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Mathematics 16 Online
OpenStudy (anonymous):

"A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 65 cells." I'm supposed to determine the rate of growth after 8 hours. What I've got: relative growth rate = k = ln(8) expression for the number of cells after t hours = P(t) = 65e^(8ln(t)) rate of growth = dP/dt = 65*8^(t)*ln(8) Thus, 65*8^(8)*ln(8) should be the rate of growth after 8 hours, unfortunately this seems to be wrong, why isn't it right?

OpenStudy (anonymous):

why is growth rate \(\ln(8)\)?

OpenStudy (anonymous):

doubles every 20 minutes, model is \[P(t)=65\times 2^{\frac{t}{20}}\]

OpenStudy (anonymous):

or if you need \[65e^{kt}\] to find \(k\) solve \[2=e^{20k}\] for \(k\)

OpenStudy (anonymous):

you get \[20k=\ln(2)\] and so \[k=\frac{\ln(2)}{20}\]

OpenStudy (anonymous):

Yep, which is the same as ln(8).

OpenStudy (anonymous):

really?

OpenStudy (anonymous):

Nope, sorry. ^_^

OpenStudy (anonymous):

i am fairly sure that \[\ln(2)<\ln(8)\] which means that \[\frac{\ln(2)}{20}\] is way less

OpenStudy (anonymous):

I solved mine differently, I just moved down stairs, lemme grab my notebook.

OpenStudy (anonymous):

ok but hope the steps are clear. you know it doubles every twenty minutes so you solve \[2=e^{20k}\] for k

OpenStudy (anonymous):

once we get \(k=\frac{\ln(2)}{20}\) you want population in 8 hours, which is 480 minutes or 24 twenty minute intervals, so your final answer should be \[65e^{24\ln(2)}\]

OpenStudy (anonymous):

But it is ln(8)... 2 = e^(k/3) k/3 = ln(2) k = 3ln(2) = ln(8) [Answer in the back of the book is given in the same form as the second.]

OpenStudy (anonymous):

Which makes sense since both = 2.07944154

OpenStudy (anonymous):

ok i get it you are using hours as unit of time, i was using minutes

OpenStudy (anonymous):

in any case the answer is the same, because we both end up with \[65e^{24\ln(2)}\]

OpenStudy (anonymous):

Um, not actually I got P(t) = 65e^(8ln(t)) which could also be represented as 65*8^t but I don't think that 65^(24ln(2)) is equivalent.

OpenStudy (anonymous):

actually now i am a bit confused, because your question actually states "I'm supposed to determine the rate of growth after 8 hours." presumably the rate of growth is the same. doubles every 20 minutes

OpenStudy (anonymous):

you want population in 8 hours. if you use hours as your unit of time, then you have \[65e^{8\times \ln(8)}\] and that is the same as \[65e^{24\ln(2)}\]

OpenStudy (anonymous):

ln(8) is the relative growth rate which is different than the specific growth rate for a given time period. (Oh, you're right... natural logarithms are funny, aren't they...) I think it's supposed to be the derivative of the function at the specific point.

OpenStudy (anonymous):

The point being 8 of course... since that's the time in hours we're looking for.

OpenStudy (anonymous):

so the question is "what is the derivative of \(65e^{3\ln(2)t}\) at \(t=8\)"?

OpenStudy (anonymous):

The question is what isn't dP/dy = ln(8)*8^t*65 at t = 8 correct, and why,,,

OpenStudy (anonymous):

Because I followed the chain rule, don't have issue with derivatives or anything and it looks, for all intensive purposes, right. But I get this huge number and it's wrong.

OpenStudy (anonymous):

derivative of \[65e^{3\ln(2)t}\] is \[65\times 3\ln(2)e^{3\ln(2)t}\] by the chain rule

OpenStudy (anonymous):

it is a huge number, so maybe i do not understand the question

OpenStudy (anonymous):

At 2.27 hours there are 20,000 cells, so there should be a reasonable number of cells, but the rate of change couldn't possibly be 2267670593.7688146, it's an insane answer.

OpenStudy (anonymous):

Yeah, I don't understand what I'm doing wrong, I guess it's an understanding issue.

OpenStudy (anonymous):

do you know what the answer is supposed to be?

OpenStudy (anonymous):

The specific question, word for word says... "Find the rate of growth after 8 hours. (Round your answer to three decimal places.)"

OpenStudy (anonymous):

it seems like you have a solution, because you know \(k=\ln(8)\) or \(3\ln(2)\)

OpenStudy (anonymous):

Yeah, it says that it's 15873694156.382

OpenStudy (anonymous):

But I don't know where they're getting that from...

OpenStudy (anonymous):

hmm let me use a calculator

OpenStudy (anonymous):

ok i think i see it

OpenStudy (anonymous):

nope, but closer

OpenStudy (anonymous):

nope, i was wrong, damn

OpenStudy (anonymous):

sorry i tried a bunch of different things to get that number and i cannot get it. i guess i do not understand the question. is there a similar example in the book to mimic?

OpenStudy (anonymous):

Yeah, don't worry about it. I'm going to let my brain re-boot on it and I'll talk to my teacher about it in class later.

OpenStudy (anonymous):

If I don't get it, that is...

OpenStudy (anonymous):

good luck

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