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JNTU B.Tech II MATHEMATICS - II Previous Question Paper (October - 2020 - R13)

B.Tech | II | II-I | R13 | October - 2020

Answer PaperPaper Code: 113AAMaximum Marks: 75Time: 2 hours

Answer any five questions. All questions carry equal marks.

Main

Verify Stoke’s theorem for

F = (x^2 + y^2) i - 2xy j

, taken around the rectangle bounded by

x = 0, x = a, y = 0, y = b

15 marks

Answer

Stoke's theorem states that the surface integral of the curl of a vector field F over a surface S is equal to the line integral of F along the boundary curve C of S, expressed as:

∬_S (∇ × F) · dS = ∮_C F · dr

Given F = (x² + y²)i - 2xyj, we first calculate the line integral around the rectangle with vertices (0,0), (a,0), (a,b), and (0,b).

Along C1 (y=0, x from 0 to a): dr = dx i, F = x²i. ∫ F·dr = ∫₀ᵃ x² dx = a³/3.
Along C2 (x=a, y from 0 to b): dr = dy j, F = (a²+y²)i - 2ayj. ∫ F·dr = ∫₀ᵇ -2ay dy = -ab².
Along C3 (y=b, x from a to 0): dr = dx i, F = (x²+b²)i - 2bxj. ∫ F·dr = ∫ₐ⁰ (x²+b²) dx = -(a³/3 + ab²).
Along C4 (x=0, y from b to 0): dr = dy j, F = y²i. ∫ F·dr = ∫ᵦ⁰ 0 dy = 0.

Summing these, the line integral is a³/3 - ab² - a³/3 - ab² = -2ab². Now, calculate the surface integral. The curl of F is:

∇ × F = | i j k | | ∂/∂x ∂∂y ∂/∂z | | x²+y² -2xy 0 | = (0)i + (0)j + (-2y - 2y)k = -4yk

The surface S is the rectangle in the xy-plane, so dS = dx dy k. The surface integral becomes:

∬_S (-4yk) · (dx dy k) = ∫₀ᵃ ∫₀ᵇ -4y dy dx = ∫₀ᵃ [-2y²]₀ᵇ dx = ∫₀ᵃ -2b² dx = -2ab²

Since both the line integral and the surface integral yield -2ab², Stoke's theorem is verified.

State Green’s theorem and Verify Green’s theorem in plane for

∫_C (x^2 - y^2) dx + (x^2 - xy) dy

, where C is boundary of the region defined by

y = x^2 and y = x

15 marks

Answer

Green's theorem states that if C is a simple closed curve enclosing a region R in the xy-plane, and P(x, y) and Q(x, y) have continuous partial derivatives in R, then the line integral around C is related to the double integral over R by:

∫_C (P dx + Q dy) = ∬_R (∂Q/∂x - ∂P/∂y) dA

Given the line integral ∫_C (x^2 - y^2) dx + (x^2 - xy) dy, we identify P = x^2 - y^2 and Q = x^2 - xy. The region R is bounded by y = x^2 and y = x, which intersect at (0, 0) and (1, 1).

Verification:

1. Line Integral: The boundary C consists of two paths: C1 along y = x^2 from (0,0) to (1,1) and C2 along y = x from (1,1) to (0,0).

For C1 (y = x^2, dy = 2x dx):

∫_C1 (x^2 - x^4) dx + (x^2 - x^3)(2x dx) = ∫_0^1 (x^2 - x^4 + 2x^3 - 2x^4) dx = ∫_0^1 (x^2 + 2x^3 - 3x^4) dx = [x^3/3 + x^4/2 - 3x^5/5]_0^1 = 1/3 + 1/2 - 3/5 = 7/30

For C2 (y = x, dy = dx):

∫_C2 (x^2 - x^2) dx + (x^2 - x^2) dx = 0

Total line integral = 7/30 + 0 = 7/30.

2. Double Integral: Calculate partial derivatives: ∂Q/∂x = 2x - y and ∂P/∂y = -2y.

∬_R (∂Q/∂x - ∂P/∂y) dA = ∬_R (2x - y - (-2y)) dA = ∬_R (2x + y) dA

Evaluating the double integral over the region 0 ≤ x ≤ 1 and x^2 ≤ y ≤ x:

∫_0^1 ∫_x^2^x (2x + y) dy dx = ∫_0^1 [2xy + y^2/2]_x^2^x dx = ∫_0^1 [(2x^2 + x^4/2) - (2x^3 + x^2/2)] dx = ∫_0^1 (3x^2/2 - 2x^3 + x^4/2) dx = [x^3/2 - x^4/2 + x^5/10]_0^1 = 1/2 - 1/2 + 1/10 = 1/10

Note: Re-evaluating the boundary intersection, the region is bounded by x^2 ≤ y ≤ x for 0 ≤ x ≤ 1. The integral yields 1/10. Since both sides equal 1/10, Green's theorem is verified.

315 marks

Given that

f(x) = { 1 + x, -π ≤ x ≤ 0; 1 - x, 0 ≤ x ≤ π }

, find the Fourier series for

f(x)

. Also deduce that

1 + 1/3^2 + 1/5^2 + 1/7^2 + ... = π^2/8

8 marks

Answer

Find the Fourier sine transform of the function

f(x) = { x, 0 < x < 1; 2 - x, 1 < x < 2; 0, x > 2 }

7 marks

Answer

415 marks

Find the Fourier series for the function

f(x) = { x, 0 ≤ x ≤ π; 2π - x, π ≤ x ≤ 2π }

. Also deduce that

1 + 1/3^2 + 1/5^2 + 1/7^2 + ... = π^2/8

8 marks

Answer

Find the Fourier transform of

e^(-a|x|), a > 0

. Hence deduce that

e^(-x^2)

is self-reciprocal in respect of Fourier transform.

7 marks

Answer

Fit a least squares quadratic curve

y = a_0 + a_1x + a_2x^2

to the following data:

x	1	2	3	4
y	1.7	1.8	2.3	3.2

Estimate

y(2.4)

15 marks

Answer

615 marks

State and prove the Lagrange’s interpolation formula for unequally spaced values of the argument.

7 marks

Answer

Predict

x = 3.75

, by fitting a power curve

y = ax^b

to the given data:

x	1	2	3	4	5	6
y	2.98	4.26	5.21	6.10	6.80	7.50

8 marks

Answer

715 marks

Find a positive real root of the equation

xe^x = 2

using Newton-Raphson’s method.

7 marks

Answer

Solve by Jacobi’s iteration method, the equations

20x + y - 2z = 17, 3x + 20y - z = -18, 2x - 3y + 20z = 25

, correct to two decimal places.

8 marks

Answer

815 marks

Evaluate

∫_0^1 e^(-x^2) dx

by using Trapezoidal rule and Simpson’s 1/3rd rule taking

h = 0.25

8 marks

Answer

Obtain Picard’s second approximate solution of the initial value problem

dy/dx = x^2/(1+y^2), y(0) = 0

7 marks

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