(a) It is given that the equation of any cone with vertex at (a, b, c) is f((x-a)/(z-c), (y-b)/(z-c)) = 0. Find the differential…

(a) Differential equation of the cone

Let u = (x−a)/(z−c), v = (y−b)/(z−c), with z = z(x, y), p = ∂z/∂x, q = ∂z/∂y. Differentiating f(u, v) = 0 partially: w.r.t. x: f_u[(z−c) − (x−a)p]/(z−c)² + f_v[−(y−b)p]/(z−c)² = 0, w.r.t. y: f_u[−(x−a)q]/(z−c)² + f_v[(z−c) − (y−b)q]/(z−c)² = 0.

So f_u[(z−c) − (x−a)p] = f_v(y−b)p and f_v[(z−c) − (y−b)q] = f_u(x−a)q. Eliminating f_u/f_v: [(z−c) − (x−a)p][(z−c) − (y−b)q] = (x−a)(y−b)pq ⟹ (z−c)[(z−c) − (x−a)p − (y−b)q] = 0.

Since z ≠ c on the cone (away from the vertex plane), (x−a)p + (y−b)q = z − c. This linear PDE is satisfied by every cone with vertex (a, b, c), since such a surface is homogeneous of degree one in (x−a, y−b, z−c).

(b) Gauss elimination

Augmented matrix, after interchanging R₁ and R₂: [1 3 2 | 13] [2 2 4 | 18] [3 1 3 | 14]

R₂ ← R₂ − 2R₁, R₃ ← R₃ − 3R₁: [1 3 2 | 13] [0 −4 0 | −8] [0 −8 −3 | −25]

R₃ ← R₃ − 2R₂: [1 3 2 | 13] [0 −4 0 | −8] [0 0 −3 | −9]

Back-substitution: z = 3, y = 2, x = 13 − 6 − 6 = 1. Check: 2+4+12 = 18, 1+6+6 = 13, 3+2+9 = 14. (x, y, z) = (1, 2, 3).

(c)(i) Number system conversions

1093 to octal: 1093 = 8·136 + 5, 136 = 8·17 + 0, 17 = 8·2 + 1, 2 = 8·0 + 2 → 2105. 0.21875 × 8 = 1.75 → 1, 0.75 × 8 = 6.00 → 6. (1093.21875)₁₀ = (2105.16)₈.

1693 to hexadecimal: 1693 = 16·105 + 13 (D), 105 = 16·6 + 9, 6 → 69D. 0.0628 × 16 = 1.0048 → 1, 0.0048 × 16 = 0.0768 → 0, 0.0768 × 16 = 1.2288 → 1, 0.2288 × 16 = 3.6608 → 3, 0.6608 × 16 = 10.5728 → A, 0.5728 × 16 = 9.1648 → 9, … (1693.0628)₁₀ ≈ (69D.1013A9…)₁₆ (non-terminating).

(c)(ii) Product of maxterms

F = xy + x′z. F = 1 when x = 1, y = 1 (rows 6, 7) or x = 0, z = 1 (rows 1, 3). Hence F = 0 at rows 0 (000), 2 (010), 4 (100), 5 (101). F = ΠM(0, 2, 4, 5) = (x + y + z)(x + y′ + z)(x′ + y + z)(x′ + y + z′).

(d) Lagrangian for the inverse-square central force

In plane polar coordinates (r, θ), for mass m: T = ½m(ṙ² + r²θ̇²). F = −dV/dr = −k/r² gives V = −k/r. L = T − V = ½m(ṙ² + r²θ̇²) + k/r.

r-equation: d/dt(mṙ) − (mrθ̇² − k/r²) = 0 ⟹ m(r̈ − rθ̇²) = −k/r².

θ-equation: θ is cyclic (∂L/∂θ = 0), so d/dt(mr²θ̇) = 0 ⟹ r²θ̇ = h (constant), conservation of angular momentum.

(e) Velocity potential and streamlines

As printed, v_θ = Mr⁻² sinθ gives ∇·v = −2Mr⁻⁴cosθ + 2Mr⁻³cosθ ≠ 0 and (∇×v)_ψ = Mr⁻⁴(2 − r)sinθ ≠ 0, so that field is neither incompressible nor irrotational. The intended (standard dipole) field is v = (2Mr⁻³cosθ, Mr⁻³sinθ, 0), which is solved below.

Irrotational: (∇×v)_ψ = (1/r)[∂(r v_θ)/∂r − ∂v_r/∂θ] = (1/r)[−2Mr⁻³sinθ + 2Mr⁻³sinθ] = 0, and the other components vanish as v_ψ = 0 and nothing depends on ψ. Also ∇·v = −2Mr⁻⁴cosθ + 2Mr⁻⁴cosθ = 0. Hence the motion is of the potential kind.

Potential (v = ∇φ): ∂φ/∂r = 2Mr⁻³cosθ ⟹ φ = −Mr⁻²cosθ + f(θ, ψ). Then (1/r)∂φ/∂θ = Mr⁻³sinθ, so f is constant. φ = −M cosθ/r² (sign reverses if v = −∇φ is used).

Streamlines: dr/v_r = r dθ/v_θ = r sinθ dψ/v_ψ. Since v_ψ = 0, ψ = constant, and dr/(2cosθ) = r dθ/sinθ ⟹ dr/r = 2cotθ dθ ⟹ r = C sin²θ, ψ = constant.