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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?

Differential Equations
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have any ideas wat to do? :)
or how to start off?
do you know an equation that works for this kind of thing?

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Other answers:

no need for one
but it's helpful
yes. i work it out until i got P = Ae^(0.15t)
since 10yrs=15% then 30yrs=45% thus the population growth in 30 yrs is 5000 x 145/100 50 x 145
the equation you can find for this problem is\[P=5000 \times 1.15^\frac{ t }{ 10 }\]
then you plug in 30 for t
wat is the denominator under t?
lol the shortcut is easier and less complicated :P ur just jelly
nevermind, @HawkCrimson the shortcut you took doesn't work
D: haha
@Peter14 i dont understand...can you show me how it starts from P = Ae^(0.15t) ?
what you actually do each 10 years is you multiply the number of people by 1.15
cause it's 15%= . 15 right? :D
where does 1.15 come from?
oh ik how to do it by the shortcut way :P but it's long and tedius
1.15 is 115% the way the problem expresses this is as an exponential growth problem.
but the question says 15% :/
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.
lol forget me, answer the girl :P
100%=5000 pop so if it has a pop increase by 15% then it will become 115% that is y
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.
okay..
have you looked at compound interest in your math class or should I try to explain that to you?
how do i calculate using the formula dP/dt = kP ?
o.o where is that from?
try this http://qrc.depaul.edu/StudyGuide2009/Notes/Savings%20Accounts/Compound%20Interest.htm
its the formula given
what math class are you in?
i mean thats what we study in application of first order differential equations
from that formula we integrate and so on
wow, you've confused me.
XD
and google is sooo slow here in china so I have to stay confused...
ok, I understand now. you were given dP/dt = kP as the equation at the beginning. I get it.
what was the exact wording of the problem? I seem to have misunderstood it before.
i just want to know how do i start from P=Ae^(0.15t) from what i integrated from the above formula
what does A represent?
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
yeaa tried that but couldnt seem to get the answer right :/
http://fym.la.asu.edu/~pvaz/mat272/ch_07.pdf
wats the answer?
7604 people
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
i got that using the shortcut method but that wont get u any marks in the exam, ill try using the formulae
maybe you need to find the annual rate of increase instead of the rate of increase for 10 years
okay i'll try
P = Ae^0.15t P = 5000e^0.15(3) P = 7841 :/
do i have to find a new rate or what?
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.
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?
i got it! thanks guys for the help :)
I think you got it less because of us and more despite us.
but you guys made it more clear so thanks :D

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