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course Mth 173
10/7 23:50
Question Write the differential equation expressing the hypothesis that the rate of change of a population is proportional to the population P. Evaluate the proportionality constant if it is known that the when the population is 2954 its rate of change is known to be 300. If this is the t=0 state of the population, then approximately what will be the population at t = 1.2? What then will be the population at t = 2.4?
Answer
Given that the rate of change of population is directly proportional to the population P.
Let dP / dt be the rate of change of population.
Thus we have dP / dt = k * P, where k is the proportionality constant.
It is also said that when the population is 2954, its rate of change of population is 300, thus 300 = k * 2954, thus k = 300 / 2954 = 0.1016
One way is by using the predictor- corrector method
The rate of the population function for population 2954 is 300. In order to calculate the population at t = 1.2, we first find the change in the population by considering the rate to be 1.2, thus the change in the population = 300 * 1.2 = 360.
Thus the new predicted population = 2954 + 360 = 3314.
The rate of change of population at 3314 population = 0.1016 * 3314 = 336.7024.
Thus the average rate of change of population in that interval = (336.7024 + 300) / 2 = 318.3512
Thus the corrected change in the population for the 1.2 time interval = 318.3512 * 1.2 = 382.02144.
Thus the final population after t = 1.2 will be 2954 + 382.02144 = 3336.02 approximately
Thus now for finding the population after t = 2.4
The rate of the population function for population 2954 is 300. In order to calculate the population at t = 2.4, we first find the change in the population by considering the rate to be 2.4, thus the change in the population = 300 * 2.4 = 720.
Thus the new predicted population = 2954 + 720 = 3674.
The rate of change of population at 3674 population = 0.1016 * 3674 = 373.2784.
Thus the average rate of change of population in that interval = (373.2784 + 300) / 2 = 336.6392
Thus the corrected change in the population for the 1.2 time interval = 336.6392 * 2.4 = 807.93408.
Thus the final population after t = 2.4 will be 2954 + 807.93408 = 3761.93 approximately
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You already have a good result for the population at t = 1.2. In this calculation you would not go back to the t = 0 state of the function, but would confine your attention to the new interval from 1.2 to 2.4.
You would average the t = 1.2 and t = 2.4 rates, and multiply this by 1.2 to get the new change in population, which you would then add to the t = 1.2 population.
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Another way is by finding the population function we need to integrate the rate of population change function. That is to find the antiderivative of the rate function.
To do that, first we need to rearrange the given function,
dP / P = kdt , on integrating both sides we get, ln(P) = kt + c ,
thus P = e ^ (kt + c) = (e ^ kt) * e^c, where c is the integration constant, and the value of k is known to us. In order to find the value of c, it is known to us that the population of 2954 is at t = 0. Thus
2954 = (e ^ 0) *(e^c) = e ^ c
Thus the population function becomes P = 2954 * ( e ^ 0.1016t )
Thus population at t = 1.2 will be
P = 2954 * ( e ^ 0.1016 * 1.2 ) = 2954 * 1.129 = 3335.07 approximately
Next the population at t = 2.4 will be
P = 2954 * ( e^0.1016* 2.4 ) = 2954 * 1.276 = 3769.30 approximately.
Thus here we also notice that the accuracy of the approximation by the predictor-corrector method decreases as the interval becomes larger and larger.
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Well explained, but you did have one conceptual error, which led to a less-accurate prediction that would be expected. Be sure to see my note.
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