I'm not sure how to do this question
7. The growth of bacteria in culture can be described by the equation
N1 = N0e'
where N is the number of bactena at any time t, No is the initial number of bacteria, and k
is a constant. The time taken for growth to double the number of bacteria of a particular
strain (the 'doubling time') was 30 mm. A culture was started with 200 bacteria of this
strain
Determine the followmg (giving appropnate units in each case)
(a) the value of k for this strain (2]
(b) the rate of increase m bacterial number when the number in the colony
was 2000 131
(c) the minimum number of bacteria in this culture (1]
for a i calculated k =0.023 or (ln 2)/30
for c i think it would be at t = 0
Nt = No.e^(k0)
Nt = No.e^0
Nt = No.1
Nt = No = 200
i don't know how to do b and im not sure if im right with a and c please help
I cannot follow what you are doing. Your original equation
N1 = N0e'
is wrong, and you quote a doubling time im millimeters.
To obtain help, you need to be more careful posting your questions.
First of all your opening equation should have been
N1 = N0 e^(kt), where t is in minutes.
Does 30 mm mean "30 minutes" ? (I assumed that)
I did get the same value of k.
so your equation is
N = 200 e^(.023105t)
the rate of increase is the derivative of your function, so
dN/dt = 200(.023105) e^(.023105t)
So we now have to find the value of t when N = 2000131
2000131 = 200 e^(.023105t)
for that I got t = 398.6 minutes
(you better check my arithmetic here)
Now sub that t value back into our derivative to find the rate of change at that time.
c) is a strange question. Since it is exponential growth, the minimum value would be at the start, namely 200
sorry it was a copy and paste job should checked it thanks for the help
To solve part (a) of the question, you correctly used the equation N1 = N0e^(kt) to determine the value of k for this strain. The doubling time is given as 30 minutes, and you correctly calculated k = (ln 2)/30 or approximately 0.023. This value represents the growth rate of the bacteria.
For part (b) of the question, you are asked to determine the rate of increase in the bacterial number when the number in the colony is 2000.
To find the rate of increase, we need to calculate the derivative of N with respect to t. Taking the derivative of N = N0e^(kt) with respect to t, we get:
dN/dt = kN0e^(kt)
Substituting the given values, we have k = 0.023, N0 = 200, and N = 2000. Plugging these values into the equation, we get:
dN/dt = 0.023 * 200 * e^(0.023t)
Now, since we want to find the rate of increase when N = 2000, we set N = 2000 in our equation and solve for dt:
2000 = 0.023 * 200 * e^(0.023t)
Dividing both sides by 0.023 * 200, we have:
e^(0.023t) = 2000 / (0.023 * 200)
Now, we take the natural logarithm of both sides to solve for t:
ln(e^(0.023t)) = ln(2000 / (0.023 * 200))
0.023t = ln(2000 / (0.023 * 200))
Finally, we solve for t:
t = ln(2000 / (0.023 * 200)) / 0.023
Using a calculator, t is approximately equal to 110.28 minutes.
Therefore, the rate of increase in the bacterial number when the number in the colony is 2000 is given by dN/dt = 0.023 * 200 * e^(0.023 * 110.28).
For part (c) of the question, you correctly determined that the minimum number of bacteria in the culture would be at t = 0. At the start (t = 0), the number of bacteria is Nt = No = 200, as you correctly calculated.
To summarize:
(a) The value of k for this strain is approximately 0.023.
(b) The rate of increase in bacterial number when the number in the colony is 2000 is dN/dt = 0.023 * 200 * e^(0.023 * 110.28).
(c) The minimum number of bacteria in this culture is 200.