Math 4791/5791 Solutions 3
Fall 1998

    1. The fact that the percentage growth rate of the cost of living is constant tells you immediately that exponential growth is taking place. Letting C(t) be the cost of living at time t, where t=0 represents 1990, we can write immediately that tex2html_wrap_inline91 , where the rate constant k must be determined. It does not follow that k=0.034! This rate constant will not give exactly 3.4% growth per year. The condition that C increases by 3.4% per years means that tex2html_wrap_inline99 and also that tex2html_wrap_inline101 . Equating these two expressions for C(1), we have that tex2html_wrap_inline105 , or tex2html_wrap_inline107 ; the correct value of k is close to the wrong value of 0.034. This discrepancy is small when the growth rate is relatively small. Thus the growth law for the cost of living is tex2html_wrap_inline111 .
    2. Note that the doubling time, tex2html_wrap_inline113 , is constant for all times. Thus the doubling time can be found by finding the time t at which tex2html_wrap_inline117 . This leads to the expression that tex2html_wrap_inline119 years. The approximate Rule of 70 says that the doubling time is tex2html_wrap_inline121 years (P is the percentage growth rate).
    3. Note that the tripling time, tex2html_wrap_inline125 , is also constant for all times. It can be found by finding the time t at which tex2html_wrap_inline129 . This leads to the condition tex2html_wrap_inline131 . Solving for tex2html_wrap_inline125 , we have that tex2html_wrap_inline135 years. In general, the time for an exponentially growing quantity to increase m-fold is tex2html_wrap_inline139 .
    4. Notice that in the two calculations above, the quantity tex2html_wrap_inline141 cancels out. This says that the doubling time or tripling time is independent of tex2html_wrap_inline141 . It means that the doubling time or tripling time is constnat for all time.
    5. We have already written that tex2html_wrap_inline111 , using two different log bases. To express C(t) using base 10, either
      • write tex2html_wrap_inline149 and find that tex2html_wrap_inline151 , or
      • write directly that tex2html_wrap_inline153
      In either case we have that tex2html_wrap_inline155 .
    1. The phase portrait for the ODE tex2html_wrap_inline157 is shown below and left. The equilibrium points are tex2html_wrap_inline159 . Using the arrows of the flow, it is clear that the equilibrium point x = -1 is stable, x=0 is unstable, and x=1 is stable. A few typical solution curves are shown below and right.

      See hard copy for figure.

    2. The phase portrait for the ODE tex2html_wrap_inline167 is shown below and left. The equilibrium points occur at tex2html_wrap_inline169 , for all integers n. The equilibrium points alternate in stability as shown. A few typical solution curves are shown below and right.

      See hard copy for figure.

  3. We must design a vector field f with zeros at x=-1, 0, 2. The zero at x=-1 must be a double zero because the equilibrium point there is half-stable. Keeping the function as simple as possible, any function of the form tex2html_wrap_inline179 with A>0 will do the job.
  4. From the graph of the solution curves, we see that x=1 is an unstable equilibrium point and x=0 is a stable equilibrium point. The point x=-1 is not an equilibrium point. The simplest function with the required properties is f(x) = x(x-1).
    1. Because the graviatational term mg has a positive sign, we have chosen to measure position and velocity positive in the downward direction. We can start by writing the ODE in the form tex2html_wrap_inline193 . The vector field, shown below and left has two equilibrium points at tex2html_wrap_inline195 . Only the positive equilibrium point is physically maeningful, and we see that it is stable. A solution starting at the initial condition v(0)=0 will approach this stable equilibrium point: tex2html_wrap_inline199 . It make s sense to denote this limiting value of the velocity tex2html_wrap_inline201 and call it the terminal velocity.

      See hard copy for figure.

    2. The ODE is separable and can be written

      displaymath77

      where tex2html_wrap_inline203 . Integrating by partial fractions, simplifying, and using the initial condition v(0)=0 leads to the solution

      displaymath78

    3. This solution shows the expected behavior that tex2html_wrap_inline207 . Notice that, with all other factors equal, tex2html_wrap_inline201 increases with the mass of the object, m, and decreases with the air resistance, k. The graph of the velocity function is shown above and middle.
    4. Either of the above expressions for v(t) can be integrated once more, either directly or using tables, to find the position function, assuming that s(0)=0:

      displaymath79

      This form is useful because it shows that as tex2html_wrap_inline219 , tex2html_wrap_inline221 which corresponds to motion with a uniform terminal velocity (see figure above and right). With a little algebra, the position function can also be written

      displaymath80

    1. The ODE x'(t) = f(x) = x(1-x) has equilibrium points at x=0 and x=1. To analyze the stability of x=0, we assume the solution consists of small perturbations around the point x=0; thus tex2html_wrap_inline233 , or simply tex2html_wrap_inline235 , where tex2html_wrap_inline237 . Substituting into the ODE we find that tex2html_wrap_inline239 ; we have neglected nonlinear terms in tex2html_wrap_inline241 because if tex2html_wrap_inline237 , then tex2html_wrap_inline245 . Thus the ODE for the perturbations is approximately tex2html_wrap_inline247 which has the solution tex2html_wrap_inline249 . We see that perturbations grow and the equilibrium point is unstable. We could have found the same result using the derivative test: f'(0) = 1 which implies instability.

      To analyze the stability of x=1, we assume the solution consists of small perturbations around the point x=1; thus tex2html_wrap_inline257 , where tex2html_wrap_inline237 . Substituting into the ODE we find that tex2html_wrap_inline261 , where we have neglected nonlinear terms in tex2html_wrap_inline241 . Thus the ODE for the perturbations is approximately tex2html_wrap_inline265 which has the solution tex2html_wrap_inline267 . We see that perturbations decay and the equilibrium point is stable. We could have found the same result using the derivative test: f'(1) = -1 which implies instability.

    2. The ODE tex2html_wrap_inline271 must be considered in three different cases. When a>0 there are three equilibrium points (see right figure below): tex2html_wrap_inline275 . We find that tex2html_wrap_inline277 , f'(0) = a>0 tex2html_wrap_inline279 , and tex2html_wrap_inline281 . Therefore the equilibrium point x=0 is unstable, while the points tex2html_wrap_inline285 are stable.

      With a=0, the ODE reduces to x'(t) = tex2html_wrap_inline289 . The only equilibrium point is at x=0. The fact that f(0)=f'(0) = f''(0)=0 does not necessarily mean that x=0 is half-stable. The figure below center shows clearly that this equilibrium point is stable. With a<0, the ODE has only x=0 as an equilibrium point ( tex2html_wrap_inline285 cease to be real-valued roots). We see that f'(0) = a<0, so x=0 is stable (as confirmed by the phase portrait below and right.

      See hard copy for figure.


Fri Sep 25 14:04:33 MDT 1998