Question

Hi OpenStudy. :) I'm working on a logistic growth equation where I'm being asked in the first step to solve for k and M. I'll attach what I've done so far. What I'm looking for is a reality check on my approach. Screenshot and problem attached.

Answer 1

The growth follows this equation:\[\frac{\text{dP}}{\text{dt}}=k\cdot P(M-P) \] My work is in the attached image. P is the population in millions, carrying capacity, M, is 10,000,000, and the peak growth rate is 3,000,000 per day. From this, I need to find M and k in the equation.

Answer 2

I think that I probably should have put in the given values for M and P, and set them equal to 0 to solve for k. Might try that next.

Answer 3

Here's what that looks like.

Answer 4

What you have in that red box to the left is no different from the initial setup for your DE.

Your solution for \(C\) is correct.

Since \(P\) has units in millions, you can use say \(P=1\) in the case that the population is 1 million. Since the carrying capacity is 10 million, then you know that \(P(t)\to10\) as \(t\to\infty\).
\[P(t)=\frac{M}{\color{red}{\frac{M-P_0}{P_0}}e^{-tkM}+1}~~\implies~~10=M\text{ as }t\to\infty\]
The peak growth rate is 3 million, so \(\max\left\{\dfrac{dP}{dt}\right\}=3\). The second derivative is equal to 0 for whatever \(t\) gives this maximum growth rate. I would think you're expected to solve \(P'(t^*)=3\) for \(t^*\) (the time at which population growth attains its maximum), then use the population at that time, \(P(t^*)\), to solve
\[kP(t^*)\bigg(10-P(t^*)\bigg)=3\]

Answer 5

I'm taking a short diversion on this to wrap up some other work for the same class. I'll be back shortly.

Answer 6

Ah, I missed the million bit. :P That's making more sense now. I'll post my response once I'm through. Thanks so much for your help!

Answer 7

That was what I needed, thanks! It took me a little while to to sort out, but I think I've got it now. :)

Answer 8

Oh, and here's the new and improved.

Answer 9

If
\[\frac{dP}{dt}=kP(10-P)\]
then
\[\begin{align*}\frac{d^2P}{dt^2}&=k\frac{dP}{dt}(10-P)-kP\frac{dP}{dt}\\\\
&=k^2P(10-P)^2-k^2P^2(10-P)\\\\
&=k^2P(10-P)(10-2P)\end{align*}\]
Notice that you're taking the derivative with respect to \(t\), not \(P\), so chain rule is a must.

Answer 10

Ah, yes. It's always the little details that pile up. Or in this case, the infinitely small details. :)

Answer 11

Why is k squared?

Answer 12

Just examining the first term in the first line:
\[k\frac{dP}{dt}(10-P)=k\left(kP(10-P)\right)(10-P)=k^2P(10-P)^2\]

Answer 13

Since k is a constant, wouldn't I do this?

Answer 14

Then I get get P=5.

Answer 15

We get the same result in the end, I just use the product rule right away, whereas you expanded and performed the derivative term-by-term.

Answer 16

Ok, that's good. I'll see where I get from here.

Answer 17

Also note that since you're solving \(\dfrac{d^2P}{dt^2}=0\), you'll get more than one solution.
\[\begin{align*}
\frac{d^2P}{dt^2}&=10k\frac{dP}{dt}-2kP\frac{dP}{dt}\\\\
0&=2k\frac{dP}{dt}(5-P)
\end{align*}\]
which means the second derivative is also zero when \(\dfrac{dP}{dt}=0\), and not just when \(P=5\).

Answer 18

Ah, right. That makes sense. Ok, here's what it looks like so far.

Answer 19

And there we have it. Thanks again. :)