Exponential Growth of BacteriaDate: 7 Aug 1995 08:19:31 -0400 From: Greg Sharpe Subject: Growth & Decay A colony of N bacteria increases with time according to the formula dN -- = kN where K is constant. dT If initially there are N bacteria show that the number at any time, N, 0 kt is given by N = N e 0 If the number of bacteria doubles every 2 hours, how long (to the nearest minute) will the colony take to treble? Date: 8 Aug 1995 22:35:26 -0400 From: Dr. Ken Subject: Re: Growth & Decay Hello there! This is an example of a differential equation. To solve it, we'll use integration: dN -- = kN dt dN = kN * dt 1/N * dN = k * dT Then integrate: log[N] = k*t + C N = e^(k*t + C) = e^C * e^kt To find C (or rather e^C), we plug in known simultaneous values of N and t: N = e^C * e^0 = e^C 0 So if you plug back in, you get the desired equation. >If the number of bacteria doubles every 2 hours, how long (to the nearest >minute) will the colony take to treble? The information "doubles every 2 hours" lets us actually find k. We'll plug in k=0 and k=1. Since we know N = N e^kt, we know 2*N = N , i.e. 2*N e^0 = N e^k, so 0 0 1 0 0 e^k = 2. Therefore k = Log[2]. Right? So then we have N = N e^(Log[2] * t). We can find out how long it will take to treble by 0 figuring out when the amount of N is 3*N . So we have the equation 0 3N = N e^(Log[2] * t), which you can tackle with a calculator. 0 0 Hope this helps! -K |
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