(a)(i) Find the orthogonal trajectories of the family of confocal conics (x²)/(a²+λ) + (y²)/(b²+λ) = 1; a > b > 0 are constants…

(a)(i) Orthogonal trajectories of confocal conics

Let p = dy/dx and c² = a² − b². Write A = a² + λ, B = b² + λ, so A − B = c². The family gives x²/A + y²/B = 1 and, on differentiation, x/A + yp/B = 0.

Solving these two linear equations: 1/A = p/[x(xp − y)] and 1/B = −1/[y(xp − y)]. Hence A − B = (xp − y)(x/p + y) = c², that is,

xy p² + (x² − y² − c²)p − xy = 0.

For orthogonal trajectories replace p by −1/p: xy/p² − (x² − y² − c²)/p − xy = 0. Multiplying by −p² gives xy p² + (x² − y² − c²)p − xy = 0, the same equation.

So the differential equation is unchanged, the trajectories belong to the same family, and the family is self-orthogonal. The orthogonal trajectories are the confocal conics x²/(a² + μ) + y²/(b² + μ) = 1 (ellipses for μ > −b², hyperbolas for −a² < μ < −b²).

(a)(ii) x²y″ − 2x(1 + x)y′ + 2(1 + x)y = x³

Homogeneous equation: y₁ = x is a solution. Put y = xv: the equation reduces to x³(v″ − 2v′) = 0, so v′ = Ce²ˣ and y₂ = xe²ˣ. So

y_c = C₁x + C₂xe²ˣ.

For the non-homogeneous equation, write it as y″ − (2(1 + x)/x)y′ + (2(1 + x)/x²)y = x, so R = x. The Wronskian is W = y₁y₂′ − y₁′y₂ = 2x²e²ˣ.

y_p = −y₁∫y₂R/W dx + y₂∫y₁R/W dx = −x∫(1/2)dx + xe²ˣ∫(e⁻²ˣ/2)dx = −x²/2 − x/4.

The term −x/4 is absorbed in C₁x, so the general solution is

y = C₁x + C₂xe²ˣ − x²/2.

(b) Projectile motion and locus of vertices

Take the point of projection as origin, x horizontal and y upward. The motion is uniform horizontally with velocity u cos θ and uniformly accelerated vertically with acceleration −g: x = ut cos θ, y = ut sin θ − gt²/2. Eliminating t: y = x tan θ − gx²/(2u²cos²θ), a parabola.

Time of flight 2u sin θ/g; range u² sin 2θ/g; the vertex (greatest height) is at X = u² sin θ cos θ/g = u² sin 2θ/(2g), Y = u² sin²θ/(2g).

For u = 4√g, u² = 16g, so X = 8 sin 2θ and Y = 8 sin²θ. Then X² = 64 sin²2θ = 256 sin²θ cos²θ = 256(Y/8)(1 − Y/8) = 32Y − 4Y².

So X² + 4Y² − 32Y = 0, that is,

X²/64 + (Y − 4)²/16 = 1,

an ellipse with centre (0, 4), semi-axes 8 along X and 4 along Y.

(c) Stokes' theorem for the paraboloid

By Stokes' theorem, ∬_S (∇ × F)·n̂ dS = ∮_C F·dr, where C is the boundary circle x² + y² = 4, z = 0, traversed anticlockwise as seen from above, matching the outward (upward) normal.

On C, z = 0 and dz = 0, so only F_x and F_y matter: F·dr = (x² + y − 4)dx + 3xy dy. With x = 2cos t, y = 2sin t:

F·dr = [(4cos²t + 2sin t − 4)(−2sin t) + 12cos t sin t(2cos t)]dt = (16cos²t sin t − 4sin²t + 8sin t)dt.

Over 0 to 2π, ∫cos²t sin t dt = 0, ∫sin t dt = 0 and ∫sin²t dt = π. Hence

∬_S (∇ × F)·n̂ dS = −4π.