# P3 Differentiation!!!Watch

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Thread starter 13 years ago
#1
Hi....
I'm having trouble with with q34, reveiw exercise 1, book p3 so i neeed your help plz... :

The number N of bacteria in a certain culture at time t hours is given by N=600e^ct, where c is a constant. SHow that at any instant the number of bacteria is increasing at a rate propertional to the number of bacteria present at that instant. The number of bacteria increases from 600 when t =0 to 1800 when t = 2.

(I) Show that c = 0.5 ln 3
(II) Show that the number of bacteria present at t hours is 600B^0.5t, where B is a constant and state the value of B.

Thanks.....
0
13 years ago
#2
N = 600e^(ct)
dN/dt = 600ce^(ct) = cN

(i)
t=2, N=1800
1800 = 600e^(2c)
3 = e^(2c)
ln3 = ln[e^(2c)] = 2c.lne = 2c
c = 0.5 ln3

(ii)
N = 600e^[(0.5t)ln3] = 600e^[ln(3^{0.5t})] = 600.3^(0.5t)
B = 3
0
Thread starter 13 years ago
#3
thanks mate...
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