snowplow problem

[Albert Chan]

(03-18-2021 08:38 PM)Albert Chan Wrote:  First hour snowplow distance traveled = doubled of 2nd hour (due to more snow)

Extending the problem to distance traveled ratio = k, k > 1

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

ln((x+1)/x) = k * ln((x+2)/(x+1))
ln(x/(x+1)) = -k * ln(2 - x/(x+1))

Let z = x/(x+1), 0 < z < 1

z = (2-z)^(-k)

f(z) = z - (2-z)^(-k)
f'(z) = 1 - k*(2-z)^(-k-1)

Solve by Newton's method, and we need a rough guess.
(estimate both integrals with single rectangle)

1/(x+1/2) = k /(x+3/2)
x = (3-k)/(2k-2)             -- rough guess for x
z = x/(x+1) = (3-k) / ((3-k)+(2k-2)) = (3-k)/(k+1)

Code:

function snowplow(k, verbal)
    local z = max(0,(3-k)/(k+1))
    repeat
        if verbal then print(z/(1-z)) end
        local y = (2-z)^-k
        local eps = (y-z) / (1 - k*y/(2-z))
        z = z + eps         -- newton correction
    until z == z + eps*eps
    return z/(1-z)          -- time when started snowplow
end

lua> snowplow(1.5, true)
1.4999999999999998
1.566890844473769
1.568114464087791
1.5681148594401135
1.5681148594401568       --> 94 minutes

lua> snowplow(2, true)
0.49999999999999994
0.6136363636363635
0.6180278644242209
0.6180339887380153
0.618033988749895        --> 37 minutes