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JNTU B.Tech II MATHEMATICS - II Previous Question Paper (August/September - 2022 - R13)

B.Tech | II | I | R13 | August/September - 2022

Answer PaperPaper Code: 113AAMaximum Marks: 75Time: 3 Hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

Show that the vector 𝐹̅ = (π‘₯ + 3𝑦)𝑖̅+ (𝑦 βˆ’ 3𝑧)𝑗̅+ (π‘₯ βˆ’ 2𝑧)π‘˜Μ… is solenoidal and also find 𝐹̅ βˆ™ π‘π‘’π‘Ÿπ‘™ 𝐹̅.

7 marks

Answer

A vector field F is solenoidal if its divergence is zero, denoted by div F = βˆ‡ Β· F = 0. Given F = (x + 3y)i + (y - 3z)j + (x - 2z)k, the divergence is calculated as:

βˆ‡ Β· F = βˆ‚/βˆ‚x(x + 3y) + βˆ‚/βˆ‚y(y - 3z) + βˆ‚/βˆ‚z(x - 2z) = 1 + 1 - 2 = 0

Since the divergence is zero, the vector field is solenoidal. To find F Β· curl F, we first compute curl F = βˆ‡ Γ— F:

βˆ‡ Γ— F = | i j k | | βˆ‚/βˆ‚x βˆ‚/βˆ‚y βˆ‚/βˆ‚z | | (x + 3y) (y - 3z) (x - 2z) | = i(βˆ‚/βˆ‚y(x - 2z) - βˆ‚/βˆ‚z(y - 3z)) - j(βˆ‚/βˆ‚x(x - 2z) - βˆ‚/βˆ‚z(x + 3y)) + k(βˆ‚/βˆ‚x(y - 3z) - βˆ‚/βˆ‚y(x + 3y)) = i(0 - (-3)) - j(1 - 0) + k(0 - 3) = 3i - j - 3k

Now, calculate the dot product F Β· curl F:

F Β· curl F = (x + 3y)(3) + (y - 3z)(-1) + (x - 2z)(-3) = 3x + 9y - y + 3z - 3x + 6z = 8y + 9z
b

In what direction from (3, 1, βˆ’2) is the directional derivative of πœ™ = π‘₯³𝑦² + 𝑦𝑧 maximum? Find also the magnitude of this maximum.

8 marks

Answer

The directional derivative of a scalar field Ο† is maximum in the direction of the gradient vector βˆ‡Ο†. Given Ο† = xΒ³yΒ² + yz, the gradient is:

βˆ‡Ο† = (βˆ‚Ο†/βˆ‚x)i + (βˆ‚Ο†/βˆ‚y)j + (βˆ‚Ο†/βˆ‚z)k = (3xΒ²yΒ²)i + (2xΒ³y + z)j + (y)k

At the point (3, 1, -2):

βˆ‡Ο†(3, 1, -2) = (3(3)Β²(1)Β²)i + (2(3)Β³(1) + (-2))j + (1)k = 27i + 52j + k

The direction of maximum increase is the vector 27i + 52j + k. The magnitude of this maximum directional derivative is the magnitude of the gradient vector:

|βˆ‡Ο†| = √(27Β² + 52Β² + 1Β²) = √(729 + 2704 + 1) = √3434 β‰ˆ 58.6
2

State Stokes theorem. Verify it for the vector field 𝐹̅ = (2π‘₯ βˆ’ 𝑦)π‘–Μ…βˆ’ π‘¦π‘§Β²π‘—Μ…βˆ’ π‘¦Β²π‘§π‘˜Μ… over the upper half surface of the sphere π‘₯Β² + 𝑦² + 𝑧² = 1, bounded by its projection on the π‘₯𝑦-plane.

15 marks

Answer

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315 marks
a

If 𝑓(π‘₯) = sin π‘₯ for 0 ≀ π‘₯ ≀ πœ‹/4 and cos π‘₯ for πœ‹/4 ≀ π‘₯ ≀ πœ‹/2, obtain the half-range sine series.

7 marks

Answer

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b

Find the Fourier transform of 𝑓(π‘₯) = 1 for |π‘₯| ≀ π‘Ž and 0 for |π‘₯| > π‘Ž. Hence prove that βˆ«β‚€^∞ (sinΒ² π‘Žπ‘₯)/π‘₯Β² 𝑑π‘₯ = πœ‹π‘Ž/2.

8 marks

Answer

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415 marks
a

Determine the Fourier series for the function πœ‹-π‘₯Β² in βˆ’πœ‹ < π‘₯ < πœ‹.

6 marks

Answer

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b

Find the Fourier sine transform of 𝑓(π‘₯) = 𝑒⁻|π‘₯|. Hence prove that βˆ«β‚€^∞ (π‘₯ sin π‘šπ‘₯)/(1+π‘₯Β²) 𝑑π‘₯ = (πœ‹/2)π‘’β»π‘š, π‘š > 0.

9 marks

Answer

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515 marks
a

Use Newton’s formula of interpolation to find cos(0.17) given that cos(0) = 1, cos(0.05) = 0.9988, cos(0.10) = 0.9950, cos(0.15) = 0.9888, cos(0.20) = 0.9801, cos(0.25) = 0.9689 and cos(0.30) = 0.9553.

7 marks

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b

Using Lagrange’s interpolation, find the form of the function 𝑓(π‘₯) and hence evaluate 𝑓(0) from the data: (βˆ’1, 7), (1, 5), (2, 15) and (3, 19).

8 marks

Answer

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615 marks
a

Find 𝑦₂₅ given that 𝑦₂₀ = 24, 𝑦₂₄ = 32, π‘¦β‚‚β‚ˆ = 35, 𝑦₃₂ = 40, using Gauss forward difference formula.

7 marks

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b

Fit a parabolic curve to the following data and evaluate 𝑦(10):
π‘₯: 1, 2, 3, 4, 5, 6, 7
𝑦(π‘₯): 14, 17, 22, 35, 67, 89, 102

8 marks

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715 marks
a

Find a real root of the equation π‘₯Β³ βˆ’ 2 ln π‘₯ βˆ’ 4 = 0 correct to three decimal places using the method of false position.

7 marks

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b

Apply LU decomposition method to solve the system of equations: π‘₯ + 𝑦 + 2𝑧 = 4; 3π‘₯ + 𝑦 βˆ’ 3𝑧 = βˆ’4; 2π‘₯ βˆ’ 3𝑦 βˆ’ 5𝑧 = βˆ’5.

8 marks

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815 marks
a

Evaluate the integral βˆ«β‚€ΒΉ 𝑑π‘₯/(1+π‘₯Β²) using Gauss-Legendre one-point, 2-point and 3-point formulas. Also compare with the exact solution.

8 marks

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b

Solve the Initial Value Problem: 𝑑𝑦/𝑑π‘₯ = π‘₯ + 𝑦, 𝑦(0) = 1 using Taylor’s series method and compute 𝑦(1).

7 marks

Answer

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