Show that the vector field given by A = 3x2y i + (x3 - 2yz2) j + (3z2 - 2y2z) k is irrotational but not solenoidal. Also find its scalar potential ∅(x, y, z).

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JNTU B.Tech II MATHEMATICS-II Previous Question Paper (March - 2022 - R15)

B.Tech | II | II-II | R15 | March - 2022

Answer PaperPaper Code: 124DDMaximum Marks: 75Time: 3 Hours

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115 marks

Show that the vector field given by A = 3x²y i + (x³ - 2yz²) j + (3z² - 2y²z) k is irrotational but not solenoidal. Also find its scalar potential ∅(x, y, z).

8 marks

Answer

A vector field A is irrotational if the curl of A is zero. Calculating the curl:

∇ × A = (∂/∂y(3z² - 2y²z) - ∂/∂z(x³ - 2yz²)) i - (∂/∂x(3z² - 2y²z) - ∂/∂z(3x²y)) j + (∂/∂x(x³ - 2yz²) - ∂/∂y(3x²y)) k

∇ × A = (-4yz - (-4yz)) i - (0 - 0) j + (3x² - 3x²) k = 0 i + 0 j + 0 k

Since ∇ × A = 0, the field is irrotational. A field is solenoidal if the divergence is zero.

∇ · A = ∂/∂x(3x²y) + ∂/∂y(x³ - 2yz²) + ∂/∂z(3z² - 2y²z) = 6xy - 2z² + (6z - 2y²)

Since ∇ · A = 6xy - 2y² + 6z - 2z² ≠ 0, the field is not solenoidal. To find the scalar potential φ, we solve ∇φ = A:

∂φ/∂x = 3x²y ⇒ φ = x³y + f(y, z)

∂φ/∂y = x³ + ∂f/∂y = x³ - 2yz² ⇒ ∂f/∂y = -2yz² ⇒ f = -y²z² + g(z)

∂φ/∂z = -2y²z + g'(z) = 3z² - 2y²z ⇒ g'(z) = 3z² ⇒ g(z) = z³ + C

Thus, φ(x, y, z) = x³y - y²z² + z³ + C.

Evaluate

∫∫_s F · n dS

where F = 6z i - 4 j + y k and S is the portion of the plane 2x + 3y + 6z = 12 in the first octant.

7 marks

Answer

We evaluate the surface integral using the projection method on the xy-plane. The surface is z = (12 - 2x - 3y)/6 = 2 - x/3 - y/2. The unit normal n is given by ∇(2x + 3y + 6z - 12) / |∇(2x + 3y + 6z - 12)| = (2i + 3j + 6k) / 7. The differential surface element dS = (dx dy) / |n · k| = (dx dy) / (6/7) = (7/6) dx dy.

F · n = (6z i - 4 j + y k) · (2i + 3j + 6k) / 7 = (12z - 12 + 6y) / 7

Substituting z:

F · n = (12(2 - x/3 - y/2) - 12 + 6y) / 7 = (24 - 4x - 6y - 12 + 6y) / 7 = (12 - 4x) / 7

The integral becomes:

∫∫_R ((12 - 4x) / 7) * (7/6) dx dy = ∫₀⁶ ∫₀^(12-2x)/3 (2 - 2x/3) dy dx = ∫₀⁶ (2 - 2x/3) * ((12 - 2x)/3) dx

= 2/9 ∫₀⁶ (6 - x)² dx = 2/9 [-(6 - x)³ / 3]₀⁶ = 2/9 * (6³ / 3) = 2/9 * 72 = 16.

215 marks

Using Green’s theorem evaluate

∫_C (x dy - y dx)

, where C is a closed curve in the xy-plane enclosing an area A.

7 marks

Answer

State the Stoke’s theorem and divergence theorem. If F is irrotational, then find

∫∫_S F · n dS

where S is an open surface.

8 marks

Answer

315 marks

Find Fourier sine transform of f(x) =

1 / (x(x² + a²))

and hence deduce cosine transform of

1 / (x² + a²)

7 marks

Answer

Find the fourier series of f(x) = e^x in (-π, π).

8 marks

Answer

415 marks

Find the Fourier series expansion of the function f(x) = 0 for -π ≤ x ≤ 0 and f(x) = sin x for 0 ≤ x ≤ π, and deduce

1/(1·3) - 1/(3·5) + 1/(5·7) ... = (π - 2)/4

10 marks

Answer

State and prove change of scale property of Fourier transform.

5 marks

Answer

515 marks

Find sin 52° using Newton’s interpolation formula from the data points (45, 0.7071), (50, 0.7660), (55, 0.8192), (60, 0.8660).

8 marks

Answer

Write the normal equations to fit a curve of the form y = ab^x from the given data points (x_i, y_i) for i = 1, 2, ...

7 marks

Answer

615 marks

Fit a curve y = a + bx + cx² to the following data: x: 1, 3, 5, 7, 9; y: 2, 7, 10, 11, 9.

7 marks

Answer

Fit a straight line to the following data giving weights to x as 1, 1, 2, 1, 1 by the method of least square. x: 0, 1, 2, 3, 4; y: 1, 1.8, 3.3, 4.5, 6.5.

8 marks

Answer

715 marks

Find the inverse of the matrix [[2, 3, 1], [1, 2, 3], [3, 2, 1]] using LU decomposition method.

8 marks

Answer

Find the root of 2x - log₁₀x = 7, correct to five places of decimal using iteration method.

7 marks

Answer

Using the shooting method, solve the first boundary value problem u'' = u + 1, 0 < x < 1; u(0) = 0; u(1) = e - 1.

15 marks

JNTU B.Tech II MATHEMATICS-II Previous Question Paper (March - 2022 - R15)

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