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!
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.
The answer of 33 was correct.
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!
Ah, the wonderful world of amoeba population growth! Let me try to help you with this in a lighthearted and simplified way.
So, you know that the amoeba population is increasing according to the law of exponential growth. And you're given that there were 100 amoeba after the second day and 300 amoeba after the fourth day. Great!
To get this clown party started, let's assume that the original sample size was "x" amoeba. We're trying to figure out what "x" is.
Since we're dealing with exponential growth, we can use the formula: P(t) = P0 * e^(kt), where P(t) is the population at time "t", P0 is the initial population, "e" is Euler's number (approximately 2.71828), and "k" is the growth constant.
Now, let's plug in the given values. After the second day (t = 2), the population is 100, so we have the equation 100 = x * e^(2k). After the fourth day (t = 4), the population is 300, so we have the equation 300 = x * e^(4k).
Now, we have two equations and two unknowns (x and k). To solve this, we can divide the second equation by the first to eliminate x: 300/100 = (x * e^(4k))/(x * e^(2k)). Simplifying, we get 3 = e^(2k).
Taking the natural logarithm of both sides, ln(3) = 2k. Finally, we can solve for k: k = ln(3)/2.
Now that we know "k", we can substitute it back into either of the original equations to solve for "x". Let's use the first one: 100 = x * e^(2 * ln(3)/2).
Simplifying that equation, 100 = x * (e^ln(3))^2/2 = x * 3.
Dividing both sides by 3, we find x = 100/3, which is approximately 33.33.
So, my dear friend, the approximate number of amoeba in the original sample is approximately 33. And that, my friend, is no small number of amoeba! I hope this has brought some clownish clarity to your problem!
To determine the approximate amoeba population in the original sample, we need to set up the differential equation for exponential growth.
The differential equation for exponential growth is given by:
dP/dt = kP
Where:
- dP/dt represents the rate of change of the population over time.
- P represents the population size.
- k represents the growth rate constant.
Now, let's start by using the given information:
After the second day, there were 100 amoeba.
After the fourth day, there were 300 amoeba.
We can use these data points to find the growth rate constant (k).
Let P2 be the population size after the second day (100) and P4 be the population size after the fourth day (300).
We can use the formula for exponential growth to relate these two data points:
P4 = P2 * e^(kt)
Substituting the given values:
300 = 100 * e^(k * 4)
Simplifying, we get:
3 = e^(4k)
Now, we need to take the natural logarithm of both sides to solve for k:
ln(3) = ln(e^(4k))
ln(3) = 4k
Next, divide both sides by 4 to isolate k:
k = ln(3)/4
Now that we have the growth rate constant (k), we can use it to solve for the original population size (P0). The formula for exponential growth is:
P = P0 * e^(kt)
Letting P0 represent the original population size, P2 represent the population size after the second day, and k represent the growth rate constant, we have:
P2 = P0 * e^(2k)
Substituting the known values:
100 = P0 * e^(2 * ln(3)/4)
Simplifying, we get:
100 = P0 * e^(ln(3)/2)
Using the property of logarithms, we know that e^(ln(3)/2) is equal to the square root of e:
100 = P0 * sqrt(e)
To solve for P0, divide both sides by the square root of e:
P0 = 100 / sqrt(e)
Now, we can approximate the value using a calculator:
P0 ≈ 100 / 1.6487 ≈ 60.717
Therefore, approximately 60.717 amoeba were in the original sample.
Since the answer choices are whole numbers, we round our answer to the nearest whole number, giving us:
Approximately 61 amoeba were in the original sample.
Thus, the answer is not given in the options provided.