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

@&

@&
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.
*@

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"

@&
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.

*@

Self-critique (if necessary):

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

Self-critique rating:

________________________________________

#$&*

@&
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?
*@

@&

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.

week4quiz3

@&

@& 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. *@

@& 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? *@

@&

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).Be sure to include the entire document, including my notes.

@&
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.
*@

@&
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?
*@

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.