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Need help with a diff. eq?

  1. Feb 18, 2008 #1
    the problem is this:

    Early one morning it began to snow at a constant rate. at 7 am a snowplow set off to clear a road. By 8 am it had traveled 2 miles, but it took two more hours(untill 10 am) for the snowplow to go an additional 2 miles.
    a. let t=0 when it began to snow, and let x denote the distance traveled by the snowplow at time t. assuming that the snowplow clears snow from the road at a constant rate (in cubic feet per hour, say) show that

    k* dx/dt=1/t ------------(1)

    where k is a constant

    b. what time did it start snowing?

    well i do not actually understand what the part a. is asking. Is it just asking us to solve the diff eq (1) and see whether the solution function satisfies the conditions
    x(1)=2, and x(3)=4? or??? If this is the case, then i am able to solve eq. (1), i actually did, and it satisfies the requirements mentioned.
    As for the part b. i have no idea how to determine when it started to snow?

    Any hints would be appreciated??
  2. jcsd
  3. Feb 19, 2008 #2
    OK, the very nice way to deal with these, is the convolution but... Since texing the formulas are hard and this guy can definitely way better, please watch the video lecture 21 on this page.


    in the end you will be surprised about the problem he solves. I know it is not the nicest way to point fingers but I hope you won't regret this. Then you can take the functions as constant and proceed with ease and joy.
  4. Feb 19, 2008 #3
    Well thankyou for the link, but it cannot be of any help to me for the moment! I cannot actually follow what that guy is saying. But i do not think one would need such mathematical tools to do the particular problem i posted,it should be done with much more simpler diff. eq tools.
    I am only a freshman, but i am just trying to make a start in diff. eq. so when the time comes to take the course i might already know some things.

    Is there any simpler method of doing it?
  5. Feb 19, 2008 #4
    I thought id have a go at it as im doing an introduction to applied maths this semester.

    I think its asking you to make a basic model of the situation.

    If you say the speed of the plow dx/dt is inversely related to the depth of the snow D such that.

    eq1. dx/dt = u/D

    where u is a constant
    then make an expression for D say (d(0) + kt - z) where d(0) is the depth of snow at time 0 k is a constant for snow fall rate t = time and z is the plows constant ability to shovle the stuff out the way.

    Then we sub this in to eq1 and rearrange to get

    (k/(u+z-d(0)))x(dx/dt) = 1/t

    And bobs your uncle theres the answer.

    For the second part you can just use the information you are given to find the constant and then work from there.

    Hope this is right ;)

  6. Feb 20, 2008 #5
    Well how did you get from this part:dx/dt = u/D, where you let D=d(0)+kt-z, to

    (k/(u+z-d(0)))x(dx/dt) = 1/t
  7. Feb 20, 2008 #6
    OR, wouldn't it be better if we modified a little bit what you did, like this:
    The depth of the snow will be D=kt, where k is the constant at which it snows, since, the problem says that t=0 is the time when it starts to snow, then it means that at t=0 there is no depth of snow at all, so d(0)=0,(in your equation). ALso i do not see how the snowplow's ability to shovel the snow out of the road affects the depth of the snow, as far as i am concerned it only affects the amount of the snow that is on the road. SO, is it safe not to consider the z constant-the ability of the snoplow to shovel the snow out of the road at all, when describing the depth of the snow at any time?? I think this way we would come to the desired result, since now we would have:

    dx/dt=u/kt so from here
    (k/u)dx/dt=1/t, now let b=k/u, so finally we have


    Am i at least close to understanding this problem or just too stupid???
  8. Feb 21, 2008 #7
    yes for this case we do not need to consider the z constant as it can be merged with the other constant u?

    For the t=0 condition in your case you take t=0 when it starts to snow wheras I take it to be when the plow starts moving.This is why I added the factor d(0) it just depends on your preference if you worked out d(0) you could work out how many hours the snow had been falling previous to setting off by saying d(0)=kt

  9. Feb 21, 2008 #8
    Yeah i guess i figured out the last part, the part under b. Here is a very short outline of how i performed it
    i solved for the diff eq k*dx/dt=1/t, first
    after that i let t_1 denote the time when the snowplow set off to clean the roads, so t_1 corresponds to 7 a.m after that we have x(t_1)=0, x(t_1 + 1)=2, and x(t_1 + 3)=4, so i plugged all these values in the solution of the diff eq, wich as i remember was

    x(t)=ln(t)/k + M so i got three eq. with three unknowns, t_1, M, k and from here i found t_1 to be 1, which corresponded to 7 am, so the from here i concluded that it started to snow at 6 am.
  10. Feb 21, 2008 #9
    Congratulations ;)

    Im thinking youre probably at the same level of maths as me when it comes to this kind of thing. Im also in my first year doin an introduction to pure and applied maths.
    Anyway if you have any more problems post em up then I can have a crack at em as I need the practice.

  11. Feb 21, 2008 #10
    Well i do not actually know. I am just a freshman and this is my first semester ever, since i started on the spring semester. I am in general math though. I am not actually officialy taking diff eq. but i am just trying to make a start on it so when it comes the time to take it officialy i will have a feeling to what diff equation are about!
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