(a) Find the orthogonal trajectories of the family of curves r = c(sec θ + tan θ), where c is a parameter. (10 marks) (b) Solve…

(a) Let S = sec θ + tan θ. Then

dS/dθ = sec θ tan θ + sec² θ = sec θ(sec θ + tan θ) = sec θ S.

Given r = cS, differentiating with respect to θ gives

dr/dθ = c sec θ S = r sec θ.

For a polar curve r = r(θ), the differential equation of its orthogonal trajectories R = R(θ) is

dR/dθ = -R²/(dr/dθ),

where dr/dθ is evaluated at the intersection point. Hence

dR/dθ = -R²/(R sec θ) = -R cos θ.

Separating variables,

dR/R = -cos θ dθ.

Integrating,

ln R = -sin θ + ln C.

Therefore the orthogonal trajectories are

R = C e^(-sin θ), equivalently r e^(sin θ) = C.

(b) Let Y(s) = L{y(t)}. By the convolution theorem,

L{∫₀ᵗ y(x) cos(t-x) dx} = Y(s) · s/(s²+1).

Also L{cos t} = s/(s²+1). Thus

Y(s) = s/(s²+1) + Y(s) · s/(s²+1).

So

Y(s)[1 - s/(s²+1)] = s/(s²+1).

Since 1 - s/(s²+1) = (s² - s + 1)/(s²+1),

Y(s) = s/(s² - s + 1).

Complete the square:

s² - s + 1 = (s - 1/2)² + 3/4.

Write s = (s - 1/2) + 1/2. Then

Y(s) = (s - 1/2)/[(s - 1/2)² + (√3/2)²] + (1/2)/[(s - 1/2)² + (√3/2)²].

Taking inverse Laplace transforms,

y(t) = e^(t/2) cos(√3 t/2) + (1/√3)e^(t/2) sin(√3 t/2).

Hence

y(t) = e^(t/2)[cos(√3 t/2) + (1/√3) sin(√3 t/2)] = (2/√3)e^(t/2) sin(√3 t/2 + π/3).

(c) Write α = (t, t+1, 2t+1), β = (2t, 3t-1, t), γ = (1, 3t, 1). The vector area of the parallelogram with edges α and γ is A = α × γ:

A = [ (t+1)(1) - (2t+1)(3t) ]î - [ t(1) - (2t+1)(1) ]ĵ + [ t(3t) - (t+1)(1) ]k̂

= (-6t² - 2t + 1)î + (t+1)ĵ + (3t² - t - 1)k̂.

Therefore

dA/dt = (-12t - 2)î + ĵ + (6t - 1)k̂.

At t = 1,

dA/dt = -14î + ĵ + 5k̂, and |dA/dt| = √222 units²/s.

For volume V = α · (β × γ). First,

β × γ = (-3t² + 3t - 1)î - tĵ + (6t² - 3t + 1)k̂.

Thus

V = t(-3t² + 3t - 1) + (t+1)(-t) + (2t+1)(6t² - 3t + 1)

= 9t³ + 2t² - 3t + 1.

Hence

dV/dt = 27t² + 4t - 3.

At t = 1,

dV/dt = 27 + 4 - 3 = 28 units³/s.

(d) Let R be the radius and a = 3R/8 the distance of the centre of mass of the solid hemisphere from its base centre. Assume rolling without slipping. Stability is decided by the potential energy V = Mg z_G.

(i) Curved surface rests on the sphere. At equilibrium the line joining the centre O of the fixed sphere to the centre C of curvature of the hemisphere is vertical, with OC = 2R. For a small turn φ of OC, rolling without slipping between two equal spherical surfaces makes the hemisphere rotate through 2φ. The height of C is 2R cos φ, and the centre of mass G is at distance a below C along the symmetry axis. Thus

z_G = 2R cos φ - a cos 2φ.

So

V(φ) - V(0) = Mg[2R cos φ - a cos 2φ - (2R - a)].

Using a = 3R/8, expansion gives

V(φ) - V(0) ≈ Mg(-R/4)φ².

This is negative for small φ, so V is maximum at equilibrium.

Hence the equilibrium is unstable.

(ii) Flat surface rests on the sphere. Let e be the unit normal from O to the contact point, making a small angle φ with the vertical. The flat face is tangent to the sphere. Let C be the centre of the flat face. Rolling without slipping gives the tangential shift s = -Rφ. Hence

z_C = R cos φ + Rφ sin φ.

Since the centre of mass is at distance a above C along e,

z_G = (R + a)cos φ + Rφ sin φ.

Therefore

V(φ) - V(0) = Mg[(R + a)cos φ + Rφ sin φ - (R + a)]

≈ Mg[(R - a)/2]φ².

With a = 3R/8,

V(φ) - V(0) ≈ Mg(5R/16)φ² > 0.

Thus V is minimum at equilibrium.

Hence the equilibrium is stable.

(e)(i) Let r(t) = f(t)î + g(t)ĵ. Then

r' = f'î + g'ĵ, r'' = f''î + g''ĵ.

The speed is

|r'| = ([f']² + [g']²)^(1/2).

The curvature is given by

κ = |r' × r''| / |r'|³.

In the plane,

r' × r'' = (f'g'' - g'f'')k̂,

so |r' × r''| = |f'g'' - g'f''|. Hence

κ = |f'(t)g''(t) - g'(t)f''(t)| / ([f'(t)]² + [g'(t)]²)^(3/2).

For a plane curve the binormal is constant, so dB/ds = 0. By the Frenet formula dB/ds = -τN, hence

τ = 0 at every point of the plane curve.

(e)(ii) Let s be arc length, and let P(s) be a point on the curve. The principal normal at P(s) is the line P(s) + λN(s). At the consecutive point s + ds, the principal normal is

P(s) + ds T + μ[N(s) + N'(s)ds].

If these two lines intersect, then to first order

λN = ds T + μN + μN' ds.

Using the Frenet formula N' = -κT + τB, this becomes

λN = ds T + μN + μ(-κT + τB)ds.

Comparing coefficients of T, N, B:

• From T: 0 = ds - μκ ds, so μ = 1/κ.
• From N: λ = μ.
• From B: 0 = μτ ds, so τ = 0.

Thus if τ ≠ 0, the consecutive principal normals are skew and do not intersect. They can intersect only when

τ = 0.