# P3 VERY EASY differentiation questionWatch

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#1
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 proportional to the number of bacteria present at that instant.

This is what I've done so far:

dN/dt = 600ce^ct

I'm just confused how to EXACTLY show the two are proportional...can someone please show me how you would do it in an exam? Thx!
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14 years ago
#2
sorry but what youve done so far is wrong lol
u should have dN/dt=600e^ct
then you split the variable to get int-1 dn= int- 600e^ct dt
which gives you cN=600e^ct
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#3
(Original post by RObTRIP)
sorry but what youve done so far is wrong lol
u should have dN/dt=600e^ct
then you split the variable to get int-1 dn= int- 600e^ct dt
which gives you cN=600e^ct
sorry i didnt get this step:

-1 dn= int- 600e^ct dt

plz explain...doesnt look familiar...
0
14 years ago
#4
i meant im intergrating the left side which is 1 with respect to dn and the right side with respect to dt lol this plain text stuff makes things look so complicated
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14 years ago
#5
(Original post by RObTRIP)
sorry but what youve done so far is wrong lol
u should have dN/dt=600e^ct then you split the variable to get int-1 dn= int- 600e^ct dt
which gives you cN=600e^ct
why is this not dN/dt = 600c.e^ct
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#6
(Original post by manps)
why is this not dN/dt = 600c.e^ct
thats wht i said first! lol!
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14 years ago
#7
whoops im stupid sorry!!! i read it wrong i thought what you first wrote was the first part /me cowers in shame!
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14 years ago
#8
(Original post by devesh254)
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 proportional to the number of bacteria present at that instant.
N = 600e^ct
---> dN/dt = c.600e^ct
---> dN/dt = cN
---> dN/dt is proportional to N.

QED
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