ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: sophie.tran999 on August 01, 2011, 05:47:57 pm
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Can anyone help me with this question, i get stuck with a)
Q: in a certain bacterial culture, the rate of increse is proportional to the number of bacteria present.
a) if the number doubles in 3 hours, find the hourly growth rate.
i can deal with b) and c).
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Let N represent the number of bacteria present
let t represent time
then 
so at t=0 lets start of with n Bacteria
then at t=3, N=2n
solve the differential equation
flip it 

+c)
solve for c, plug in t=0, and N=n +c\Rightarrow c=-\frac{1}{k}log_{e}(|n|))
so )
so the hourly growth rate will be k
so plug in t=3 and N=2n
)
)
So the hourly growth rate is )
Wait I'm still wrong
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thx, thats wat i did afetr readin the question 1st time, but look, the question askes to find the GROWTH RATE, not find t in term of N :D
the anwser is : dN/dt = 0.23N
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1/3loge(2)=0.23 so that means that dN/dt=0.23N, I think that answers it right?
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since growth rate =dN/dt
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i think so. damn me lol
thx
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Yeh it's right and works, I had myself stuck in a loop for a min there.
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solve for c, plug in t=0, and N=n, this is important part. i missed this one
instead od plugging t=0 and N=o first, i just pluged t=3 and N=2N. :(