Equivalence of growth rate found through derivation of growth equation and rate found through $A = Pe^{rt}$.

So I had a question regarding bacteria growth where I was asked to find the growth rate of $500$ bacteria given this equation: $500\big(1 + (\frac{4t}{50+t^2})\big)$ where $t =$ time in hours and is equal to $2$.

I was supposed to use derivation to find the answer but I elected to use the continuous compounding function $A = Pe^{rt}$ where $A =$ final pop., $P =$ initial pop., $e$ is euler's number, $r =$ rate, and $t =$ time. I plugged $2$ into the original equation to get $574.074$. This should equal $A$. So I next did $574.074 = 500e^{2r}$, divided the $500$ out, took the natural $\log$ of both sides, and divided by $2$ to get finally $r$.

Now, my real questions: is the $r$ I found using $Pe^{rt}$ equal to the $r$ I would have found using derivation? If so why? How are the two functions equivalent? How can I transform the derivation so that its similar and equal to the growth function?

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I'm not sure why are you trying to use the "compounding function".

In terms of population, what you call compounding function (whos name comes from interest rate calculation I believe) comes from what's called the Malthusian Growth Model, wich states that the rate of change of a population is proportional to the current population number, in other words:

$$N'(t) = k N(t),$$

where $N(t)$ is the population at time $t$ and $k$ is the rate of growth.

$$N(t) = 500\left(1 + \frac{4t}{50 + t^2}\right).$$

Taking the derivative, the rate of change is

$$N'(t) = 500\left(\frac{4}{50+t^2} - \frac{8t^2}{(50+t^2)^2}\right)$$

wich hardly is a malthusian growth.

A visual study

One can see the way the population of bacteria behaves by looking the plot of $t\,vs.\,N(t)$

We can observe that the population at $t=0$ is $N(0) = 500$ and starts to grow. There is a time $t_c$ where the population can't sustain the growth (lack of space, food, you name it), and the population starts to diminish, only to stabilize again, with the original number of bacteria.

The maxmimum of population occurs when $N'(t) = 0$, wich can be calculated straightforward, giving $t_c = \sqrt{50}$ and $N_{max} \approx 641$ bacteria.

This means that, at time $t=2$ the growth must be positive, and given by the rate of change of the population in time, i.e. the derivative, wich was computed above. Hence, at $t=2$, the rate of growth is

$$N'(2) \approx 31.55,$$

meaning that at $t=2$ the population is growing roughly by 31 bacteria per time unit.

At the end, one can suppose that the function $N$ is modeling the growth of bacteria in a petri dish, where some nutrient was placed, giving a suddent growth, and when the nutrient has been finished, the extra population becomes unsustainable returning to it's original equilibrium.

Malthusian Growth

In the case of malthusian growth, it is supposed that the change in population is proportional to the current population, i.e.

$$N'(t) = k N(t),$$

where $k$ is the rate of growth, and depends on the type of population (different for ants and whales).

The solution of the ode is

$$N(t) = N_0 e^{k t},$$

where $N_0$ is the population at time $t = 0$.

If $k > 0$, then the population grows exponentially (blue), if $k < 0$ it decreases exponentially (red).

When $k > 0$, $k$ is also called birth rate, while $k < 0$ means mortality.

Malthus model is unrealistic first of all because it predicts that populations grow indeterminately, wich is false (in the times of Malthus this was sort of an accepted truth), but most importantly because it neglects factors as competition, deceases, food, space, etc.

What you calculated

The $r$ you calculated is the rate of growth a malhtusian population must possess in order to achive the same number of individuals as the bacteria colony at time $t=2$.

If we plot the malthusian population (blue), against the bacteria (red) using your parameters we see

The intersection is in $t=2$.

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So in other words, I have failed? Was using the Malthusian Growth Model completely wrong? To me at least the logic was sound; the variables and substitutions fit quite nicely. – Stumbleine75 Oct 4 '12 at 0:43
I think you are confusing the growth rate with the rate of change. Think of it this way: the grow rate is a constant that describes the capacity of a population to grow, when it behaves as a malthusian population. I might be wrong though, as I don't know the hole context of your lectures. Maybe if you post the exact question, we can be sure. – Pragabhava Oct 4 '12 at 0:48
A population of 500 bacteria is introduced into a culture and grows in number according to the equation P(t)=500(1+(4t/50+t^2)), where t is measured in hours. Find the rate at which the population is growing when t=2. – Stumbleine75 Oct 4 '12 at 0:50
@ThroatOfWinter57 I was correct then. See my full edit for an explanation. If you like the answer, please consider accepting it. – Pragabhava Oct 4 '12 at 1:27