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Decay and growth (1 Viewer)

Aysce

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Just learnt this topic and I don't really get how to do this question :/

12. A population in a certain city is growing at a rate proportional to the population itself. After 3 years the population increases by 20%. How long will it take for the population to double?

Thank you.
 

Aesytic

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the rate of the growth of the population being proportional to itself just means that dP/dt = kP where k is some constant. since the population is growing, k is positive as the rate of growth must always be positive in order for the population to grow

now, P=Ae^kt, where A is the starting population of the city. normally, you'd be told to show that this equation satisfies the differential equation by differentiating, but to derive this equation, you can also integrate dP/dt:

dP/dt = kP
taking reciprocals,
dt/dP = 1/(kP)
integrating both sides with respect to P,
integral dt/dP * dP = (1/k)*integral dP/P
t = lnP/k + C
when t=0, P=A
0 = lnA/k + C
C = -lnA/k
.'. t = (lnP-lnA)/k
kt = lnP - lnA
= ln(P/A)
P/A = e^kt
P = Ae^kt

anyway, onto the question, when t=3, the population will be 120% of its initial size, or 1.2*A
subbing these values in,
1.2A = Ae^3k
1.2 = e^3k
3k = ln(1.2)
k= ln(1.2)/3

now we have a value for k, and we have to find out what t equals for the population to double, that is, when P=2A (i assume it means double the initial population, and not the population after 3 years):
2A = Ae^[(tln(1.2))/3]
2 = e^[(tln(1.2))/3]
ln2 = (tln(1.2))/3
t = (3ln2)/(ln(1.2))

round off to the nearest year to get your answer
 

Aysce

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Thank you very much Aesytic (Dayuumm similar names ;) ). I appreciate your effort and good luck for your trials :)
 

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