A cage can hold up to 1000 birds, and 200 birds are initially moved into the cage. Suppose that the population of birds grows according to the logistic model. If after two months there are 400 birds in the cage, after how many months will the population reach 800 birds? Give your answer to the nearest month.
This question concerns the rate of growth of a population, so we’re going to answer it using differential equations. We’re not just going to use the model for simple population growth. We’re told in the question that we need to use the logistic model. Now, the reason for this is apparent in the question. We’re told that the cage can hold up to 1000 birds. There is a finite number of birds that this cage can support.
The logistic model takes account of this by incorporating a maximum carrying capacity for the population. Whereas the simple model for population growth assumes that the population can continue to grow at an exponential rate. The logistic model is as follows. It tells us the rate of change of the population d𝑝 by d𝑡 is equal to 𝑘𝑃 multiplied by one minus 𝑃 over 𝐿. Here, 𝑘 is the growth rate of the population and 𝐿 is the maximum population that can be supported, the carrying capacity, which in this question is 1000.
We can quote the standard solution to this differential equation. The population as a function of the time 𝑡 is equal to 𝐿 over one plus 𝐴𝑒 to the negative 𝑘𝑡, where 𝐴 is equal to 𝐿 minus 𝑃 nought over 𝑃 nought. And here, 𝑃 nought represents the initial population. Let’s think about what we know for this problem then.
The cage can hold up to 1000 birds, so the carrying capacity for this problem or 𝐿 is equal to 1000. We also know that 200 birds are initially moved into the cage. So, our value, 𝑃 nought, is equal to 200. We can therefore work out the value of the constant 𝐴. It’s equal to 𝐿 minus 𝑃 nought over 𝑃 nought. So, 1000 minus 200 over 200 or 800 over 200, which is equal to four.
Notice that we haven’t explicitly been told the growth rate in this problem. So, the value 𝑘 remains an unknown constant for now. But substituting 𝐿 equals 1000 and 𝐴 equals four, we have that 𝑃 is equal to 1000 over one plus four 𝑒 to the negative 𝑘𝑡. We are given another key piece of information in the question, which is that after two months, there are 400 birds in the cage. So, we know that when 𝑡 is equal to two, 𝑃 is equal to 400. And so, we have a pair of values for 𝑡 and 𝑃 that we can use to determine this constant 𝑘.
Substituting 400 for 𝑃 and two for 𝑡 gives the equation 400 equals 1000 over one plus four 𝑒 to the negative two 𝑘. Multiplying both sides by one plus four 𝑒 to the negative two 𝑘 and then dividing by 400 gives one plus four 𝑒 to the negative two 𝑘 equals 1000 over 400. Which simplifies to 10 over four or five over two, which is 2.5. We can then subtract one from each side and divide by four to give 𝑒 to the power of negative two 𝑘 is equal to one-quarter of 2.5 minus one, which is three-eighths or 0.375.
To continue solving for 𝑘, we take the natural logarithm of each side of the equation, giving the natural logarithm of 𝑒 to the power of negative two 𝑘 equals the natural logarithm of 0.375. But of course, raising 𝑒 to a power and then taking the natural logarithm of this are inverses of one another. So, the left-hand side just simplifies to negative two 𝑘. Finally, we divide both sides of the equation by negative two, giving 𝑘 equals negative a half the natural logarithm of 0.375 or 0.490 as a decimal to three significant figures. So, we’ve determined the growth rate, 𝑘.
What the question wants to know, though, is how long it will take or after how many months the population of birds in the cage will be 800. So, we need to substitute 800 for the population, 𝑃, and solve the resulting equation to find 𝑡. We substitute 800 for the population, 𝑃, then and the value of the constant 𝑘 we’ve just found. You can either use the exact value or the three-significant-figure value to give 800 equals 1000 over one plus four 𝑒 to the negative 0.490𝑡. And now, we solve in a way very similar to when we solved for 𝑘.
First, we multiply by one plus four 𝑒 to the negative 0.490𝑡 and also divide by 800. 1000 over 800 simplifies to five over four or 1.25. We subtract one and divide by four to give 𝑒 to the power of negative 0.490𝑡 is equal to one sixteenth. Next, we’ll take the natural logarithm of each side of this equation, knowing that it will cancel out the exponential on the left-hand side. Leaving us with negative 0.490𝑡 equals the natural logarithm of one over 16. Finally, we divide both sides of the equation by negative 0.490. Or, we can use the exact value of negative one-half times the natural logarithm of 0.375 in place of the value of 0.490.
If you do that, though, remember 𝑘 is actually a positive value despite appearing negative when we express it as negative a half the natural logarithm of 0.375. So, if you’re substituting 𝑘 in its exact form, it will be one over negative negative a half times the natural logarithm of 0.375. Evaluating on a calculator will give 𝑡 equals 5.658 if using 0.490 for 𝑘, or 5.653 if using the exact value. In either case, rounding to the nearest month will give an answer of six months.
So, by recalling the standard solution to the logistic model for population growth, we were able to find the value of the constant 𝐴. Then use the given information about the population at a time 𝑡 equals two to find the value of our constant 𝑘, the growth rate. And finally substitute a population of 800 as well as unknown constants to find the value of 𝑡 at which this population occurred. And we found that to the nearest month, the population will reach 800 birds after six months.