Calculus
posted by Ben Klotz
Suppose bacteria is growing on a pizza that has been taken out of the refrigerator at a rate that is proportional to the number of bacteria. Suppose there were 50 bacteria when the pizza was removed from the refrigerator and one hour later there were 200 bacteria. How many bacteria would there be 6 hours after the pizza was removed from the refrigerator?

Steve
first, it's "bacteria are" growing.
You are told that the for a population p of these bacteria,
dp/dt = kp
so, that means that
dp/p = k dt
ln p = kt + c
p(t) = c e^(kt)
p(0) = c = 50
So,
p(t) = 50e^(kt)
p(1) = 200, so
50e^k = 200
k = ln 4
p(t) = 50e^(ln4 t)
Or, since it is clear that the population quadruples every hour, (and since e^ln4 = 4):
p(t) = 50*4^t
So, now find p(6)
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