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# Differential equation watch

1. Wonder how to solve this problem...

Snow is falling at a constant rate and the snowplow's speed was inversely proportional to the depth of snow. So although it was able to clear 2 km of road in 1st hour, it was only able to clear 1 km in the second hour.
(i) If it was snowing for T hours before the snowplow started, set up a differential equation for dx/dt, where x km is the total distance that the snow plogu has traveled after t hours.
(ii) By solving this differential equation and elininating the constants, show that (T+1/T)cube = (T+2/T) squared.
(iii) To the nearest min, find the time that it started snowing.

Thanks!
2. (i)
Depth of snow = A(T + t)

dx/dt
= B/(Depth of snow)
= C/(T + t)

(ii)
x = C ln(T + t) + D

When t = 0, x = 0. So C ln(T) + D = 0.
When t = 1, x = 2. So C ln(T + 1) + D = 2.
When t = 2, x = 3. So C ln(T + 2) + D = 3.

So we have three equations relating the three unknowns C, D and T.

C(ln(T + 1) - ln(T)) = 2
C(ln(T + 2) - ln(T)) = 3

3(ln(T + 1) - ln(T)) = 2(ln(T + 2) - ln(T))
3ln[(T + 1)/T] = 2ln[(T + 2)/T]
ln{[(T + 1)/T]^3} = ln{[(T + 2)/T]^2}
[(T + 1)/T]^3 = [(T + 2)/T]^2

(iii)
Multiplying both sides of [(T + 1)/T]^3 = [(T + 2)/T]^2 by T^3 gives

(T + 1)^3 = T(T + 2)^2
T^3 + 3T^2 + 3T + 1 = T^3 + 4T^2 + 4T
T^2 + T - 1 = 0

T = (1/2)(-1 sqrt(5))

Since T > 0, we take the larger root: T = (1/2)(-1 + sqrt(5)) = 0.618 hours = 37 minutes.

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