im stuck on this question, can someone help me?
The rate of increase of a population (p) of micro-organism at time (t) is given by dp/dt = kp
where k is a positive constant. Given that at t=0 the population was of size 8, and at t=1 the population is 56, find the size of the population at time t=2.
what i have done so far
1/kp dp=1 dt
1/k. Ln(kp)= t+c
when t=0 p=8
when t=1 p=56
1/k. Ln(56k) -1 =c
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- 29-05-2016 19:11
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- 29-05-2016 19:15
If you know the initial population and that the population at t=1 is 56, you can use this to find the constant and formulate an exponential equation in terms of P and t. Then, just substitute t=2 into your equation to find P
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- 29-05-2016 19:18
I'd probably be dividing just by p at the start? 1/p dp = kdt?
This gives c = ln(8) and works out nicer?
NB there is a 95% chance this is wrong as I am awful at C4...
- 29-05-2016 19:21
- 29-05-2016 19:24
can i cancel both the Ln 7 ?