==> analysis/snow.s <==
11:22:55.077 am.
Method:
Let b = the depth of the snow at noon, a = the rate of increase in the
depth. Then the depth at time t (where noon is t=0) is at+b, the
snowfall started at t_0=-b/a, and the snowplow's rate of progress is
ds/dt = k/(at+b).
If the snowplow starts at s=0 then s(t) = (k/a) log(1+at/b). Note that
s(2 hours) = 1.5 s(1 hour), or log(1+2A/b) = 1.5 log(1+A/b), where
A = (1 hour)*a. Letting x = A/b we have (1+2x)^2 = (1+x)^3. Solve for
x and t_0 = -(1 hour)/x.
The exact answer is 11:(90-30 Sqrt[5]).
_American Mathematics Monthly_, April 1937, page 245
E 275. Proposed by J. A. Benner, Lafayette College, Easton. Pa.
The solution appears, appropriately, in the December 1937 issue,
pp. 666-667. Also solved by William Douglas, C. E. Springer,
E. P. Starke, W. J. Taylor, and the proposer.
See R.P. Agnew, "Differential Equations," 2nd edition, p. 39 ff.