# Math Help - rate of growth in a population

1. ## rate of growth in a population

Using the following formula, I have to calculate the rate of growth of a population of rabbits after 3 years. The formula is
P(t)=500/1+e^-t
there are no brackets
I know I can't use my calculator to solve for e when the exponent is negative so I am guessing I have to some how alter the equation in order to solve it. If anyone can give me an idea of what to do to solve for e, I would really appreciate it. Thank you

2. Originally Posted by surffan
Using the following formula, I have to calculate the rate of growth of a population of rabbits after 3 years. The formula is
P(t)=500/1+e^-t
there are no brackets
I know I can't use my calculator to solve for e when the exponent is negative so I am guessing I have to some how alter the equation in order to solve it. If anyone can give me an idea of what to do to solve for e, I would really appreciate it. Thank you
Hi surffan,

Your formula is ambiguous. What's the 500 divided by? Are you trying to write $P(t)=\frac{500}{1+e^{-t}}$

If you're talking about exponential growth, shouldn't it be $P(t)=P_0e^{kt}$.

You're not solving for e. e is an irrational constant approximately equal to 2.718281828.

3. that is the equation given in my text for the specific question, i have to sub in 3 for t so its 3 years, and I know that i can't use a negative exponent. i was wondering if i could alter it to 500(1+e^3) or something but it doesn't seem to make sense.

4. Originally Posted by surffan
that is the equation given in my text for the specific question, i have to sub in 3 for t so its 3 years, and I know that i can't use a negative exponent. i was wondering if i could alter it to 500(1+e^3) or something but it doesn't seem to make sense.
I'm still not clear on what you're trying to do, but the fact is that you can use negative exponents in what you just wrote. Do you have an e button on your calculator? If not, use the approximation I gave you in my first post.

$500(1+2.718281828^{-3}$

Just do the arithmetic. Comes out to about 525. Does that make sense?

5. Originally Posted by surffan
Using the following formula, I have to calculate the rate of growth of a population of rabbits after 3 years. The formula is
P(t)=500/1+e^-t
there are no brackets
I know I can't use my calculator to solve for e when the exponent is negative so I am guessing I have to some how alter the equation in order to solve it. If anyone can give me an idea of what to do to solve for e, I would really appreciate it. Thank you
I get the impression you're being asked to find $P(3)$. From what I can gather the question would read $P(t) = \frac{500}{1+e^{-t}}$

In this case sub 3 for t:

$P(3) = \frac{500}{1+e^{-3}}$.

I get an answer of $P(3) \approx 476

$

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Bonus Material...

$e$ is a constant, much like $2$ or $\sqrt7$. It is the base of the natural logarithm ( $\ln(e) = 1$). It's unique in calculus for being the only function equal to it's derivative.

As $e$ is a constant the normal rules of exponents apply - $e^{-a} = \frac{1}{e^a}$.

6. what the question is asking for is the rate of growth, so in order for me to find that i need to first find the derivative which I believe can be found if I alter the equation to become P(t)=500(1+e^-t)^-1 , the thing is I don't really know how to find the derivative for this equation since its exponential but it also contains e which makes it tricky, I know I can use a combination of the chain rule for exponential functions as well as the product rule, but I can't really figure it out using e.

7. Originally Posted by surffan
what the question is asking for is the rate of growth, so in order for me to find that i need to first find the derivative which I believe can be found if I alter the equation to become P(t)=500(1+e^-t)^-1 , the thing is I don't really know how to find the derivative for this equation since its exponential but it also contains e which makes it tricky, I know I can use a combination of the chain rule for exponential functions as well as the product rule, but I can't really figure it out using e.
Use the chain rule and the fact that $\frac{d}{dx} e^{kx} = ke^{kx}$. All the chain rule is in this case is to multiply the derivative of e^(-t) by the derivative of the main expression using the power law.

$-500(1+e^{-t})^{-2} \cdot -e^{-t} = \frac{500}{e^t(1+e^{-t})^2}$

8. Originally Posted by surffan
what the question is asking for is the rate of growth, so in order for me to find that i need to first find the derivative which I believe can be found if I alter the equation to become P(t)=500(1+e^-t)^-1 , the thing is I don't really know how to find the derivative for this equation since its exponential but it also contains e which makes it tricky, I know I can use a combination of the chain rule for exponential functions as well as the product rule, but I can't really figure it out using e.
I agree. The rate of population growth is the derivative of the population equation:
$P(t)=\frac{500}{(1+e^{-t})}$ can be rewritten as
$P(t)=500(1+e^{-t})^{-1}$. Using the chain rule we get
$\frac{dP}{dt}=500e^{-t}(1+e^{-t})^{-2}$, and simplifying gives
$\frac{dP}{dt}=\frac{500e^{-t}}{(1+e^{-t})^{2}}$ or
$\frac{dP}{dt}=\frac{500e^{-t}}{(1+e^{-t})^{2}}$.

Plugging 3 in makes the resultant growth rate 22.588.

The rules for differentiating $e^x$ are very simple, since $\frac{d}{dx}(e^x)=e^x$ (which means $e^x$ is its own derivative and all you have to worry about is the chain rule). For example, the derivative of $e^{-x}$ is $-e^{-x}$ and the derivative of $e^{2x^2}$ is $4x*e^{2x^2}$.

One comment though - shouldn't this be in the calculus forum and not the precalc forum?

9. thank you so much, i was really confused about what e stood for. This is actually a highschool question. grade 12 uni course

10. Originally Posted by masters
Are you trying to write $P(t)=\frac{500}{1+e^{-t}}$

If you're talking about exponential growth, shouldn't it be $P(t)=P_0e^{kt}$.

Actually, $P(t)=\frac{500}{1+e^{-t}}$ is a standard form for logistic growth, not exponential growth. The function behaves exponentially initially, but growth eventually slows and becomes asymptotic. This is the way many populations actually grow since increasing growth rates associated with true exponential growth usually cannot be sustained (i.e. some environmental factor like lack of food, water, or space will eventually limit growth).