(a) Show that if f and g are arbitrary functions of their respective arguments, then u = f(x - kt + iαy) + g(x - kt - iαy), is a…

(a) Let ξ = x - kt + iαy, η = x - kt - iαy, u = f(ξ) + g(η), where f and g are twice differentiable arbitrary functions.

Differentiate u with respect to x: u_x = f′(ξ) + g′(η) u_xx = f″(ξ) + g″(η).

Differentiate u with respect to y. Since ξ_y = iα and η_y = -iα, u_y = iα f′(ξ) - iα g′(η) u_yy = (iα)² f″(ξ) + (-iα)² g″(η) = -α² f″(ξ) - α² g″(η).

Therefore ∂²u/∂x² + ∂²u/∂y² = (1 - α²)[f″(ξ) + g″(η)].

Now differentiate with respect to t. Since ξ_t = -k and η_t = -k, u_t = -k f′(ξ) - k g′(η) u_tt = k² f″(ξ) + k² g″(η) = k²[f″(ξ) + g″(η)].

Using α² = 1 - k²/C², we get 1 - α² = k²/C². Hence ∂²u/∂x² + ∂²u/∂y² = (k²/C²)[f″(ξ) + g″(η)].

But (1/C²)∂²u/∂t² = (1/C²)k²[f″(ξ) + g″(η)] = (k²/C²)[f″(ξ) + g″(η)].

Thus ∂²u/∂x² + ∂²u/∂y² = (1/C²)∂²u/∂t². So u is a solution.

(b) The given system is 2x + 3y - z = 5 4x + 4y - 3z = 3 2x - 3y + 2z = 2.

Write the augmented matrix: [2 3 -1 | 5] [4 4 -3 | 3] [2 -3 2 | 2]

Apply Gauss-Jordan elimination.

R2 → R2 - 2R1, R3 → R3 - R1: [2 3 -1 | 5] [0 -2 -1 | -7] [0 -6 3 | -3]

R3 → R3 - 3R2: [2 3 -1 | 5] [0 -2 -1 | -7] [0 0 6 | 18]

R3 → R3/6: [2 3 -1 | 5] [0 -2 -1 | -7] [0 0 1 | 3]

Eliminate z from R1 and R2: R1 → R1 + R3, R2 → R2 + R3: [2 3 0 | 8] [0 -2 0 | -4] [0 0 1 | 3]

R2 → R2/(-2): [2 3 0 | 8] [0 1 0 | 2] [0 0 1 | 3]

R1 → R1 - 3R2: [2 0 0 | 2] [0 1 0 | 2] [0 0 1 | 3]

R1 → R1/2: [1 0 0 | 1] [0 1 0 | 2] [0 0 1 | 3]

Hence x = 1, y = 2, z = 3.

(c)(i) In sign-magnitude representation, the most significant bit is the sign bit: 0 means positive and 1 means negative. The remaining bits give the magnitude.

For (8D)₁₆: (8D)₁₆ = 1000 1101₂. The sign bit is 1, so the number is negative. The magnitude is 000 1101₂ = 13₁₀. Therefore (8D)₁₆ ≡ -13.

For (FF)₁₆: (FF)₁₆ = 1111 1111₂. The sign bit is 1, so the number is negative. The magnitude is 111 1111₂ = 127₁₀. Therefore (FF)₁₆ ≡ -127.

(c)(ii) Expand (9B2.1A)₁₆ in powers of 16: (9B2.1A)₁₆ = 9×16² + 11×16¹ + 2×16⁰ + 1×16⁻¹ + 10×16⁻².

Now 9×16² = 9×256 = 2304, 11×16 = 176, 2×16⁰ = 2, 1×16⁻¹ = 1/16, 10×16⁻² = 10/256.

Thus (9B2.1A)₁₆ = 2304 + 176 + 2 + 1/16 + 10/256 = 2482 + 1/16 + 5/128 = 2482 + (8 + 5)/128 = 2482 + 13/128 = 317709/128 = 2482.1015625.

Therefore (9B2.1A)₁₆ = 2482.1015625₁₀.

(d) Let the board move a distance s in the direction in which the man walks. The board is uniform, so its centre of mass is at its midpoint.

Initially, take the centre of the board at x = 0. The man is at one end, so his initial coordinate is x = -a.

Finally, the man reaches the other end of the board. Since the board has shifted by s, the man’s final coordinate is x = s + a, while the board’s centre has coordinate x = s.

The horizontal plane is smooth. Hence there is no external horizontal force on the system consisting of the man and the board. The system is initially at rest, so its centre of mass remains fixed horizontally.

Using conservation of the centre of mass: M(-a) + m(0) = M(s + a) + ms.

That is, -Ma = Ms + Ma + ms -Ma - Ma = (M + m)s -2Ma = (M + m)s.

Therefore s = -2Ma/(M + m).

The negative sign shows that the board moves opposite to the direction in which the man walks. The distance covered by the board is 2Ma/(M + m).

(e) For a velocity potential φ, the velocity components are u = ∂φ/∂x, v = ∂φ/∂y, w = ∂φ/∂z.

Given φ = (1/2)(x² + y² - 2z²), we get u = ∂φ/∂x = x, v = ∂φ/∂y = y, w = ∂φ/∂z = -2z.

The differential equations of streamlines are dx/u = dy/v = dz/w, so dx/x = dy/y = dz/(-2z).

From dx/x = dy/y, integrating gives ln|x| = ln|y| + constant, hence x/y = c₁, where c₁ is a constant.

From dx/x = dz/(-2z), we have 2 dx/x = -dz/z. Integrating, 2 ln|x| = -ln|z| + constant, so x²z = c₂, where c₂ is a constant.

Thus the streamlines are given by the pair of equations x/y = c₁, x²z = c₂. Equivalently, one may also write y²z = c₃, since this follows from the same set of relations.