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course Mth 173
216 8:18 PM
week4quiz3#$&*
course Mth 173
2/15 3:07 AM
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 2627 its rate of change is known to be 400. If this is the t=0 state of the population, then approximately what will be the population at t = 1.4? What then will be the population at t = 2.8?
Proportionality
y= kP
differential is
'dy= slope *'dx
but now im thinking back to the trapezoidal graph, and the area was the change in quantity, and area can be a proportion....for instace say a circle.. A=PIr^2
it is saying the rate of chnage is 400 at 2627 P.
???how do i know what P is, x, x^2, x^3?????????? it sounds like x to me
it said rate of change is proprtional to P
so 'dy/'dx = kP
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Very good, but `dy/`dx indicates an average rate of change and 'the' rate of change is an instantaneous rate (again treading near some muddy philosophical waters).
And the population changes in time, not with respect to some unspecified variable x.
So your equation should be
dy/dt = k P.
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400=k2627
k= .15226494
dy/dx=.15226494P
???what is t? where is that coming from?
'dy=.15226494 * 'dx is the only thing I could think of...not sure if that is right...
the other was at t=0, and t=1.4 that would be a 1.4 change so
'dy=.15226494 * 1.4
'dy=.213170917
I am a little lost....
at this rate everything i get would sem to equal 1""
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Your equation would be
dy/dt = .152 P.
Now if P = 2627, you get dy/dt = 400 (provided your .152 was calculated correctly, as I believe is was).
If this rate applies as t changes from 0 to 1.4, then the approximate change `dy will be related to the change `dt by
`dy / `dt = 400,
with `dt = 1.4. This implies that
`dy = 400 `dt = 400 * 1.4 = 560.
So the new population would be approximately
2627 + 560 = 3187.
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I am having trouble remembering what part of the equation relies on the change being a small increment. At least from looking at the equation.
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When you assume that the rate remains constant while t changes by 1.4, you're subject to error, because for this function (as for most) the rate changes during this interval.
The shorter the interval the less the rate will tend to change, so the less error is likely.
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So the new population would be approximately
2627 + 560 = 3187.
dy/dx=.15226494P
485.26836378
at t=1.4, to 2.8 that is 1.4
'dy=485.26836378*1.4=679.375709292 is the change that you asked for
3187 + 679.375709292=3866.375709292 is the total new population.
where this is a population, not sure if the decimal matters to reality, but for changes in growth i guess it would
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Since there is an error inherent in assuming a constant rate for the entire interval, three significant figures would be more than enough. Consider that the rate used for the first was 400, and the rate for the second was closer to 500. There is at least a 20% change in the rate during the interval, which would lead to something like a 10% difference in the average rate (the rate having been correct at the beginning of the interval) so even a 2-significant figure result would be optimistic.
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Now the population is 3187. Using
dy/dt = .152 P
what is the rate of change of the population, and how much would the population therefore change between t = 1.4 and t = 2.8?
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Good. Check my notes for more.
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