query 20

#$&*

course

12/6 4

course Mth 174

As I stated in my response, I did not understand what was being asked for in the added problem (old 11.7.10). The book gives examples using graphs of dP/dt v. t, but not dP/dt / P v. t.

??????E??Y??]?????assignment #020020. `query 20

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

Cal 2

*********************************************

Question:

Query problem 11.7.21 was 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

We first solve diff equation,

1000 / P dP/dt = 100 - P

=dP/dt = (100 – P)/1000/P = P(100 - P)/1000, dividing thru by P(100 – P) and mult thru by dt

=dP/P(100 – P) = 1/1000 dt

=Int(dP/P(100 – P)) = using partial fractions

1/P(100-P) = A/P +B/(100 – P) = (A(100 – P) + BP)/P(100 – P) no P in numerator

so P = 0

= A(100 – 0) + B*)/P(100 – P) = 100A = 1, so A = 1/100

Now for Int(dP/P(100 – P)) we have 1/100Int(dP/P) – Int(100 – P)

= .01ln|P| + ln|100 – P|

 1/1000 dt = 1/1000t

.01ln|P| + ln|100 – P| = 1/1000t + C

=.01ln|P(100 – P)| = 1/1000t + C

 ln|P(100 – P)| = .1t + C

 |P(100 – P)| = e^(.1t + C) = Ce^(.1t)

 P = e^(.1t)/(100 – P)

???I am not sure how I went wrong, but I can’t figure out my next step?????

confidence rating #$&*:

1000 dP/dt = P ( 100 - P ) so
dP / (P ( 100 - P ) = dt/1000.

Using partial fractions the integral on the left is .01 ( ln | P | - ln | 100 - P | ).

Accepting that P can't exceed 100, we have | 100 - P | = 100 - P.

So you end up with

ln ( P / (100 - P) ) = .1 t + C so that
P / (100 - P) = A e^(.1 t), where A = e^c. Since c is arbitrary, A is an arbitrary positive number.

If you multiply both sides by 100 - P you get

P = 100 * A e^(.1 t) - P * A e^(.1 t) .

The existing equation can be rearranged so that all P terms are on the left. P can then be factored, and the equation finall solved for P:

P - P * A e^(.1 t) = 100 A e^(.1 t)
P ( 1 - A e^(.1 t)) = -100 A e^(.1 t)
P = 100 A e^(.1 t) / (1 - A e^(.1 t) ).

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

RESPONSE -->

1000/P dP/dt = 100 - P

dP/dt = (P/1000)*(100-P)

dP/dt = 0.1P*[(100-P)/100]

Therefore, L = 100; k = 0.1

P(0) = 200

A = (L - P(0))/P(0) = (100 200)/200 = -1/2

P = L/(1+Ae^(-kt) = 100/[1-(1/2)e^(-0.1t)

Graphing this function with P(0) = 200 results in a curve that is concave up,at t=0 intercepts the y-axis (P) at P=200, and has L=100 as a horizontal asymptote. Therefore, P cannot be greater than 200 or less than 100.

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.

.................................................

......!!!!!!!!...................................

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

????So do we not attempt to solve this problem with the same approach we have been trying to solve with. By that I mean should we try to solve for L and k and not set up and solve the Integrals which are set equal to each other. I believe the answer to this is yes we can use the integrals(because we worked thru an equation like this in the notes) but this method is a more efficient approach to the problem?????????

Solution makes sense all the same>

I've inserted a complete solution for the specific differential equation in this exercise.

------------------------------------------------

Self-critique Rating:2

*********************************************

Question:

Query problem 11.7.17 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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

10:54:48

what are your values of a and b?

......!!!!!!!!...................................

RESPONSE -->

I don't understand what this question is asking for. I plotted P v. t, and I used the calculator to fit a logistic equation to this data:

P = 469.245/[1+46.432e^(-0.021t)]

But this is as far as I got. The items requested in the question don't match what's in the text.

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.

The given solution plots 'dP / P vs. t for the four intervals corresponding to the five given times.

Using the values of P and t at the midpoint of each interval we get the first table.

The second table changes the t scale so it dates from the common reference point 1800. That is, t values are now interpreted as the number of years since 1800.

When we construct a graph of ( `dp / `dt ) / P vs. t, we find that it is a nearly straight line with slope -.0001 and y intercept about .028, which we replace with the slightly more accurate .0275. That is, our graph is well fit by the straight line y = -.0001 t + .0275. Here, y is (dP/dt) / P, so our straight line represents the equation

(dP/dt) / P = -.0001 t + .0275 or
1 / P dP/dt = -.0001 t + .0275. Separating variables we get
dP / P = (-.0001 t + .0275) dt. Integrating and simplifying we get the model

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

We could still evaluate A by plugging in a set of known values of P and t. For example, if we plug in the 1990 values we get

248 = A e^(.0275 * 190 - .00005 * 190^2) and
A = 248 / e^(.0275 * 190 - .00005 * 190^2) = 8.11.

Our function would therefore become

P = 8.11 * e^(.0275 * t - .00005 * t^2).

Evaluating this function for t = 0, 50, 100 and 150 we get populations of about 8, 28, 77 and 162 million, close to the given populations for 1800, 1850, 1900 and 1950.

.................................................

......!!!!!!!!...................................

10:55:00

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

......!!!!!!!!...................................

RESPONSE -->

See previous response.

.................................................

......!!!!!!!!...................................

10:55:09

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

......!!!!!!!!...................................

RESPONSE -->

See previous response.

.................................................

......!!!!!!!!...................................

10:55:17

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

......!!!!!!!!...................................

RESPONSE -->

See previous response.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

*********************************************

Question:

Query problem 11.7.28 was 11.7.14 (was 10.7.18) dP/dt = P^2 - 6 P

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The graph of dP/dt against P for positive P of dP/dt = P^2 – 6P is clearly a quadratic fn where when using our quadratic formula plugging in the values a = 1, b = -6, c = 0 we find the zeros of the equation to be 6 and 0. So as our fn passes thru our orgin(starting pt is (0, 0) because we are only dealing with pos values of P) our rate of change is neg until t reaches 6 at which rate of change is zero then continues to grow exponentially.

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

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

......!!!!!!!!...................................

RESPONSE -->

dP/dt on y-axis, P on x-axis

Graph is a portion of a parabola. From (0,0) graph is decreasing and concave up with a minimum at (3,-9), then increasing, passing (6,0) and continuing to increase without bound.

.................................................

......!!!!!!!!...................................

11:20:55

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

......!!!!!!!!...................................

RESPONSE -->

For P(0)=5: I plotted a portion of the slope field from the the previous graph to determine the shape of the solution curve, beginning with slope = -5 for Pzero=5. Since the slopes are all negative from P = 5 to P = 0, with increasingly negative slopes until P = 3 (slope -9), and then decreasingly negative slopes to P = 0, P decreases towards 0 as a limit (where the slope is 0).

The curve is concave down from t = 0 until P reaches 3, which occurs at t = 0.5 (approx.) (this point corresponds the the minimum of (3,-9) on the dP/dt v. P curve--slope of P'=0). The curve is then concave up as t increases without limit and the curve approaches P = 0.

For P(0)=8, the slope of the curve is positive and rapidly increasing, thus P increases without bound as t>>infinity. The curve is concave up throughout, since dP/dt is increasing without bound as P increases.

** 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. **

.................................................

......!!!!!!!!...................................

11:43:16

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

......!!!!!!!!...................................

RESPONSE -->

If P(0) < 6, then then dP/dt is initially negative, P is decreasing, and dP/dt remains negative as P approaches 0 as a limit. If P(0) > 6, dP/dt is initially positive, P is increasing, and dP/dt remains positive as P increases without bound. If P(0) = 6, dP/dt = 0 and the population remains 6 as t increases.

Threshold population is the minimum initial population above which the population will grow as t increases. In this case the threshold population is 6, and the population will grow as long as P(0) > 6.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I left out minimum and maximum pt but understand where they are located.

------------------------------------------------

Self-critique Rating:3

*********************************************

Question:

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

At first the robin Population begins to increase but as soon as this happens the worm pop instantly starts to decrease and decreases faster and faster until a max rate of decrease in the pop is hit. Once this max is hit the pop of the robins is now beginning to be affected, meaning the robin pop starts to decrease. as the robin starts to decrease after a small interval of time the worm pop starts to increase again and continues to grow which causes an increase in the robin pop. This completes the cycle which continues infinitely until the rate of change of the worms or the robins is affected by some outside source related to something other than the effects their growth have on each other.

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

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

......!!!!!!!!...................................

RESPONSE -->

Using the slope field to create the closed curve, the maximum robin population is approximately 2,500 with a corresponding worm population of approximately 1,000,000. The minimum robin population is approximately 400 with a corresponding worm population also approximately 1,000,000. Since dr/dt = -r +wr, with w(0)=2 and r(0) = 2, dr/dt = -2 + (2)(2) = +2. Therefore, r is intially increasing and the population point is moving counterclockwise around the closed curve.

Good. For reference:

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.

.................................................

......!!!!!!!!...................................

11:57:08

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

......!!!!!!!!...................................

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

This is the first example of a dependencey between two populations that that has really opened my eyes and made clear a lot of the ideas of rate of change and how you can work out formulas or solutions so that you are not dealing in terms of time or dist. alone. That sveral other factors can influence and be accounted for in terms of reasoning of these equations we use to find how fast(or slow) our sample is effected by change in time or dist. or whatever. Really brought it home for some reason, not so sure why. I once was blind but now I see, at least a little better ten I did.

Very good.

------------------------------------------------

Self-critique Rating:3

??????????????I suppose these are notes you want us to look over, because I did not see a question or a place to place an answer????????????????????????????????????

There were extraneous questions in the Query, which repeated questions that had already been asked. I deleted them here and in the original document.

See my notes. You're doing quite well here. Let me know if you have questions.

#$&*

*&$*&$