mr.luna
how do i set this up?! a bacteria culture initially contains 100 dells and grows at a rate proportional to its size. after an hour the population has incresed to 420.
a)find an expression for the number of bacteria after t hours.
b)find the number of bacteria after 3 hours
c)find the rate of growth after 3 hours
d)when will the population reach 10,000
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mr.luna
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i have trouble converting it to numbers. here what i have so far and i working on more. i use the P(t)=P(0)e^kt
mr.luna
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so P(0)=100 P(3)=420
mr.luna
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i set the equation P(3)=100e^3k=420 i solved and got k=ln42/3
mr.luna
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is a [100e^[(ln42/3)(t)]
mr.luna
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b) i just plug in 3 for t to the equation in a) ?
mr.luna
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for d) i put equation a) equal to 10,000 right?
mr.luna
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im stuck on c) though. how do you do that?
victorarana
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I guess that you can derive the expression you got before to obtain dP/dt. And then you sustitute the value of 3 on it.
.
mr.luna
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i got it down to 100(42^t/3) i dont how to dervie?
victorarana
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I think your initial conditions are wrong. You wrote P(3) = 420, and on the problem says P(1) = 420
victorarana
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\[420 = 100e^{k(1)}\]
victorarana
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so, \[k = \ln4.2\]
mr.luna
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ok crap .yeah your right! good catch
victorarana
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\[P = 100e^{\ln(4.2)t}\]
victorarana
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if you derive you get: \[\frac{d}{dt}100e^{(\ln 4.2)k}= 100(\ln4.2)e^{(\ln 4.2)k}\]
victorarana
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sorry its t instead of k
mr.luna
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im trying to get waht you got for the derviative right now. its not coming out for me.
victorarana
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what did you get?
mr.luna
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instead of ln4.2 i get 4.2k
victorarana
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remember that \[\frac{d}{dx}e^{ax} = ae^x\]
victorarana
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Sorry i missed an "a"
\[\frac{d}{dx}e^{ax}=ae^{ax}\]
mr.luna
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oh. ok. got it now. thanks
victorarana
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You're welcome.