One morning it starts to snow at a constant rate. Later, at 6:00am, a snow plow sets out to clear a straight street. The plow can remove a fixed volume of snow per unit time.

If the plow covered twice as much distance in the first hour as the second hour, what time did it start snowing?

We can assume that the snow piles up at 1 unit per hour, as it's arbitrary what units are being used. We need to find a c (number of hours before the beginning of plowing) such that

∫{0 to 1}(a/(x+c))dx = 2∫{1 to 2}(a/(x+c))dx

a/(x+c) = a(x+c)^-1, so the integral involves the natural log function, and we get

a [ln(x+c)]{0 to 1} = 2a [ln(x+c)]{1 to 2}

ln(1+c) - ln(c) = 2 ln(2+c) + ln(c)

Taking antilog:

(1+c)³ = c (2+c)²

c³ + 3c² + 3c + 1 = c³ + 4c² + 4c

0 = c² + c - 1

c = (-1 ±√(1+4))/2 = -1/2 ±√(5/4)

The lower solution is negative and thus spurious. The upper solutin is .6180339887498949, which is the reciprocal of the golden ratio, aka phi or tau depending on the text.

Subtracting from 6 and converting to h:m:s gives us 5:22:55.0776... but that's already too accurate for the start of a snowfall.

Posted by Charlie on 2003-06-05 10:33:03