snowplow problem

[Albert Chan]

Let snow started at time 0, time x = noon-time, when snowplow was started.
Assumed speed of snowplow inverse-proportional to snow thickness.

First hour snowplow distance traveled = doubled of 2nd hour (due to more snow)

∫(1/t, t=x .. x+1) = 2 * ∫(1/t, t=x+1 .. x+2)

ln(x+1) - ln(x) = 2 * (ln(x+2) - ln(x+1))
3*ln(x+1) = 2*ln(x+2) + ln(x)

(x+1)³ = (x+2)² * x
x² + x - 1 = 0
(x + φ) * (x - (φ-1)) = 0

Since x > 0, x = φ-1 ≈ 0.6180 hour (37 minutes)

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We can estimate integrals with mid-point rules.


∫(1/t, t=x .. x+1) ≈ (1/(x+1/4) + 1/(x+3/4)) / 2   // estimate with 2 rectangles
∫(1/t, t=x+1 .. x+2) ≈ 1/(x+3/2)                         // 1 rectangle OK, since the left is flatter

(x+1/2) / ((x+1/4) * (x+3/4)) = 2 * 1/(x+3/2)
(x+1/2) * (x+3/2) = 2 * (x+1/4)*(x+3/4)            // scale away denominator
x² + 2x + 3/4 = 2x² + 2x + 3/8
x² = 3/8
x = √(3/8) ≈ 0.6124 hours (37 minutes)