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Trouble with The Snowplow Problem 
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#1
Jan2911, 02:49 PM

P: 441

"One morning it began to snow very hard and continued snowing stadily throughout the day. A snowplow set out at 8:00AM to clear a road, clearing 2 miles by 11:00AM and an additional mile by 1:00 PM. At what time did it start snowing?"
This my my approach. In the problem it states that we can assume the rate of snowfall is constant and that the speed at which the plow goes is inversely proportional to the height of the snow. Using this information, the differential equation I come up with is.. [tex]\frac{dx}{dt} = \frac{k}{S}[/tex] Where k = proportionality constant, and S = the depth of the snow. Since we can assume the rate of snowfall is constant, it yields another differential equation... [tex]\frac{dS}{dt} = C[/tex] Solving for S.. [tex]\frac{dS}{dt} = C \rightarrow dS = Cdt \rightarrow S = Ct+D [/tex] Substituting this into the first equation and solving.. [tex]\frac{dx}{dt} = \frac{k}{Ct+D} \rightarrow x = \frac{k}{C}lnCt+D+E[/tex] My next guess would use (t=8 , x=0) (t=11 , x=2) and (t=13 , x=3) as initial values to solve for C, D and E. I get VERY messy results however. With the help of my calculator I get an numerical answer for t to be equal to 1.7085 or 11:18PM of the previous day. This doesn't make sense considering it states it starts to snow in the MORNING. if the time were positive it would translate to 1:42AM. Any help would be appreciated. ~Matt EDIT: The chapter this is in is about solving first order differential equations. 


#2
Jan2911, 03:04 PM

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Your equation has four parameters, k, C, D, and E, but you have only three conditions. How did you get specific values for all four?



#3
Jan2911, 03:12 PM

P: 441

Yeah thats a good point, I basically know that my answer is wrong. Haha.
I am looking for some extra insight as to what I should try next. Would this be the correct way to go about this problem? 


#4
Jan2911, 04:02 PM

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Trouble with The Snowplow Problem
You don't really have 4 indepedent parameters.
Take your expression (k/C) ln (Ct +D) + E =(k/C) ln (C (t + D/C)) + E = (k ln C/ C) ln (t + D/C) + E which is the same as = A ln(t + B) + E Note, you want to find when Ct + D = 0, so you only need the value of D/C This makes sense in real life. If it had snowed twice as fast, and the snowplow was twice as powerful, then the distance cleared in a given time would have been the same. Now you have 3 equations in 3 variables 0 = A ln (8+B) + E 2 = A ln (11+B) + E 3 = A ln(12+B) + E Subtract pairs of equations to elminate E Then divide the two equations you get to eliminate A You will get an equation connecting ln (12+B), ln (11+B), and ln(8+B). Using the properties of logs, you can turn that equation into ln[something] = 0 or [something] = 1 And you will then have a equation you can solve for B (without using a calculator!) 


#5
Jan2911, 06:10 PM

P: 441

Thanks for the help! This is what I got, tell me if it looks right.
(I did it in MathType, it would have taken WAY too long using the Latex code. http://img202.imageshack.us/img202/6...lowproblem.gif 


#6
Jan2911, 06:58 PM

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That's the right idea, except you made one mistake.
"Pascal's triangle" was a good thing to say, but (11+B)^3 is not 11^3 + 11^2 B + 11B^2 + B^3 Hint: (11+B)^2 is not 11^2 + 11B + B^2. You expanded (13+B)^2 correctly, so you should be able to see what is wrong. If you can't see it, just multiply out (11+B)^3 the long way. 


#7
Jan2911, 07:06 PM

P: 441

Oh! I forgot to multiply the two middle terms by 3! I redid the calculations and got 12:15AM as the result.
Thanks a lot :D 


#8
Jan3011, 10:31 AM

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A neater way to sort out the constants is like this: You have the differential equation dx/dt = k/S You know the rate of snowfall is constant starting at time T, and you want to find T, so make T one of your parameters right from the beginning. The depth of snow S is a straight line graph through the point (T,0), which is S = c(t  T) for some value of c Now you have dx/dt = (k/c) / (tT) and you can see that k and c are not really independent parameters. So let k/c = a and the DE becomes dx/dt = a/(tT) for some value of a. From then on, everything is same as your solution, but you got rid of the unwanted parameters right from the start. 


#9
Jan3011, 10:53 AM

P: 441

Makes sense! Thanks :)



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