Exponential Bacteria Growth Problem Solution: Finding k and Doubling Time

[nokia8650]

Question:

The number of bacteria present in a culture at time t hours is modeled by the
continuous variable N and the relationship

N = 2000e^(kt)

where k is a constant.

Given that when t = 3, N = 18 000, find

(a) the value of k to 3 significant figures

(b) how long it takes for the number of bacteria present to double, giving your
answer to the nearest minute

k is found to be 0.73241 to 5s.f

For the next part, i simply let N=18000*2=36000, and solved to find t=3.94 therefore it takes 3.94-3 = 0.94 hours for the number to double. This obviously yields the correct answer, however the markscheme let N=4000 subsequently solved to give 0.94 immediately (without having to subtract 3). Can someone please explain the logic behind the markschemes method?

Thanks

 

[HallsofIvy]

Question:

The number of bacteria present in a culture at time t hours is modeled by the
continuous variable N and the relationship

N = 2000e^(kt)

where k is a constant.

Given that when t = 3, N = 18 000, find

(a) the value of k to 3 significant figures

(b) how long it takes for the number of bacteria present to double, giving your
answer to the nearest minute

k is found to be 0.73241 to 5s.f

For the next part, i simply let N=18000*2=36000, and solved to find t=3.94 therefore it takes 3.94-3 = 0.94 hours for the number to double. This obviously yields the correct answer, however the markscheme let N=4000 subsequently solved to give 0.94 immediately (without having to subtract 3). Can someone please explain the logic behind the markschemes method?

Thanks

The "time to double" for any exponential function (or "half life" if k is negative) is a constant so it doesn't matter what t you start with. You chose to start with t= 3 while the "markscheme" started with t= 0. When t= 0, N= 2000 and when the number has doubled, N= 2*2000= 4000. Since N= 2000 when t= 0, the "time to double" is just t so that N(t)= 4000. 4000= 2000e^(kt) so e^(kt)= 2.

You could as easily have said: "Taking T= time to double, N(3)= 18000= 2000e^{3k} and N(T+3)= 36000= 2000e^{k(T+3)}. Dividing the second equation by the first, 2= e^{k(T+3)}/e^{3k}= e^{kT+ 3k- 3k}= e^{kT}, again getting e^{kT)= 2.

 

[nokia8650]

Thank you!