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\centerline{\huge\sf 'Sno Problem}
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In order to find out what time it began snowing, you must engage in a bit of calculus.
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Let $h$ denote the number of hours that it had been snowing before noon,
so that the snow started at hour $12-h$.
Let $s$ be the constant rate of snowfall (in inches per hour), and let
$r$ the constant rate of removal (in cubic feet per hour).
At time $12+t$, that is, $t$ hours after noon, the depth of the
snow is $s \times (t+h)$.
The speed of the snowplow (in miles per hour) is inversely proportional
to the depth of the snow, in order to maintain a constant rate of
removal.
So the snowplow is travelling at speed
$\frac{c r}{s \times (t+h)}$,
where the constant $c$ converts
(inches times miles times width of the snowplow blade) to (cubic feet).
Fortunately we don't need to know any of the constants $c$, $r$ or $s$.
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Between 12:00 and 1:00 the snowplow travels two miles, so that
$$ \int_0^1 \frac{c r}{s (t+h)} dt = 2 .$$
Between 1:00 and 2:00, with deeper snow, it travels only one mile:
$$ \int_1^2 \frac{c r}{s (t+h)} dt = 1 .$$
Integrating $1/(t+h)$ gives a logarithm, so that we find
$$\frac{c r}{s} \left( \log (1+h) - \log h \right) =
2 \frac{c r}{s} \left( \log (2+h) - \log (1+h) \right) $$
which yields
$$\frac{1+h}{h} = \left( \frac{2+h}{1+h} \right) ^2 $$
$$(1+h)^3 = (2+h)^2 h $$
$$h^2 + h - 1 = 0$$
$$h \in \left\{ \frac{\sqrt{5} -1}{2} , \frac{-\sqrt{5}-1}{2} \right\}
= \{ 0.618 , -1.618 \}$$
Because it started snowing before noon, $h$ is positive.
$h=0.618$ hours translates to 37 minutes, so it started snowing
at 11:23am.
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