calc problem help!!!?
The doubling period of a baterial population is 15 minutes. At time t=110 minutes, the bacterial population was 70000. With t representing minutes, the formula for the population is p(t) = Ae^kt.
a) what does k =?
b)the initial population at t=0 is
c)size of population after 4 hours is
PLEASE HELP ME IM SO LOST thanks!!!
Comments
I see how it could be easy to get lost. This is one of most problems where you just need to be creative.
a) Start with what you know. t=110, right? Then p(110)=70000=A*e^(110*k).
Since you know that it doubles every 15 minutes, you know that at 95 minutes (15 minutes previous), there were 35000 bacteria. Thus, p(95)=35000=A*e^(95*k).
70000=A*e^110k and 70000=2*A*e^95k. Thus, Ae^110k=2Ae^95k.
Solve this by dividing: 2=e^15k -> k = ln(2)/15 = 0.0462.
b) We need to figure out what A is because at t=0, the e part is 1, so p(0)=A*1.
To find A, you can use either of the equations above. I'll use the 110-minute one. 70000=A*e^0.0462*110 = A*161.09. Thus, A=70000/161.09 = 434.52.
c) Now we have the whole equation. p(0)=434.52*e^0.0462t.
At t=240 minutes, p(240)=434.52*65381=28,409,722. That's a lot.
(a.)
70,000 = Ae^(110k)
140,000 = Ae^(125k)
2 = e^(15k) => k = ln(2)/15
(b.)
A = 434.05
p(0) = A = 434 approx.
(c.)
p(240) = 434.05*(2^16) = 28,445,901