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# is there a difference? :repost

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I have the equation:

$$P(t)=\frac{K}{9e^{-rt}+1}$$

this equation was derived from solving the logistic growth equation:

$$\frac{dP}{dt}=rP\big(1-\frac{P}{K}\big)$$

where,

r is the rate of growth = 1.13

K is the maximum population = 10,000

I've been asked to change the growth rate to:

$$r= 0.83 + \frac{0.3}{1+0.3t}$$

My question is,

Do I have to solve the entire logistic growth equation again, that is,

or can I just substituted r into the derived equation and rearrage, that is,

I wanted to get some clarification as I've tried doing both process with a simpler equation, however the answer I get quite is complex and I'm not sure If I've done it correctly.

In my opinion I don't think it matters if I substitute the new r straight into P(t) ... but i'm not overly confident about that

Thank you.

Mar 30, 2018

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I don't see how you can avoid doing the integration again, since r is now a function of t.  The result I get when so doing is as follows:

Mar 30, 2018
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thanks for your reply Alan. I was thinking that might be the case.

I see your image has been blocked. Would you be able to unblock it please?

Mar 30, 2018
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It's blocked for me as well!

However, if you scroll through the Answers pages (try https://web2.0calc.com/answers/page/8179) you can see the image.  It's only when you select the specific question (or answer) that the image is blocked!

Alan  Mar 31, 2018
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great! thank you. yes I got the same with:

vest4R  Mar 31, 2018