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course Mth 173
6/23/13~around 11PM
Week 4 Quiz 3 Version 1Write 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.
300 = k(2954)
K = .1016
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Good, but you haven't written the differential equation, which would be
rate of change of P = k P
written in mathematical notation as
dP / dt = k P.
dP/dt = 300 when P = 2954, so you have correctly solved for k.
Your equation is therefore
dP / dt = .1016 * P.
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If this is the t=0 state of the population, then approximately what will be the population at t = 1.2?
--> plugging in values for the model below
(P - 2954) / (1.2 - 0) = .1016P
P - 2954 = .1219P
P = .1219P + 2954
0 = .1219P + 2954 - P
0 = -.8781P + 2954
.8781P = 2954
P = 3364.0815
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Good, but your solution was pretty roundabout and might have obscured what's going on here. The use of P for both the old and the new population also caused a discrepancy between your approximation and the standard approximation.
More to the point:
dP / dt = .1016 P
so when P = 2954 you have
dP/dt = .1016 * 2954 = 300.
Changing at this rate for a time interval of 1.2, the change will be
dP = 300 * dt = 300 * 1.2 = 360
resulting in population a
P = 2954 + 360 = 3314.
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This standard first-approximation model assumes that the rate of population change stays constant for the entire interval, which since the population is continuously changing is not actually the case.
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What then will be the population at t = 2.4?
-->
(P - 2954) / 2.4 - 0) = .1016P
P - 2954 = .2439P
P = .2439P + 2954
0 = .2439P + 2954 - P
0 = -.7561P + 2954
.7561P = 2954
P = 3906.8906
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Good, but you do want to use the standard approximation, as illustrated in my notes.
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