(a) Obtain the partial differential equation by eliminating arbitrary function f from the equation f(x+y+z, x²+y²+z²) = 0. (10…

(a) PDE from f(x + y + z, x² + y² + z²) = 0

Let u = x + y + z, v = x² + y² + z², p = ∂z/∂x, q = ∂z/∂y. Differentiating f(u, v) = 0 with respect to x and y:

f_u(1 + p) + f_v(2x + 2zp) = 0, f_u(1 + q) + f_v(2y + 2zq) = 0.

These are homogeneous in f_u, f_v, which are not both zero, so the determinant vanishes:

(1 + p)(2y + 2zq) − (1 + q)(2x + 2zp) = 0.

Expanding, the pq terms cancel, leaving (y − z)p + (z − x)q = x − y. This is the required Lagrange linear PDE.

(b) Root of 3x = 1 + cos x

Let f(x) = 3x − 1 − cos x. Since f′(x) = 3 + sin x > 0, f is increasing and has exactly one real root. f(0) = −2 < 0 and f(π/2) = 3π/2 − 1 ≈ 3.712 > 0, so the root lies in (0, π/2).

Regula falsi, x = (af(b) − bf(a))/(f(b) − f(a)), keeping the interval with a sign change:

x₁ = 0.549961, f = −0.20266; x₂ = 0.602804, f = −0.015336; x₃ = 0.606787, f = −0.001125.

Take x₀ = 0.6068. Newton–Raphson, xₙ₊₁ = xₙ − (3xₙ − 1 − cos xₙ)/(3 + sin xₙ):

x₀ = 0.6068 gives x₁ = 0.607101659; then x₂ = 0.607101648; a further step leaves 0.607101648.

To 8 significant figures the root is x = 0.60710165.

(c)(i) (3798.3875)₁₀ in octal and hexadecimal

Integer part 3798. Octal: 3798 = 8·474 + 6, 474 = 8·59 + 2, 59 = 8·7 + 3, 7 = 8·0 + 7, giving 7326. Hexadecimal: 3798 = 16·237 + 6, 237 = 16·14 + 13, 14 = E, giving ED6.

Fraction 0.3875. Octal, multiplying repeatedly by 8: 3.1 → 3; 0.1 × 8 = 0.8 → 0; 0.8 × 8 = 6.4 → 6; 0.4 × 8 = 3.2 → 3; 0.2 × 8 = 1.6 → 1; 0.6 × 8 = 4.8 → 4; then 0.8 recurs, so the block 6314 repeats: 0.30(6314 repeating). Hexadecimal, multiplying by 16: 6.2 → 6; 0.2 × 16 = 3.2 → 3; and 0.2 recurs, so 0.63333….

(3798.3875)₁₀ = (7326.3063146314…)₈ = (ED6.6333…)₁₆.

(c)(ii) Principal conjunctive normal form of (¬P → R) ∧ (Q ⇔ P)

¬P → R is equivalent to P ∨ R, and Q ⇔ P is true when P and Q have equal truth values. The formula is therefore true only for (P, Q, R) = (T, T, T), (T, T, F) and (F, F, T), and false for the other five rows: (T, F, T), (T, F, F), (F, T, T), (F, T, F), (F, F, F).

Each false row gives a maxterm in which a variable appears negated if it is T in that row:

PCNF = (¬P ∨ Q ∨ ¬R) ∧ (¬P ∨ Q ∨ R) ∧ (P ∨ ¬Q ∨ ¬R) ∧ (P ∨ ¬Q ∨ R) ∧ (P ∨ Q ∨ R).

(d) Equation of motion by D'Alembert's principle

Take the centre of the circle (radius a) as origin, x horizontal and y vertically upward; the wire is smooth, so the reaction does no virtual work. The only applied force is gravity (0, −mg).

D'Alembert's principle: the applied forces and the reversed effective forces do no virtual work, Σ(F − mr̈)·δr = 0:

(−mẍ)δx + (−mg − mÿ)δy = 0, that is, ẍδx + (ÿ + g)δy = 0.

The constraint x² + y² = a² gives xδx + yδy = 0, so a virtual displacement is δx = −yε, δy = xε with ε arbitrary. Substituting: −ẍyε + (ÿ + g)xε = 0, hence

ẍy − ÿx − gx = 0.

(Check: with x = a sin θ, y = −a cos θ this becomes θ̈ = −(g/a) sin θ, the simple pendulum.)

(e) Sources and sinks for w = log(z − a²/z)

w = log[(z − a)(z + a)/z] = log(z − a) + log(z + a) − log z. Taking w = m log(z − c) as a source of strength m at c (if the opposite sign convention is used, sources and sinks interchange), the flow is due to unit sources at (a, 0) and (−a, 0) and a unit sink of equal strength at the origin.

For z = re^(iθ): z − a²/z = (r − a²/r)cos θ + i(r + a²/r) sin θ, so the stream function is

ψ = arg(z − a²/z) = tan⁻¹[(r² + a²) sin θ/((r² − a²) cos θ)].

Streamlines are the curves (r² + a²) sin θ = k(r² − a²) cos θ. For ψ = π/2 the real part (r² − a²)cos θ must vanish, so r = a or cos θ = 0, that is, the circle r = a and the y-axis. Hence these two together form one dividing streamline. The sketch shows fluid leaving the sources at (±a, 0), part going to the sink at O inside the circle and the rest flowing out to infinity, with no fluid crossing r = a or the y-axis.