The population of a town grows at a rate proportional to the population present at time t. The initial population of 5000 people is increased by 15% in the period of 10 years. What will the estimated population be in 30 years time?

- anonymous

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- anonymous

have any ideas wat to do? :)

- anonymous

or how to start off?

- anonymous

do you know an equation that works for this kind of thing?

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## More answers

- anonymous

no need for one

- anonymous

but it's helpful

- anonymous

yes. i work it out until i got P = Ae^(0.15t)

- anonymous

since 10yrs=15% then
30yrs=45%
thus the population growth in 30 yrs is
5000 x 145/100
50 x 145

- anonymous

the equation you can find for this problem is\[P=5000 \times 1.15^\frac{ t }{ 10 }\]

- anonymous

then you plug in 30 for t

- anonymous

wat is the denominator under t?

- anonymous

lol the shortcut is easier and less complicated :P ur just jelly

- anonymous

nevermind, @HawkCrimson the shortcut you took doesn't work

- anonymous

D: haha

- anonymous

@Peter14 i dont understand...can you show me how it starts from P = Ae^(0.15t) ?

- anonymous

what you actually do each 10 years is you multiply the number of people by 1.15

- anonymous

cause it's 15%= . 15 right? :D

- anonymous

where does 1.15 come from?

- anonymous

oh ik how to do it by the shortcut way :P but it's long and tedius

- anonymous

1.15 is 115%
the way the problem expresses this is as an exponential growth problem.

- anonymous

but the question says 15% :/

- anonymous

the shortcut you took would work for an arithmetic sequence (meaning you add a certain amount each time) but this is a geometric sequence (meaning you multiply by a certain amount each time.

- anonymous

lol forget me, answer the girl :P

- anonymous

100%=5000 pop so if it has a pop increase by 15% then it will become 115% that is y

- anonymous

the question says the population increases by 15% each time. The number you get shouldn't decrease at all. Let's look at the first case. If you multiply 5000 by 15% you get the increase in population: 750 people. You can then add that to 5000 to get 5750.
or on the other hand, you can multiply 5000 by 115% (because 115% is 100%+15%) and that gives you 5750 with only one operation.

- anonymous

okay..

- anonymous

have you looked at compound interest in your math class or should I try to explain that to you?

- anonymous

how do i calculate using the formula dP/dt = kP ?

- anonymous

o.o where is that from?

- anonymous

try this http://qrc.depaul.edu/StudyGuide2009/Notes/Savings%20Accounts/Compound%20Interest.htm

- anonymous

its the formula given

- anonymous

what math class are you in?

- anonymous

i mean thats what we study in application of first order differential equations

- anonymous

from that formula we integrate and so on

- anonymous

wow, you've confused me.

- anonymous

XD

- anonymous

and google is sooo slow here in china so I have to stay confused...

- anonymous

ok, I understand now. you were given dP/dt = kP as the equation at the beginning. I get it.

- anonymous

what was the exact wording of the problem? I seem to have misunderstood it before.

- anonymous

i just want to know how do i start from P=Ae^(0.15t) from what i integrated from the above formula

- anonymous

what does A represent?

- anonymous

I think T would be 3, because it seems from the description at the top that T is in intervals of 10 years
and then P looks like your final population

- anonymous

yeaa tried that but couldnt seem to get the answer right :/

- anonymous

http://fym.la.asu.edu/~pvaz/mat272/ch_07.pdf

- anonymous

wats the answer?

- anonymous

7604 people

- anonymous

you get that using the formula for compound interest, but I think they want you to show your work from using the methods they give you

- anonymous

i got that using the shortcut method but that wont get u any marks in the exam, ill try using the formulae

- anonymous

maybe you need to find the annual rate of increase instead of the rate of increase for 10 years

- anonymous

okay i'll try

- anonymous

P = Ae^0.15t
P = 5000e^0.15(3)
P = 7841
:/

- anonymous

do i have to find a new rate or what?

- anonymous

yes, you do
and I know by doing some graphing tricks that the rate ends up approximately 0.013975, but I don't know exactly how you get that numerically. It should have something to do with ln something.

- anonymous

so in the textbook chapter you posted the final equation is P=A(e^kt)
you are trying to find k given t=30 and a=5000
do they give you P? I guess what i'm trying to say is, what exactly do they give you at the start of the problem?

- anonymous

i got it! thanks guys for the help :)

- anonymous

I think you got it less because of us and more despite us.

- anonymous

but you guys made it more clear so thanks :D

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