Differentiation

[kedaman]

One day it started to snow and it snowed at a constant speed. at 12:00 a snowplough started, it's speed reverse proportional to the thickness of the snow. The first hour it went twice longer than the second hour. When did it start to snow?

[Juriaan Borst]

quick estimation with a spreadsheet: 9:30

[[Digital-X-Treme]]

kedaman, is there an actual solution you have worked out, as it seems once again, (as with the previous differentiation question about a cone inside a sphere), that there isn't enough information to even form a decent solution...?

Laterz

[kedaman]

As i remember it, i found the solution to the sphere problem, but i did use some radical methods with my Ti83 Equation solver. Although i now know i could solve it another way.

Yes i've calculated this one (without any radical methods) and the answer is 15:18 (i could be wrong though)

[HarryW]

Well I just tried to work it out with algebra, and I got:

t = c/k - 2

where:

t is the number of hours it was snowing for when the snowplough started;
c is the initial speed of the snowplough in unit distance per hour
k is the rate at which the snow rises, in unit height per hour

Of course, I was assuming that by 'reverse proportional' you meant that v = -kt + c

I suppose you may have meant 'inversely proportional', ie v = c/kt

[kedaman]

yeah sorry that was what i meant, inversery proportional

[kedaman]

and there's an error, it's 8:42 not 15:18, i calculated the time forward intead of backward from 12:00. What counts is that it is 3.30 hours.

[Guv]

At first glance, it seems intuitively obvious that there is not enough information to solve this problem. However, my intuition has been wrong before, so a few more glances seemed worthwhile. The following looks like a good start.

Velocity = 1 / SnowRate * Time

Distance is the integral of velocity with respect to time, resulting in

• Distance = Integral( 1 / SnowRate * Time )
  
• Distance = IntegrationConstant + Log(Time) / SnowRate

At this point, it is not obvious how to precede. It looks like one of the above two should be used with the relation Distance1 = 2 * Distance2, where Distance1 is the plow's Noon to 1PM progress, and Distance2 is the 1PM to 2PM progress. Now express the two distances as definite integrals. An alternative is to forget the definite integral and work with the ordinary distance equation containing the constant of integration.

Perhaps we have enough information to solve this problem after all. Both approaches are likely to work, since they are fundamentally equivalent to each other.

There is a bit of a problem because it started snowing prior to Noon, but the above equations are applicable to the plow starting at Noon, and are meaningless prior to Noon.

The key seems to be deciding when time starts, and what units to use.

• Hours look reasonable for time. If minutes, we must deal with Log(60 or more). Perhaps the choice of units merely makes the SnowRate bigger or smaller, and this decision is not important. Still, it seems strange to not really know what distance and time units are suitable. Assume hours for time (this cannot really be wrong), and hope that distance units need not be assigned.
  
• Time = zero at noon? This will not work because Log(0) is not well defined at zero. I hesitate to say it is infinite, because that might result in more posts than I would like to see here. Let us say that Log(X) gets bigger as X approaches zero, and perhaps we should avoid this value.
  
• Time = zero when it started to snow? We do not have to worry about Log(0) because the plow started after it began to snow. Then assign Time = X to Noon. If we can solve for X, it started to snow at (Noon - X). There is at least one subtle assumption here. It is not obvious to me that this approach is valid, but it seems plausible.
  
• Perhaps work with distance = zero at Noon and investigate the implications.

At this point, higher priority issues turned up, and I quit work on the problem. Just decided to formalize these ideas and post. Maybe somebody else will know what to do next. If not, I will try to take some more glances.