Question

A point moves at a constant velocity v = 0, 5 m / s on a sphere along a meridian from the north pole to the south pole. The sphere radius grows according to


`r(t) = ut + r_(0)`
(u = 2 m / s and r0=2 m at t1 = 0 s)

a) After which time t2 does the point reach the South Pole? (Hint: ∫a／(ax+b) dx = ln(ax+b)+c )

b) At the time t1 = 0, a particle starts at the same speed from the south pole to the north pole with the same velocity v on the same meridian. When are they going to meet ?​

Answer 1

Step-by-step explanation:

a)

After a short time dt, the point moves a short distance ds.

ds = v dt

The distance is equal to the arc length:

r dθ = v dt

Radius is a function of time:

(ut + r₀) dθ = v dt

Separate the variables:

1/v dθ = 1 / (ut + r₀) dt

Multiply both sides by u:

u/v dθ = u / (ut + r₀) dt

Integrate both sides:

u/v θ [θ₁ to θ₂] = ln(ut + r₀) [t₁ to t₂]

(θ₂ − θ₁) u/v = ln(ut₂ + r₀) − ln(ut₁ + r₀)

(θ₂ − θ₁) u/v = ln[(ut₂ + r₀) / (ut₁ + r₀)]

e^((θ₂ − θ₁) u/v) = (ut₂ + r₀) / (ut₁ + r₀)

ut₂ + r₀ = (ut₁ + r₀) e^((θ₂ − θ₁) u/v)

Plug in values:

2t₂ + 2 = (0 + 2) e^((π − 0) 2/0.5)

2t₂ + 2 = 2 e^(4π)

t₂ + 1 = e^(4π)

t₂ = -1 + e^(4π)

t₂ ≈ 286750

b)

This time, a second point from the south pole moves up at the same speed and meets the first point. Logically, they meet in the middle, when θ₂ = π/2.

ut₂ + r₀ = (ut₁ + r₀) e^((θ₂ − θ₁) u/v)

2t₂ + 2 = (0 + 2) e^((π/2 − 0) 2/0.5)

2t₂ + 2 = 2 e^(2π)

t₂ + 1 = e^(2π)

t₂ = -1 + e^(2π)

t₂ ≈ 534.5