Math 4791/5791 - Solution Set 2
Fall 1998

  1. Two population models. The point of the problem is to note that
    constant absolute rate of change tex2html_wrap_inline87 linear growth

    constant percent rate of change tex2html_wrap_inline87 exponential growth.

    The elk population grows at a rate of 300 elk/year which a constant rate of change of absolute numbers; therefore the elk population is growing linearly. Letting t=0 represent 1990 and p(t) be the population after tex2html_wrap_inline95 years, we have that p(t)=4500+300 t. It follows that in 1995, p(5)=6000 elk and in 2000, p(10)=7500 elk.

    On the other hand, the world population is growing at a rate of 1.6% per year which is a constant percent (or fractional or relative) rate of change; therefore the world population is growing exponentially. Letting t=0 represent 1988 and p(t) be the population in billions after tex2html_wrap_inline95 years, we have that tex2html_wrap_inline109 , where the rate constant k must be determined from the percentage growth rate. Note that in one year the population grows by 1.016 of its current value. Therefore tex2html_wrap_inline113 . Solving for k, we have that tex2html_wrap_inline117 . The correct rate constant is not k=.016, but it is close since tex2html_wrap_inline121 when |r| << 1. The growth law is given by

    displaymath71

    The population will reach 6 billion (according to this model) when tex2html_wrap_inline125 . Solving for t we have that tex2html_wrap_inline129 years which means the population reaches 6 billion just prior to 2000. The net gain in population in 1990 alone was tex2html_wrap_inline131 billion or 80 million people. The net gain in 2000 alone will be p(12)-p(11)=.095 billion or 95 million people. The net gain in absolute numbers is increasing! The doubling time of the population is tex2html_wrap_inline135 years.

    To find the time of the hypothetical Eden, solve tex2html_wrap_inline137 for t (note we must include the ``billions'' in the initial population); this gives us that t= -1363 years which means 1363 years before 1988 or 625 AD. The model is inaccurate for ``retrodiction'' since the world's population growth rate has been significantly less than 1.6% for most of the history of humanity.

  2. Arguing as we did in the previous problem, the rate constant for the prairie dog population is tex2html_wrap_inline143 . The population at time t is tex2html_wrap_inline147 . Notice that the rate constant is not k = 0.15 since the function tex2html_wrap_inline151 does not increase by 15% each year. Using the properties of logarithms, tex2html_wrap_inline153 can also be written tex2html_wrap_inline155 which clearly increases by 15% each time unit.

    To find the doubling time, solve tex2html_wrap_inline157 for t. This condition results in the equation tex2html_wrap_inline161 or tex2html_wrap_inline163 . We find that the doubling time is tex2html_wrap_inline165 years and it is constant for all times. In general, the time required for the population to increase m-fold is tex2html_wrap_inline169 , so tex2html_wrap_inline171 years and tex2html_wrap_inline173 years.

  3. Using the model developed in class with S=300 gm/liter, R=10 liter/min, and V=1000 liters, the governing equation for the concentration is C'(t) +0.01 C(t) = 3. This first order linear ODE can be solved directly to give the general solution tex2html_wrap_inline183 , where k is an arbitrary constant. Using the initial condition C(0)=50 implies that k=-250. Thus the solution to the initial value problem is

    displaymath72

    Note that tex2html_wrap_inline191 . The graph of the solution begins at the point (0,50) and rises monotonically to the horizontal asymptote at C=300 which says that the solution that is initially in the mixing tank is slowly replaced by solution from the supply tank.

  4. The governing equation must be formulated by accounting for the mass that flows in and out of the tank. First note that the volume of solution in the mixing tank increases at a rate of 2 liters/minute (the difference between the inflow and outflow rates). Therefore the volume is given by V(t)=1000+2t. We see that the mixing tank reaches its capacity of 2000 liters (and the experiment ends) when t=500 minutes. Now we can do the bookkeeping on the mass of salt in the mixing tank. At any given t the change in salt in a small interval tex2html_wrap_inline201 is

    displaymath73

    where tex2html_wrap_inline203 liters/minute and tex2html_wrap_inline205 liters/minute. Using the same limit argument that we used in class and substituting the volume function, we pass to the ODE

    displaymath74

    This first order, linear, variable coefficient ODE can be solved using the integrating factor tex2html_wrap_inline207 . The solution is

    displaymath75

    Note that M(0)=5000 grams as it should. We can now recover the concentration using the relation C(t)=M(t)/V(t). We find that

    displaymath76

    Note that C(0) = 50 grams/liter, C(500)=292.2 grams/liter and tex2html_wrap_inline191 which is a good check if we imagine a mixing tank with an infinite capacity. Thus, when the experiment ends, the mixing tank has reached 97% of its steady state level. The graph of concentration begins at (0,50) and rises monotonically towards a horizontal asymptote at C=500. The graph actually stops, as does the experiment, at (500,292.2).

    We could also solve the problem directly for the concentration by carefully deriving the governing equation in terms of C(t). We see that

    displaymath77

    Rearranging terms, noting that V'(t)=2, and dividing by V(t), the ODE becomes

    displaymath78

    This first order, linear, variable coefficient ODE can be solved using the integrating factor tex2html_wrap_inline231 . The solution is identical to the concentration function given above.


Bill Briggs
Wed Sep 2 05:19:08 MDT 1998