#$&* course Mth 174 Assignment 20course Mth 174
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Given Solution: RESPONSE --> STUDENT SOLUTION: 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. INSTRUCTOR SOLUTION 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. Symbolic interpretation: 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 - ˝ e^(-t/1000) always less than 1 so that the population is always greater than 100. Specific detailed solution: We will separate variables and integrate. 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 we 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) ).
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: We use P and t values at the midpoint intervals to construct the first table. The second table consists of the number of years since 1800. We make a graph of (‘dp/ ‘dt) / P vs. t. The graph is a straight line represented by the equation dP / P = (-.0001 t + .0275) dt. The final solution is P = A e^(.0275 t - .00005 t^2) for A since the negative population is not important. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 10:54:48 what are your values of a and b?
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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.
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10:55:00 What are the coordinates of your graph points corresponding to years 1800, 1850, 1900, 1950 and 1990?
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RESPONSE --> See previous response.
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10:55:09 According to your model when will the U.S. population be a maximum, if ever?
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RESPONSE --> See previous response.
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10:55:17 Give your solution to the differential equation and describe your sketch of the solution.
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RESPONSE --> See previous response. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I could have evaluated A by plugging in the set of values of P and t. We have the function P = 8.11 * e^(.0275 * t - .00005 * t^2). We we plug in t = 0, 50, 100, and 150, we get populations of approximately 8, 28, 77, and 162 million. These are close to the populations for 1800, 1850, 1900, and 1950. ------------------------------------------------ Self-critique Rating: OK ********************************************* 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 decreases from (0,0) and is concave up. The minimum is at (3,-8) and increases. It passes through (6,0) and continues to increase. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: describe your graph of dP/dt vs. P, P>0
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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.
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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
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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.
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11:43:16 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 --> 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): ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: Query 11.8.6 (was page 570 #10) robins, worms, w=2, r=2 init YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When we use the slope field, we can see that the maximum robin population is close to 2,500. The worm population that goes along with this is approximately 1,000,000. The minimum robin population is about 400, and the corresponding worm population is also approximately 1,000,000. We know that dr/dt= -2+ (2*2) = 2 because dr/dt= -r +wr, and w(0)=2 and r(0)=2. In the beginning, r increases and the population point moves in a counterclockwise direction around the closed curve. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: what are your estimates of the maximum and minimum robin populations, and what are the corresponding worm populations?
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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:
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11:57:08 Explain, if you have not our a done so, how used to given slope field to obtain your estimates.
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