# CALCULUS

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9. Suppose an experimental population of amoeba increases according to the law of exponential growth. There were 100 amoeba after the second day of the experiment and 300 amoeba after the fourth day. Approximately how many amoeba were in the original sample?

A. 5
B. 33
C. 71
D. 10
E. Not enough information to determine

I posted this earlier, but I didn't really get a straightforward answer. I am confused as to how to set up the differential equation to represent this problem. Please help me make sense of this in a simple fashion!

• CALCULUS -

Write down the general equation for exponential growth:

N(t) = A exp(b t)

You see that there are two unknown parameters here, and you would need two data points to solve for them. You are free to chose whatever units you want to use for the time, a different choice will be compensated for by the parameter b.

Then if we take t to be the time after t days, we have:

N(0) = A

N(2) = 100

N(4) = 300

Then the equation for N(t) implies that:

N(4)/N(2) = exp(2 b)

And from the given data points, we see that this is 3. Then we can solve for A, using e.g. the data point N(2)=100:

According to the equation for N(t):

N(2) = A exp(2 b)

We know that exp(2 b) = 3, so:

N(2) = 3 A

Equating this to 100 gives:

A = 33

N(0) is equal to A, so in the original sample there were 33 amoeba.

• CALCULUS -

Thank you so much. The last person just kind of put the equation out there without really explaining it, which just confused me even more. You really simplified this out for me, thank you!

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