Saturday
March 25, 2017

Post a New Question

Posted by on .

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

  • Math - ,

    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.

  • Math - ,

    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

  • Math - ,

    sorry it was a copy and paste job should checked it thanks for the help

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question