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?
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)
To solve this problem, we need to set up a differential equation that represents the growth of bacteria on the pizza. Let's denote the number of bacteria as N(t), where t represents the time in hours.
According to the problem, the rate of growth of bacteria is proportional to the number of bacteria. This can be written as:
dN/dt = kN
Where k is the proportionality constant.
To solve this differential equation, we can use separation of variables:
dN/N = k dt
Integrating both sides:
∫(1/N)dN = k ∫dt
ln(N) = kt + C
Where C is the constant of integration.
To find the value of C, we can use the initial condition given in the problem. When the pizza was removed from the refrigerator, there were 50 bacteria. Therefore, at t = 0, N = 50:
ln(50) = k(0) + C
ln(50) = C
Now we have the equation:
ln(N) = kt + ln(50)
To find N after 6 hours, we substitute t = 6 into the equation:
ln(N) = k(6) + ln(50)
To find k, we can use another given condition. One hour after removing the pizza from the refrigerator, there were 200 bacteria. Therefore, at t = 1, N = 200:
ln(200) = k(1) + ln(50)
Now we have a system of equations with two unknowns (k and N):
ln(N) = 6k + ln(50)
ln(200) = k + ln(50)
We can solve this system of equations to find the values of k and ln(N). Once we have those values, we can calculate N by taking the "e" power of ln(N):
N = e^(ln(N))
Finally, substitute t = 6 into the equation to find the number of bacteria after 6 hours.