Assignment 20 Query

course Mth 174

?z|?????L?^??????assignment #020020. `query 20

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Cal 2

12-26-2007

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22:39:03

Query problem 11.7.6 plot (dP/dt) / P vs. t, where P is a solution to 1000 / P dP/dt = 100 - P with initial population P = 0.{}{}Can P(t) ever exceed 200?{}{}Can P(t) ever drop below 100?

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RESPONSE -->

No, P(t) will not exceed 200 and it will not drop below 100 because 100 is the equilibrium point. I found this to be the equilibrium point by using the following formula:

L = k / a = 100 / 1 = 100

This means that with an initial population of 200, the population will drop to 100 and stay there.

I think I should have solved the differential equation and then taken the limit, but I couldn't get a satisfactory answer.

First analyzing the equation qualitatively we draw the following conclusions:

If P is initially 200 then we have at t = 0 the equation

1000 / 200 * dP/dt = 100 - 200 so that

dP/dt = -20,

which tells us that the population initially decreases.

As long as P > 100 the population will continue to decrease, since dP/dt = (100 - P) * (P / 1000) will remain negative as long as P > 100.

As P approaches 100 from 'above'--i.e., from populations greater than 100—the rate of change dP/dt = (100 - P) * P / 1000 approaches 0. So the population cannot decrease to less than 100 if it starts at a value greater than 100.

** The analytical solution will confirm these conclusions: **

This equation is of the form dP/dt = k P ( 1 - P / L). Its solution is obtained by the process in the text (write in the form dP / [ P ( L – P) ] = k / L dt , integrate the left-hand side by separation of variables, etc.).

The solution is P = L / (1 + A e^(-k t) ) with A = (L - P0) / P0.

The present problem has dP/dt = P/1000 * (100 - P) has L = 100 and k = 1/1000, with P0 = 200, so A = (100 - 200) / 200 = -1/2 and its solution is

P = 100 / (1 + (-1/2) e^(-t/1000) ) = 100 / (1 - 1/2 e^(-t/1000) ).

For t = 0 the denominator is 1/2; as t -> infinity e^(-t/1000) -> 0 and the denominator approaches 1.

Thus P is initially 100 / (1/2) = 200, as we already know, and at t increases the population approaches 100 / 1 = 100, with the denominator 1 - ½ e^(-t/1000) always less than 1 so that the population is always greater than 100.

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23:36:10

Query problem THIS PROBLEM DOES NOT EXIST IN THE NEW EDITION 11.7.10 plot (dP/dt) / P vs. t for given population data and estimate a and b for 1 / P dP/dt = a - bt; solve and sketch soln. sample pop: 1800 5.3, 1850 23.1, 1900 76, 1950 150, 1990 248.7

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RESPONSE -->

I believe I first need to find the relative rates of change for each year:

1800: 6.72%

1850: 4.5%

1900: 1.9%

1950: 1.3%

These are all equal to 1 / P dP/dt = (next year pop - previous year pop) / (50 * previous year pop)

Next I need to isolate and solve for a. I believe I can use 1/P dP/dt = 6.72% at t=0:

6.72 = a - b(0)

.0672 = a

Next solve for b at t=50 with 1/P dP/dt = .045:

.045 = .0672 - b(50)

-.0222 = -b(50)

.000444 = b

So the equation becomes:

1/P dP/dt = .0672 - .000444t

Solving this equation:

dP/P = (.0672 - .000444t) dt

ln | P | = .0672t - .000222t^2 + C

P = A e^( .0672t - .000222t^2 )

At t=0, P=5.3:

5.3 = A

P = 5.3 e^( .0672t - .000222t^2 )

The graph of the solution is an exponential function which rises to a maximum of 856 then falls again.

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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23:36:16

what are your values of a and b?

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RESPONSE -->

See above.

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23:43:27

What are the coordinates of your graph points corresponding to years 1800, 1850, 1900, 1950 and 1990?

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RESPONSE -->

1800: 5.3

1850: 87.6

1900: 477

1950: 856

1990: 614

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23:44:06

According to your model when will the U.S. population be a maximum, if ever?

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RESPONSE -->

It looks like the maximum will be at t=151 or 1950 with a max of 856.

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23:44:14

Give your solution to the differential equation and describe your sketch of the solution.

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RESPONSE -->

See above.

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23:50:53

Query problem 11.7.14 (was 10.7.18) dP/dt = P^2 - 6 P

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RESPONSE -->

a) I first sketched a graph of this equation. It seems to have an unstable equilibrium point at P=6, with the values of P rising for P > 6 and falling for P < 6.

b) For P between 0 and 6, dP/dt is negative and so the graph falls away from 6. This part of the graph is concave down. This tells me that on a graph of P against t, that P would drop as t increases.

c) With P(0) = 8, the graph of dP/dt would increase exponentially upward.

d) For P > 6, dP/dt is positive and for P < 6, dP/dt is negative.

P=6 is called the threshold population because at this point any change to the population could push it over the threshold in either direction.

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23:50:58

describe your graph of dP/dt vs. P, P>0

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RESPONSE -->

See above.

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00:00:32

describe the approximate shape of the solution curve for the differential equation, and describe how you used your previous graph to determine the shape; describe in particular how you determined where P was increasing and where decreasing, and where it was concave of where concave down

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RESPONSE -->

The solution curve is broken up depending on the initial value of P.

For P > 6, the graph curves upwards exponentially. The previous graph suggested this because the rate at which the value of P increased was concave up, so the graph of P curves upward.

For P < 6, the graph curves upwards, concave up. The previous graph suggested this because it curved downward towards 0, which means that value of P levels out at some point, making the graph of P concave down.

Good. Compare with the following:

** 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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00:00:40

describe how the nature of the solution changes around P = 6 and explain the meaning of the term 'threshold population'.

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RESPONSE -->

See above.

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00:05:45

Query 11.8.6 (was page 570 #10) robins, worms, w=2, r=2 init

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RESPONSE -->

I had some difficulty with this one. I needed to find the maximum and minimum values of the robin population, so I think I need to find the points at which dr/dt = 0. So here's what I did:

dr/dt = 0 = -r + wr

0 = r(w - 1)

0 = w - 1

w = 1

So the value of the worm population must be 1 at the maximum and minimum points of r. However, I can't figure out the values of r from here.

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00:07:29

what are your estimates of the maximum and minimum robin populations, and what are the corresponding worm populations?

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RESPONSE -->

I only found the worm populations, but I estimated the robin populations from the graph.

r~2.5 at w=1

r~.5 at w=1

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00:08:03

Explain, if you have not our a done so, how used to given slope field to obtain your estimates.

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RESPONSE -->

I found the initial point given in problem 5 and drew a curve around the slope field, then drew lines to find the points.

** 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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00:08:20

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

I guess I didn't really get this section.

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"

I think you pretty much got it.

See my notes and let me know if you have questions.