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Thread: [RESOLVED] modelling problem

  1. #1

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    [RESOLVED] modelling problem

    hi

    im stuck on an additional homework problem, any ideas on how to solve appreciated

    A sphere of ice i melting so its volume decreases whilst it maintains the same shape (but not size).

    (a) Use volume balancing to derive differential equations describing the rate of change of the radius of the sphere , r, if it is melting:

    i) with a constant rate of volume loss;
    ii) at a rate of volume loss proportional to the surface area of the sphere;
    iii) at a rate of volume loss proportional to the volume itself.

    -constants of proportionality are k1, k2, and k3

    -solve each of the equations to find an expression for r as a function of time t, assuming that the initial radius of the sphere is ro.

    -ro = 2x10^-3, k1= 10^-6 m^3 s^-1
    k2 and k3 should be chosen to make the initial rate of melting the same in all 3 cases.

    thanks,

    Donna
    Last edited by hani1987; Mar 26th, 2007 at 06:02 PM. Reason: [RESOLVED]

  2. #2
    vbuggy krtxmrtz's Avatar
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    Re: modelling problem

    Quote Originally Posted by hani1987
    hi

    im stuck on an additional homework problem, any ideas on how to solve appreciated

    A sphere of ice i melting so its volume decreases whilst it maintains the same shape (but not size).

    (a) Use volume balancing to derive differential equations describing the rate of change of the radius of the sphere , r, if it is melting:

    i) with a constant rate of volume loss;
    ii) at a rate of volume loss proportional to the surface area of the sphere;
    iii) at a rate of volume loss proportional to the volume itself.

    -constants of proportionality are k1, k2, and k3

    -solve each of the equations to find an expression for r as a function of time t, assuming that the initial radius of the sphere is ro.

    -ro = 2x10^-3, k1= 10^-6 m^3 s^-1
    k2 and k3 should be chosen to make the initial rate of melting the same in all 3 cases.

    thanks,

    Donna
    Hi, welcome to the forums.

    The volume is V = 4*Pi*r3/3 so the rate of change is:

    dV/dt = d(4*Pi*r3/3)/dt = 4*Pi*r2dr/dt

    This is our basic equation and I'll call it (E0).

    Part (a) Constant rate of volume loss. This means dV/dt = constant, i.e. dV/dt = -k1
    where I have assumed k1 is positive, so I use a negative sign as the volume decreases, i.e. the rate is negative.

    From (E0):

    4*Pi*r2dr/dt = -k1 or 4*Pi*r2dr = -k1dt

    Integrating this you get,

    4*Pi*r3/3 = -k1 + C

    where C is the integration constant to be determined. If the initial radius of the sphere (when t=0) is r0 then substituting you get C = 4*Pi*r03/3

    Putting C into the equation and rearranging you finally obtain:

    r = [r03 - 3k1t/(4*Pi)]1/3

    Parts (b) and (c) are similar, all you have to do is write the adequate expressions for dV/dt, i.e.

    (b) dV/dt = -k2*4*Pi*r2
    and
    (c) dV/dt = -k3*V = -4*k3*Pi*r3/3

    and substitute in (E0).
    Last edited by krtxmrtz; Mar 23rd, 2007 at 01:37 AM.
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    Re: modelling problem

    thanks!

  4. #4
    vbuggy krtxmrtz's Avatar
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    Re: modelling problem

    Quote Originally Posted by hani1987
    thanks!
    You're welcome. Post back if you need any more help.
    Lottery is a tax on people who are bad at maths
    If only mosquitoes sucked fat instead of blood...
    To do is to be (Descartes). To be is to do (Sartre). To be do be do (Sinatra)

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    Re: modelling problem

    Hey thanks for helping out, there is only one more thing i need to know before its all solved:

    for part c, my final equation is ln r = -1/3k3t + C

    so to find c you put t= 0, which gives ln r0 = c?

    and taking exponentials, gives r = e^(-1/3k3t) + r0?

    should i work out the constant before or after integrating?

    thanks!

  6. #6
    vbuggy krtxmrtz's Avatar
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    Re: modelling problem

    Your final equation should be

    dr/r = -k3dt/3 and after integrating,

    ln(r) = -k3t/3 + C

    C = ln(r0) and all you have to do is substitute and operate:

    ln(r) = -k3t/3 + ln(r0)
    ln(r) - ln(r0) = -k3t/3
    ln(r/r0) = -k3t/3
    r/r0 = e-k3t/3
    r = r0e-k3t/3
    Lottery is a tax on people who are bad at maths
    If only mosquitoes sucked fat instead of blood...
    To do is to be (Descartes). To be is to do (Sartre). To be do be do (Sinatra)

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