Find the relative growth rate. (Assume t is measured in hours.)
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 64 cells.
do you know the formula?
\[y = 60 t ^{\frac{ 1 }{ 3 }}\]
i mean the initial would be 64
Nope, I don't know it.. totally new and haven't been taught it yet. And you sure? So the the initial is 64?
that's what your q states (at least)
Am I good at googling ? http://answers.yahoo.com/question/index?qid=20080316183032AA2lJBD
well its clear if t is the time then at t_0=0 f(0)=64 at t(20) f(20)=64^2 and so on ...
20 minutes is \[\frac{ 1 }{ 3 }\] of a hour
but im not completly sure about the exponent, it could be . .
@SolomonZelman You're good at googling but the answers are wrong xD
hmm so the population f(t) would be \(\large f(t)=f(0)e^{kt}\) f(0)=64 hmm as t in hour so yes its 1/3 for 20 min now we need to find k using \(\large f(1/3)=64 * e^{1/3k}\) ohh and f(1/3)=64*2 (not as i mintioned before ) solve for k so u would have the population function and the relative growth rate would be k !
Formula A = P (2)^(t/h) A = final population P = initial population t = time h = time it takes to double In this case, A = 64(2)^(t/20), where t is in minutes
and what is the the relative growth rate @sourwing :o
set e^r = 2^(1/20) and solve for r
r = 0.03465 or 3.465%. relative rate in different time unit will certainly be different.
in hours,rate is e^r = e^(1/(20/60)) r = 0.23105 or 23.105%
typo, it's e^r = 2^(1/(20/60))
relative rate is still correct
ive got 2.079 hmm idk maby there is somthing wrong with my answer
opps. the equation was right, i did the math wrong :DD e^r = 2^(1/(20/60)), then r = 2.079 %
so were both are correct ^_^
yep ^.-b
:D
Tried following along and makes sense now, thanks guys :D
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