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Homework Help: Homework help, differential eq

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data

    Hey i've been studying over my extension math revision and i just cant get my head around these two questions. I'm sure i've been setting up the differential equations wrong but im not sure whats right. The test is tomorow and tuesday so im kind of freaking out. heres the two questions, thanks in advance.

    1. Oil is leaking out of a barrel loaded in a cargo ship. The rate at which the oil level is dropping seems proportional to the square root of the level of oil in the barrel at that time.
    Write a differential equation for y(t), the level of oil (in cms) and time (hours).
    Find a general solution for the equation.
    If the barrel was filled up to 144cm when it started leaking and the level dropped to 80cm in four hours, how long will it take for the barrel to empty?

    2. An investigation is under way in the department of forests, land and conservation to improve our understanding of the rate at which bush fires spread. A bush fire has broken out and a spotter plane has already taken some photographs. The first photograph shows a fire area of half a hectare. Comparison with a photograph taken six hours later shows that the fire seems to be spreading uniformly in all directions and has burned an additional 1700 square metres.
    Use differential equations and find two possible models for the spread of the fire.

    3. The attempt at a solution
    i have been for q1, dr/dt = y^(1/2). fliping to dt/dr then intergrating
    just some assitance with setting up the begining equation will be enough to get me going its just the start of these i get confused about
    thankyou =D
  2. jcsd
  3. Sep 13, 2008 #2


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    What's r? Isn't the first equation just dy/dt=k*y^(1/2)? Separate the variables and integrate both sides. I.e y^(-1/2)*dy=k*dt where k is the constant of proportionality. Is that enough to get you started? Stop freaking out.
  4. Sep 13, 2008 #3


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    Well, in words the rate at which the oil drops out, is a measure of how quickly the level of oil in the tank decreases...in math that would be Rate=dy/dt. Now this rate is proportional to the square root of the level of oil in the tank,

    i.e. [tex]\frac{dy}{dt} \propto y^{1/2}[/tex]

    or [tex]\frac{dy}{dt} = Cy^{1/2}[/tex] where [tex]C[/tex] is some unknown constant of proportionality.

    Now can you use a similar line of reasoning in the second question to set up a DE for the area burned, [tex]A(t)[/tex]?
  5. Sep 14, 2008 #4
    dy= ky^1/2 dt
    y^(-1/2) dy = kdt integrate
    y^1/2 / [1/2] =kt +c
    2 y^1/2 = kt +c

    when t=0 y=144 substitute
    2[sqrt[144]] =k0+c

    2y^1/2 = kt +24

    when t=4 y=80 substitute
    2 sqrt80 =k[4] +24
    2[4sqrt5]-24 =4k
    k= 2sqrt5 -6

    2y^1/2 =[2sqrt5 -6] t +24 answer
    y^1/2 =sqrt5 -3] t +12
    when y=0 what is t ?
    set y=0
    0=[sqrt5 -3]t +12
    t= 12 / [3-sqrt5]

    thats what i had before, but the original ky^1/2 should be -?
    and for the second question i did
    A=pi r^2
    where r is the radius of th burnt area

    the rate of expansion is dr/dt. Buy dr/dt is a constant k
    dA/dt = 2 pi r k
    dA = 2pi r k dt
    A=pi R^2 k t +c

    at t=0 A(0) = 1/2 hectare
    A(0)=pi r^2 k[0]+c

    A= pi r^2 k t +A(0)
    at t=6 A(6) = 1700 metres sq. + A(0) solve for r(6) and substitute

    A(6)= pi [r(6)^2] 6 k +A(0) solve for k

    k is rate of the fire is advancing

    are they right?
  6. Sep 14, 2008 #5


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    Your solution to question (1) is correct. There is no need to worry whether to use +k or -k in the ODE, because if you use +k you get a k that is always less than zero and so the negative sign is built in.

    Question 2 is more complicated. You don't know that dr/dt is a constant, it could be almost any function of t. The question asks you to find two different solutions, so the easiest ones are probably: (a) dr/dt=k and (b) dr/dt=Ct+D, but you could choose almost any function as long as you don't have to use too many constants (you only have two initial conditions. so you can only solve for two constants).

    Anyways, good luck on your test and remember to clearly define your variables! (unless they are already defined in the question)
  7. Sep 14, 2008 #6
    thanks for your reply
    but it says the fire is spreading uniformly
    wouldnt that make dr/dt a constant

    this is how i got one of the formulas
    im still puzzled on the second haha


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  8. Sep 14, 2008 #7


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    I take "spreading uniformly" to mean that the fire isn't growing faster to the left of the center of the burned area than it is to the right of center...that just means that dr/dt won't be some complicated function of r,it can still be a function of t though: it may have burned faster in the last 3 hours than it did in the first 3, but that doesn't mean it isn't spreading uniformly out from the center.
  9. Sep 14, 2008 #8
    ah damn
    i gotta redo it then
    my classmate got for one of the equations
    im not sure how he did it though
  10. Sep 14, 2008 #9


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    he probably set dr/dt =Ae^(Bt) and then solved for A and B... this would be another valid solution, but I think there are many others aswell.
  11. Sep 14, 2008 #10
    ohh thanks for your help
    so my equation is junk?
  12. Sep 14, 2008 #11


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    No, your equation is fine; it's just one of many possible solutions, and the question asked for two different solutions so you only answered half the question.
  13. Sep 14, 2008 #12
    ohh thanks for all your help
    i worked out the dr/dt =Ae^(Bt) one, that was so much easyer then all that stuff i did
    i think i might redo mine because the r(t) part doesnt seem correct, it makes the equation only really work when you know the radius or the area
    just to bug you one more time haha, would you know of any other ways like dr/dt =Ae^(Bt) i could get another equation?
  14. Sep 14, 2008 #13


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    I could think of MANY: dr/dt=At^423876.227 or dr/dt =Bsinh(t^3/(4t-1)) for example, but the easiest two to solve are probably dr/dt=k and dr/dt=Ae^Bt.

    The question is very ambiguous, and I don't think you would see one worded like that on an exam.
  15. Sep 14, 2008 #14
    =D=D thankyou very much
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