2. The rate of change in the number of bacteria in a culture is proportional to the number present. In a certain laboratory experiment, a culture has 10,000 bacterial initially, 20,000 bacteria at time t1 minutes, and 100,000 bacteria at (t1 + 10) minutes.
a. In terms of t only, find the number of bacteria in the culture at any time t minutes, t ≥ 0,
b. How many bacteria were there after 20 minutes?
c. How many minutes had elapsed when the 20,000 bacteria were observed?
a. In terms of t only, the rate of change in the number of bacteria can be represented as: dN/dt = kN, where N is the number of bacteria at time t and k is the constant of proportionality.
To solve this differential equation, we can separate the variables:
dN/N = kdt
Integrating both sides gives: ln(N) = kt + C, where C is the constant of integration.
Using the initial condition N(0) = 10,000, we can substitute t = 0 and N = 10,000 into the equation above to solve for C:
ln(10,000) = k(0) + C
9.21034 = C
Therefore, the equation for the number of bacteria at any time t minutes (t ≥ 0) is:
ln(N) = kt + 9.21034
b. To find the number of bacteria after 20 minutes, we substitute t = 20 into the equation:
ln(N) = k(20) + 9.21034
Simplifying the equation, we can solve for N:
ln(N) = 20k + 9.21034
N = e^(20k + 9.21034)
c. When the number of bacteria is 20,000, we can substitute N = 20,000 into the equation above:
20,000 = e^(tk + 9.21034)
To solve for t, we can take the natural logarithm of both sides and solve for tk:
ln(20,000) = tk + 9.21034
Simplifying the equation, we can solve for tk:
tk = ln(20,000) - 9.21034
Once we find tk, we can solve for t by dividing both sides by k:
t = (ln(20,000) - 9.21034) / k
To solve this problem, we can start by setting up a differential equation that describes the rate of change of the number of bacteria in the culture. Let's call the number of bacteria at any time t as N(t).
a. In terms of t only, the differential equation can be written as:
dN/dt = k * N(t)
where k is the proportionality constant.
To find the number of bacteria in the culture at any time t minutes, we need to solve this differential equation. Let's assume the solution is of the form N(t) = C * e^(kt), where C is a constant to be determined.
Substituting this into the differential equation, we get:
dN/dt = k * C * e^(kt)
Since dN/dt is the rate of change of N with respect to t, and the problem states that the rate of change is proportional to the number of bacteria, we can say that the rate of change at any time t is equal to the number of bacteria at that time multiplied by some constant.
Substituting the values given in the problem, we have:
10,000 * k = k * C * e^(k * t1)
100,000 * k = k * C * e^(k * (t1 + 10))
Simplifying these equations by canceling out the k's, we get:
10,000 = C * e^(k * t1)
100,000 = C * e^(k * (t1 + 10))
Now we have a system of two equations with two unknowns (C and k). We can solve these equations simultaneously to find the values of C and k.
b. To find the number of bacteria after 20 minutes, we substitute t = 20 into the equation we found in part a. This will give us the value of N(20).
c. To find the number of minutes that have elapsed when the number of bacteria is 20,000, we substitute N(t) = 20,000 into the equation we found in part a. We can then solve for t.
To solve this problem, we need to use the concept of exponential growth/decay. Given that the rate of change of the number of bacteria is proportional to the number present, we can use the general form of exponential growth/decay:
N(t) = N₀ * e^(kt)
Where:
- N(t) is the number of bacteria at time t
- N₀ is the initial number of bacteria
- e is the base of the natural logarithm (approximately 2.71828)
- k is the growth/decay constant
- t is the time in minutes
a. To find the number of bacteria in the culture at any time t minutes, we need to find the value of k. We are given two data points: (t1, 20,000) and (t1 + 10, 100,000). Plugging these values into the exponential growth equation, we get two equations:
20,000 = 10,000 * e^(kt1) [Equation 1]
100,000 = 10,000 * e^(k(t1 + 10)) [Equation 2]
To solve for k, we can divide Equation 2 by Equation 1:
5 = e^(k(t1 + 10) - kt1)
Taking the natural logarithm of both sides:
ln(5) = k(t1 + 10) - kt1
Rearranging the equation:
k = (ln(5)) / 10
Now we have the value of k. We can substitute this value back into Equation 1 to find N(t) in terms of t:
N(t) = 10,000 * e^((ln(5)/10)t)
b. To find the number of bacteria after 20 minutes, we substitute t = 20 into the equation:
N(20) = 10,000 * e^((ln(5)/10) * 20)
Simplify the equation using a calculator to find the answer.
c. To find how many minutes had elapsed when there were 20,000 bacteria, we need to solve for t. Plugging N(t) = 20,000 into the equation, we get:
20,000 = 10,000 * e^((ln(5)/10)t)
Divide both sides by 10,000:
2 = e^((ln(5)/10)t)
Taking the natural logarithm of both sides:
ln(2) = (ln(5)/10)t
Solve for t using algebraic manipulation and a calculator.
The function must have the form
N(t) = a e^kt , where a is the intitial number and k is a constant , and t is in minutes
for t=0 , a = 10000
for t = t1
20000 = 10000 e^(kt1)
2 = e^(kt1)
for t = t1+10
100000 = 10000 e^(k(t1+10)
10 = e^(kt1 + 10k)
divide the two equations
5 = e^(10k)
ln5 = 10klne
10k = ln5
k = ln5/10
a) N(t) = 10000 e^(ln5/10 t)
b)
when t = 20
N(20) = 10000 e^(ln5/10 (20)) = 250 000
c) when N(t) = 20000
2 = e^(ln5/10 t)
ln2 = ln5/10 t
t = 10ln2/ln5 = 4.3067
so in effect we found the "doubling time" to be appr. 4.307 minutes
(we can now also check if for t = 4.307 + 10 or t=14.307 we get N = 50000
10000 e^(ln5/10 (14.307))
= 10000 e^2.302623
= 10000(10.00037..)
= 100 004 , not bad