) / 10 = 0.5

The forward difference table is:

xyΔyΔ2yΔ3yΔ4y
303111-2-1427
40429-1613
5051-7-3
6035-10
7031

Using Newton's forward interpolation formula:

y(x) = y0 + uΔy0 + [u(u-1)/2!]Δ2y0 + [u(u-1)(u-2)/3!]Δ3y0 + [u(u-1)(u-2)(u-3)/4!]Δ4y0

Substituting the values y0=42, Δy0=9, Δ2y0=-16, Δ3y0=13, Δ4y0=27 and u=0.5:

y(45) = 42 + 0.5(9) + [0.5(-0.5)/2](-16) + [0.5(-0.5)(-1.5)/6](13) + [0.5(-0.5)(-1.5)(-2.5)/24](27)
y(45) = 42 + 4.5 + 2 + 0.8125 + 0.791 = 50.1035

The number of students who obtained marks between 40 and 45 is approximately 50 - 31 = 19 students (considering the range starts from 40).

b

Using Lagrange’s interpolation, find the interpolating polynomial for the data:

X013
F(X)1355
7 marks

Answer

Given data points are (x0, y0) = (0, 1), (x1, y1) = (1, 3), and (x2, y2) = (3, 55). Lagrange's interpolation formula is:

f(x) = L0(x)y0 + L1(x)y1 + L2(x)y2

Calculating the Lagrange basis polynomials:

L0(x) = [(x-1)(x-3)] / [(0-1)(0-3)] = (x2 - 4x + 3) / 3
L1(x) = [(x-0)(x-3)] / [(1-0)(1-3)] = (x2 - 3x) / -2
L2(x) = [(x-0)(x-1)] / [(3-0)(3-1)] = (x2 - x) / 6

Substituting into the formula:

f(x) = [(x2 - 4x + 3)/3](1) + [(x2 - 3x)/-2](3) + [(x2 - x)/6](55)
f(x) = (1/6) [2x2 - 8x + 6 - 9x2 + 27x + 55x2 - 55x]
f(x) = (1/6) [48x2 - 36x + 6] = 8x2 - 6x + 1
215 marks
a

Using the least square method fit a straight line for the following data:

X-1012
Y(X)1014
8 marks

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b

Using the least square approximation fit a quadratic polynomial for the data:

X012
Y(X)1617
7 marks

Answer

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315 marks
a

Using the Newton Raphson Method, find the cube root of 12.

8 marks

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b

Solve the following system by Gauss-Seidel method:

10x1 - 2x2 - x3 - x4 = 3
- 2x1 + 10x2 - x3 - x4 = 15
- x1 - x2 + 10x3 - 2x4 = 27
- x1 - x2 - 2x3 + 10x4 = -9
7 marks

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415 marks
a

The velocity V(km/min) of a moped which starts from rest is given at fixed intervals of the time T(min) as follows:

T2468101214161820
V10182529322011520

Using the 1/3 Simpson’s rule estimate the distance covered in 20 minutes.

8 marks

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b

Using 4th order Runge Kutta Method, with h = 0.1 find y(0.1) from y' = y - x with y(0) = 2

7 marks

Answer

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5

Expand

f(x) = sin(x), 0 < x < π
as a Fourier series.

15 marks

Answer

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6

A tightly stretched string of length has its ends fastened at x=0 and x=L. The mid-point of the string is then taken to height h and then released from rest in that position. Find the expression for the displacement of the string at any point at any time.

15 marks

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7

A bar of 10 cms long with insulated sides has its ends A and B maintained at temperature 50 °C and 100 °C respectively until the steady state conditions prevail. The temperature at A is suddenly raised to 90 °C and at the same time that at B is reduced to 60 °C. Find the temperature distribution in the bar at any point at any time.

15 marks

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8

Verify the Stroke’s theorem for

F = (x2 + y2)i - 2xyj
taken around the rectangle bounded by the lines x = ±a, y = 0, and y = b.

15 marks

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This page contains the solved question paper with answers of MATHEMATICAL METHODS from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech EEE I Year I, June - 2022, R13.

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