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I've been studying a little bit of basic math with a friend of mine that has just started engineering school, and we were reading from the algebra and trigonometry book from Swokowski (12ve edition). On the fifth chapter, they introduce the law of Growth (and decay), which states that, if $q_{0}$ is the value of a quantity $q$ at the time $t=0$, and such quantity grows instantaneously at a rate proportional to its current value, then the amount of quantity at a time $t$ is:

$$ q(t) = q_{0}e^{rt} $$

Where r is the growth rate (in case of it being positive).

Then, my friend suggested an example he came up with. Imagine that there exists a group of 10 rabbits at a time $t=0$, that double their number each month, so, their growth rate would be 100%, which makes $r=1$. The model would then be:

$$ N(t)= 10e^{t}$$

But, this model does not hold at all, since at t=1, when the first month has passed, there should be 20 rabbits, but this model predicts around 27. And, it escalates even worse, when three months have passed (t=3), there should be 80 rabbits, but this model predicts 200.

Perhaps my approach to this problem is totally wrong, but I really fail to see where.

Jpmarulandas
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    the base is e, you would need base 2 for r=1 to mean doubling each time $q(t) = q_02^{rt}$ – mathreadler Jun 04 '18 at 05:54
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    The small error you made was in the definition of $r$. The growth rate, as you probably mean it, is $$ \frac{q(t+1)}{q(t)} = e^r $$ And you wanted this to be equal to $e^r = 2$. Therefore $$ r = \ln{2} \Rightarrow e^{rt} = e^{\ln{2} t} = 2^t $$ – Matti P. Jun 04 '18 at 05:59
  • Growth rate is not times how much it grow, it is a constant that depends on it, so you need to calculate using the method @MattiP. did or use a point: you know that $N(1)=20$ so $10e^{r\cdot1}=20$, and solve for $r$ – ℋolo Jun 04 '18 at 06:20

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How did you come up with $r=1$?

Note that $$ q=10e^{rt}$$ along with $q(1)=20$ implies, $$20=10e^r$$ which implies $$ e^r=2 $$ that is $r=\ln 2$

The equation is $ q=10e^{t\ln 2}$ or $$ q(t)= 10(2^t)$$

Mohammad Riazi-Kermani
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  • Thanks sir! That explains why the calculations didn't seem to fit right. Although I have another short question: following this definition on the book, an example is given in which they say that the growing rate of a population is 5% (and the initial population is provided), and they then directly replace r=0.05 on the formula I gave, ¿is this wrong, then? Thanks in advance! (if by any chance, you have the means to take a peek to this book, this definition and example of the growth and decay law is at page 303) – Jpmarulandas Jun 04 '18 at 06:33
  • @Jpmarulandas If the rate is given in advance, then you may use the given rate to complete your formula, $ q=q_0 e^{rt} $. Since we did not have the rate we used the information $q(1)=20$, to find $r=ln2.$ – Mohammad Riazi-Kermani Jun 04 '18 at 11:00
  • Oh, I see, but, if you still don't mind, there is something that is still not clear to me. On this example, the one on the book, if one calculates the population after one year, with the formula (and how it is calculated on the book itself) it yields $p=153800e^{0.05\cdot1} \approx 161685.5$, but doing it "at hand", it yields $p=153800+153800\cdot 0.05=161490$. There is a great difference between the two results, but, to me, the second appears to be the correct one. Thanks in advance, again! – Jpmarulandas Jun 04 '18 at 14:45
  • With the second method the interest is paid once on 153800 for the whole year. With the first method as you gain interest the new principal and interest gain interest continuously. Like compounding every second so the the result is much larger in the first case. It is continuously compounded against the annual compounding. – Mohammad Riazi-Kermani Jun 04 '18 at 19:09