Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I had to solve the equation

$$\frac{dN}{dt} = rN \left( 1 -\frac{N}{K} \right)$$

where $r$ and $K$ are constant and $N(0) = N_0$.

I solved it and got

$$N(t) = \frac{kAe^{rt}}{1 - Ae^{rt}}$$

where $A = \frac{N_0}{k - N_0}$. Now I am being told to sketch the graph of this solution $N(t)$ against $t$ for $t \geq 0$ at fixed values of parameters $r, K$ and for two fixed values on $N_0$, one in the range $0 < N_0 < K$ and the other in the range $N_0 > K$.

How do I do this? I haven't got a clue what the graph would look like.

EDIT: I was thinking, if I divide through by $e^{rt}$ then I get $\frac{kA}{e^{-rt} - A}$ which seems easier to sketch as it now only has $1$ function of $t$, but I'm still unsure how to sketch this.

share|improve this question
    
Is this a laser rate equation? –  Ron Gordon Feb 8 '13 at 1:27
    
@rlgordonma No, Verhulst Model (of population growth) –  Kaish Feb 8 '13 at 17:40

1 Answer 1

up vote 0 down vote accepted

It is immediate from the differential equation that the rate of change of $N$ is $0$ if either $N=0$ or $N=K$, but if $N$ is between $0$ and $K$, then the rate is positive, so $N$ is increasing. If $N$ is close to either $0$ or $K$, then $dN/dt$ is close to $0$ so $N$ is increasing slowly; if $N$ is far from both (say halfway between) then $N$ is increasing fast.

That much tells you a lot about the graph, and you can see it's true just by thinking about what the differential equation says. So it approaches $K$ more slowly as it approaches it. Therefore you've got a horizontal asymptote at $N=K$, and for the same reason a horizontal asymptote at $N=0$.

If you can show that $N\left(1-\frac KN\right)$ is biggest when $N$ is halfway between $0$ and $K$, that tells you where the curve will be steepest.

If you're a bit more ambitious, try showing you have symmetry about the point at which $N$ is halfway between $0$ and $K$, i.e. if $t_0$ is the time at which $N$ is halfway between, then at time $t_0+t$, $N$ will exceed its value at $t_0$ by the same amount by which it falls short of that value at time $t_0-t$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.