Snowplow

Reuel · 2011-02-05 02:36:52

"One morning it began to snow very hard and continued constant throughout the day. At 8:00 A.M. a plow set out to clear a road, clearing 2 miles by 11:00 A.M. and an additional mile by 1:00 P.M. Assume that the rate at which the road may be cleared is inversely proportional to the depth of the snow. What time did it begin to snow?"

So we know that it took 3 hours to clear 2 miles and then, due to the continued constant snow fall, it took another 2 hours to clear 1 mile more. We know that the rate the plow can do its thing is inversely proportional to however much snow there is and we know that that rate was "2 miles every three hours" at 10:00 A.M. (the "average" of the three hours) and "1 mile every two hours" at noon.

Because the snowfall is constant but always changing it's sort of like one of those work problems where a bucket is being pulled up with water pouring out of it, right? Though I do not know if that is the way to go.

Have I missed any information?

Last edited by Reuel (2011-02-05 02:37:36)

bobbym · 2011-02-05 02:42:53

Hi;

We will know when we see the DE. What is the DE?

Bob · 2011-02-06 01:56:22

hi Reuel and bobbym

I am not confident about this answer but how about:

Let H be the depth of snow.

H is growing at a steady rate (S) due to snowfall, but diminishing, due to plowing, at a rate that is proportional to 1/H

=>

k is the constant of proportionality.

Let me know what you think.

Bob

ps. As regards http://www.mathisfunforum.com/viewtopic.php?id=15017 post #12, the guys are looking puzzled at light bulbs that are not allowed. Perhaps it should have had the caption Great Minds ....Question mark.

bobbym · 2011-02-06 02:19:30

Hi Bob;

That is what I got. But it just does not go with the answer.

Reuel · 2011-02-06 04:28:09

Hi... sorry, distracted with other matters. Let me re-read the problem.

I would start by forming two differential equations. One to describe the rate the snow is accumulating, such as

where the rate of the snowfall is constant, and then another equation to describe the velocity of the plow:

where the rate of the plow is inversely proportional to the amount of snow on the ground.

Solving the first differential equation we get

where D is a new constant. Substituting this into the second differential equation for S we then have

And solving this differential equation yields:

But now we have four parameters - k, C, D, and E - along with our variables t and P and it needs to be simplified in some manner so as to lessen the parameters by at least 1 because we have two values for time and progress and our third equation in such a system would want to solve for t when the snow fall is 0. Or something like that.

How about

I think this could be used to create a system of equations using time t and p(t) for different values to solve for A, B, and E. Am I going in the right direction, do you think?

bobbym · 2011-02-06 04:46:29

Hi;

I am not sure about the 2 simultaneous DE's. But I would have reduced,

to

But that looks incorrect too.

Bob · 2011-02-06 04:54:27

Still thinking ............................

Back when I've got something

Bob

Reuel · 2011-02-06 04:55:10

I am working on it too. I'll be back later.

Reuel · 2011-02-06 05:38:41

Finished. But I do not know if I am correct. I have 1:42 AM.

What are you guys coming up with?

I'll post my work so you can follow how I came to this solution.

Reuel · 2011-02-06 06:41:29

Let the depth of the snow at time t be represented by D(t) or

such that

Let n represent the amount of time before the snow plow got to work so that

And the function for the depth of snow at time t, substituting, is

Now, the rate of the plow's work will be inversely proportional to the depth of the snow:

Solving for P(0) = 0...

Letting P(t) be 2 when t is 3 produces

and P(t) be 3 when t = 5...

we can set them equal...

Now solve it for n. (Many basic steps are here skipped over.)

n comes out to be about 6.2915 hours before the plow began at 8. And so that is approximately 1:42AM.

Have I made any mistakes?

bobbym · 2011-02-06 07:27:16

Hi;

Just in writing it down.

Should not that be a capital P?

That should be P(5).

By the way this problem is in Stewart's Calculus.

Bob · 2011-02-06 07:45:07

hi Reuel and bobbym,

I like the approach to the problem. Better than my DE at post #3. At least you have got a function which can have 'miles' in it!

Now to the algebra.

In

k is the constant of proportionality for the snowfall.

Then in

it is the constant for the plowing.

Should these 'k's be the same?

Thereafter I'm following ok.

Answer seems reasonable.

Bob

Reuel · 2011-02-06 07:53:42

bobbym - Yes, I meant those things. I am more interested in whether or not I am correct. And I do not have Stewart's calculus book. Do they get the same answer?

bob bundy - I do not know for sure. But yes, the answer seems reasonable, especially for how slowly the roads are getting plowed. It must have been snowing for quite some time before they got started at 8am.

bobbym · 2011-02-06 08:02:08

Hi Reuel;

Their problem is slightly different. But I think yours is okay. If you want to see it check this out:

http://docs.google.com/viewer?a=v&q=cac … tjWNtzX4Jg

Reuel · 2011-02-06 08:07:50

Yeah. That doesn't say that the rate the plow can plow is inversely proportional to the depth of the snow.

Similar though.

bobbym · 2011-02-06 08:11:36

Yes, I know but looking at some other PDF's, this is quite a popular oldtime problem, I think the inversely proportional is implied since their DE is exactly the same as the one we are looking at.

Bob · 2011-02-06 09:52:59

Thanks guys. I think I've finally understood this question.

Bob

Reuel · 2011-02-06 10:09:12

I assume my solution is confirmed correct then.

Thanks for your help.

Bob · 2011-02-06 10:28:12

Oh sorry I thought that was implied. And I think you helped me. So thanks all round.

Bob

bobbym · 2011-02-06 10:31:38

I sure learned a lot from it thanks.

Math Is Fun Forum

#1 2011-02-05 02:36:52

Snowplow

#2 2011-02-05 02:42:53

Re: Snowplow

#3 2011-02-06 01:56:22

Re: Snowplow

#4 2011-02-06 02:19:30

Re: Snowplow

#5 2011-02-06 04:28:09

Re: Snowplow

#6 2011-02-06 04:46:29

Re: Snowplow

#7 2011-02-06 04:54:27

Re: Snowplow

#8 2011-02-06 04:55:10

Re: Snowplow

#9 2011-02-06 05:38:41

Re: Snowplow

#10 2011-02-06 06:41:29

Re: Snowplow

#11 2011-02-06 07:27:16

Re: Snowplow

#12 2011-02-06 07:45:07

Re: Snowplow

#13 2011-02-06 07:53:42

Re: Snowplow

#14 2011-02-06 08:02:08

Re: Snowplow

#15 2011-02-06 08:07:50

Re: Snowplow

#16 2011-02-06 08:11:36

Re: Snowplow

#17 2011-02-06 09:52:59

Re: Snowplow

#18 2011-02-06 10:09:12

Re: Snowplow

#19 2011-02-06 10:28:12

Re: Snowplow

#20 2011-02-06 10:31:38

Re: Snowplow

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