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Differential equation problem

  1. Sep 11, 2008 #1
    1. The problem statement, all variables and given/known data
    A snow blower clears 300 cubic ft of snow per hour regardless of depth of snow. At some point before noon, it starts snowing and snows steadily all day. A man starts clearing a sidewalk with the snow blower at noon. He clears 200 feet in the first hour, and 100 feet in the second hour. When did it start snowing?


    3. The attempt at a solution

    not too sure about this. the best i could come up with was dx/dt = x - H where x is the amount of snow in cubic feet and H is the amount of snow the blower clears. then i got stuck trying to separate variables and integrate. and i don't have an initial condition to get rid of the constant from integration, only two other conditions. how do i use these?

    any help is appreciated. thank you.
     
  2. jcsd
  3. Sep 11, 2008 #2

    Defennder

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    Don't see where you got x in your DE from. I'll use dS/dt instead for clarity. So we know that the rate of increase of snow as well as the rate of removal of snow are constants, so that means the DE shouldn't contain any functions of S. It's pretty much a simple integral.
     
  4. Sep 11, 2008 #3
    i don't follow. what i'm having trouble with is formulating the actual equation. would it be dS/dt = S - 300? if that's not correct what would it be? and if it is correct then how would i go about separating it? i can't figure out how to get the dS and S on one side and the dt on the other.

    thanks.
     
  5. Sep 11, 2008 #4

    Defennder

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    It says in the question that the blower clears 300ft of snow regardless of depth and it snows steadily, so that means the dS/dt should not be a function of S, existing snow.
     
  6. Sep 12, 2008 #5
    ok so the equation would be dS/dt=x-300 where x is the amount of snow falling? then integrating would yield S=t(x-300). but i'm only given information about snow removed, not snow added (x) so i'm still confused.
     
  7. Sep 12, 2008 #6

    Defennder

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    No, you don't get (x-300)t. There is an arbitrary constant of integration. S(0) tells you what that arbitrary constant should be. Let's assume it started snowing 'h' hours before noon. That means the amount of snow at noon is 'hx'. So S(0) is hx. You should be able to contruct an equation from the information given as to how much net snow is removed after the first hour with the above. Let this equal to 100. Tweak this approach a little for the second hour. Then you can find the unknowns and solve the problem.
     
  8. Sep 12, 2008 #7
    thanks for the reply. here's what i got:

    (S=integral)

    s'=x-300
    Sds=Sx-300dt
    (1) s=t(x-300)+c

    200=1(x-300)+c
    500=x+c
    (2) c=500-x

    100=2(x-300)+c
    (3) 700=2x+c

    (2) into (3)
    700=2x+(500-x)
    x=200

    700=2(200)+c
    c=300

    (1) with constants added
    s=t(200-300)+300
    s=100t+300

    when snow started
    0=100t+300
    t=-3

    so 3 hours before noon = 9am

    is this correct?
     
  9. Sep 12, 2008 #8

    Defennder

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    I haven't worked out the answer yet, but it seems you used the information given wrongly. 200 and 100 refers to the amount of snow cleared, not the amount of snow left over after he cleared some of it.
     
  10. Sep 12, 2008 #9
    ok, i just replaced 200 with x-200 and 100 with x-100 in those equations then resolved. answer came out to 11am. i guess that makes sense because it's a linear equation so snow started at 11am, by 12pm 300ft were cleared, by 1pm 200ft were cleared, by 2pm 100 ft were cleared.
     
  11. Sep 12, 2008 #10

    Defennder

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    Why 11am? I didn't get a whole number.
     
  12. Sep 14, 2008 #11

    Integral

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    Here is my approach.

    The snow is falling at a constant rate K. x is the depth of snow at time t. So:

    [tex] \frac {dx} {dt} = K [/tex]

    This is a simple separable equation yielding:

    x = Kt + c

    We need to determine the 2 constants, but we are given 2 conditions to satisfy. We know that at t=12 we have:

    x * D = 300 (D is the distance * width cleared in an hour)

    So we have (12K + c) * 200 = 300 (Volume of snow removed at 12)
    and:
    (13K +c) * 100 = 300 (Volume of snow remove at 1 (13 on the 24hr clock)

    Solving these 2 equations yields:

    K = 3/2 and c = -33/2

    So we have x = 3/2 t - 33/2

    x = 0 at t =11
     
  13. Sep 14, 2008 #12

    Dick

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    That's only correct if you assume the snow blower operates intermittantly, once at exactly 12 and again at exactly 1. I don't read the problem that way. I think you are supposed to think of it as running continuously.
     
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