Assignment 20

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Between 1800 and 1850 we have `dP / `dt = (23.1 - 5.3) / (1850 - 1800) = 17.8 / 50 = .36, approx.., meaning that the average rate of population change was .36 million per year.

During this interval the average population was about (5.3 + 23.1) / 2 = 14.2, meaning 14.2 million. However this is a linear estimate of the average value of a nonlinear function; the actual average is probably closer to 11 million or so, which would be a geometric mean.

The midpoint of the time interval is 1825. So (with the caveat that we are using a linear approximation on an interval where the nonlinearity is significant) we would say that the rate .36 million per year corresponds to population 14.2 at clock time 1825. This gives us (`dP / `dt) / P = .025, meaning .025 million per year per million of population. This could be interpreted as a birth rate or fertility rate of 2.5% (note that a million per year per million of population is the same a the number per year per individual).

Similar calculations for the four intervals defined by the data give us the following, where P_mid and t_mid are the midpoint population and clock time as they would be estimated by a piecewise linear graph (i.e., a trapezoidal graph) of P vs. t:

`dP/`dt P_mid t_mid (`dP/`dt) / P_mid

0.356 14.2 1825 0.025070423

1.058 49.55 1875 0.02135217

1.48 113 1925 0.013097345

2.4675 199.35 1970 0.012377728

Measuring time from the reference point 1800 we obtain

`dP/`dt P_mid t_mid (`dP/`dt) / P_mid

0.356 14.2 25 0.025070423

1.058 49.55 75 0.02135217

1.48 113 125 0.013097345

2.4675 199.35 170 0.012377728

The graph of (`dP / `dt) / P_mid vs. t_mid is not perfectly linear, but the linear best-fit will clearly have vertical intercept near .027 or .028, and a slope near -.0001. In fact the best-fit line is given by Excel as -.0001 t + .0275.

Our population model would therefore be the equation

(dP / dt) / P = -.0001 t + .0275.

This equation could be solved for P by first rearranging it to give us

dP / P = (-.0001 t + .0275) dt. Integrating gives us

ln | P | = .0275 t - .00005 t^2 + c so that

| P | = e^( .0275 t - .00005 t^2 + c ) and, if A = e^c,

| P | = A e^(.0275 t - .00005 t^2) for A > 0.

Since negative population is not relevant, our final solution is just

P = A e^(.0275 t - .00005 t^2) for A > 0.

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** dP/dt = P^2 - 6 P is a quadratic function a P^2 + b P + c of P, with a = 1 and b = -6. The function opens upward, with its vertex at P = -b / (2a) = -(-6) / (2 * 1) = 3. At this point P^2 - 6 P = 9 - 18 = -9.

The graph has zeros where P^2 - 6 P = 0, or P(P-6) = 0. The zeros are therefore at P = 0 and P = 6.

Thus between P = 0 and P = 6 the value of dP/dt = P^2 - 6 P is negative. dP/dt = 0 at P =0 and at P = 6 and reaches its minimum at P = 3.

If P(0) = 5 then dP/dt = 5^2 - 6 * 5 = -5 so the population initially decreases. As P decreases it comes closer to P = 3, at which value dP/dt is minimized so that the rate of decrease is greatest, so the P vs. t curve will become steeper and steeper in the downward direction. Up to the P = 3 point the graph of P vs. t will therefore be concave downard.

After P decreases to less than 3 the values of dP/dt begin to increase toward 0, however still remaining negative. The graph of P vs. t will become concave upward, with P approaching zero for large t.

If P(0) = 8 then dP/dt = 8^2 - 6 * 8 = 16 so the population initially increases. As P increases dP/dt also increases rate of increase is increasing, so the P vs. t curve will become steeper and steeper in the upward direction. The graph of P vs. t will therefore remain concave upward. The graph of P vs. t therefore continues to increase at an increasing rate, exceeding all bounds as t -> infinity. **

** This equation is of the form dP/dt = k P (1 - P / L):

P^2 - 6 P = P ( P - 6) = -P(6 - P) = -6 P ( 1 - P/6). So the right-hand side has form k P ( 1 - P / L) with k = -6 and L = 6.

You should know how to derive the solution to dP/dt = k P (1 - P / L), which is P = L / (1 + A e^(-k t) ) with A = (L - P0) / P0.

The solution in this case has A = (6 - P0) / P0.

For P0 = 5 we have A = 1/5 and P = 6 / (1 + 1/5 e^(-(-6) * t) ) = 6 / (1 + 1/5 e^(6 t) ).

The denominator increases without bound as t -> infinity, so that P approaches zero as t increases. **

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** dw/dt = w - wr and dr/dt = -r + wr so that dr/dw = r(w-1) / w(1-r).

For r = 0 (the horizontal axis) this gives us dr/dw = 0, so slope lines along the horizontal axis are horizontal.

As w -> 0 we see that dr/dw -> -infinity so the slope lines are vertical along the vertical axis, with the understanding that the vertical slopes are vertical downward.

As we approach r = 1, w = 1 we see that dr/dw approaches the form 0 / 0 so that the slope near (1, 1) will depend very sensitively on whether we are slightly to the right, slightly to the left, slightly above or slightly below (1, 1). This results in the 'circling' behavior we observe around this point.

Looking at the slope field starting at (3, 1) we see that we move upward and to the left, with the worm population decreasing and the robin population increasing. This continues but with less and less upward movement and more and more leftward movement, indicating a slowing of the growth of the robin population while the worm population continues to decrease quickly.

After awhile the graph is moving directly to the left, perhaps near the point (1.5, 2.5), just after which it begins moving downward and to the left, indicating a decreasing robin population as the worm population continues to be depleted. The movement becomes more and more downward and less and less to the left, indicating that as the robin population declines the worm population is not being decimated as quickly as before.

The graph eventually reaches a vertical downward direction, perhaps around the point (.3, 1), after which it continues moving downward but also to the right. This indicates an increasing worm population as the robins continue to decline.

The graph descends less and less rapidly as the increasing worm population begins to provide food for the robins, who stop dying off as quickly. The graph reaches its 'low point' around maybe (1, .2) before it begins to move upward and to the right once more. This indicates that while the robin population is still small enough to permit an increase in the worm population,there are enough worms to feed a growing robin population.

This increase in both populations continues until the graph returns to the initial point (3, 1), at which point there are enough robins to again begindepleting the worm population and the cycle begins to repeat. **

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Very good. I've inserted a couple of detailed solutions for your reference.