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LOI'Work: Practice Problems #1-21 A population of mosquitoes is increasing by 7.0% per day: How long will it take the population to double?2 A population of co...

Question

LOI'Work: Practice Problems #1-21 A population of mosquitoes is increasing by 7.0% per day: How long will it take the population to double?2 A population of corn borers in a large corn field is growing exponentially and has a growth rate (r) of 1.5 per week. When there are 200 corn borers, how fast (in corn borers per week) is the population growing (dN/dt)?

LOI' Work: Practice Problems #1-2 1 A population of mosquitoes is increasing by 7.0% per day: How long will it take the population to double? 2 A population of corn borers in a large corn field is growing exponentially and has a growth rate (r) of 1.5 per week. When there are 200 corn borers, how fast (in corn borers per week) is the population growing (dN/dt)?



Answers

The size $P$ of a certain insect population at time $t$ (in days) obeys the law of uninhibited growth $P(t)=500 e^{0.02 t}$ (a) Determine the number of insects at $t=0$ days. (b) What is the growth rate of the insect population? (c) What is the population after 10 days? (d) When will the insect population reach $800 ?$ (e) When will the insect population double?

So in this question, we want to know how long it's gonna take for a population to double. And this is our equation. To figure that out. The rate in this problem is 12%. So this will be Ellen of 2/0 120.12 and in her calculator that will then give us a doubling time of 5.8 years.

Okay, so we're going to start off with p groups. P 18 equals 500 e 5000.2 teeth. So the T equals time. Okay, So what is asking for of the number of insects at zero days? So we're gonna do is we're gonna plug in zero to t. Thank you. So 7500 e to the 0.2 times zero, which will equal 506 uh, zero days. Okay, Now be B is asking for the girl three. So the girl three equals K. So if you replace the point several, too with the K K equals 0.2 And if you move the decimal two to the right equals 2%. So the girl three, it's 2%. No. All right. See, see, is a population after 10 dates. So this is similar to a We're going to plug in 10 days to t still be 500 e to the twin 02 times 10. And if you put that in the calculator, that should give you 610.70 in sex at 10 days, Right? Not be de is asking for the population that one with the population you take 100 sex. So for this one, we're trying to find a T. How long will it take to get it here? So eight hunger equals P 32 So 800 equals 500 E prime zero to team. So we're going to divide out 500. You okay? You good? 1.6 cause into the prime zero to t. So now we're going to get the hell and bring a little 16 What the Ellen does to a that council That out. So off 0.0 to t. We're going to divide out the 0.2 you left with t equals to e cools 20 dream print. Five days to reach 100 in six. Okay, My last one. E Try and see when the population doubles. So we started off at 500 when you double it. You 2000. So 1000 maybe the p t. So some color too de we'll get 1000 equals 500 e to the 0.0 to t. So they were trying to find team who were going to divide out 500. You kept too cozy. Clean zero to t Did you get the Poland my life too. Oh, cancel up to even get from zero to t delight out 0.2 Your T equals 34.626 days carriage 2000 and six.

This question. We want to know how long it will take the fish population to double. It tells us the rate at which these fish are this population is growing is 25%. So this will be Ellen too. Over 0.25 And so how long will it take? This fish population to double is going to be 2.8 years.

Missing area. We have a colony of mosquitoes with uninhibited growth, so they're allowed to increase in size the population for as long as possible, as long as on the problem says you can't. So, in part, we were asked to form create a function, um, expressing end with teeth where n is the population, the colony. And he is the time, uh, in days so we can just use the logistic of formula for natural growth for population. So a off T to the population is equal to, uh, the initial amounts of the population a population times E, which is a number race to the K times T power. Where K is the rate of growth, he is the rate of growth or decay or, in this case, growth because the colonies growing and, um, t is time and a zero is starting population. So this is the answer to part A and for part B were asked, were asked how long they were given that there are 1000 mosquitoes as a zero a zero So initial population and after one the day there is 18 1800. So if 18 to 1000 mosquitoes 1800 mosquitoes after one day. So if zeros equals 1000 and we can substitute time T for one So we just are left with a K. And now we can actually just solve this out because you want to try to find the rates were not given the right where we have to find it with this information. And then we're gonna apply the right to a different problem. So ah, we can divide both sides by 100 and we actually 1800 divided by 100. Sequel to 18 and it's equal to eat. OK, we can take the Ellen of both sides. So Ellen of U two cages, Lisa's K and L of 18 is equal to approximately 2.89 2.89 So K is equal to 2.89 and, um, this actually weaken apply in the scenario that we're asked to where we're ask defying the size of the colony after three days so we can use K here we get a of T, which is what we're solving for is equal to starting population, which is still 1000. Oh, sorry. I actually forgot 1000 over here, so uh, going back. We This must be 1.8. So l and of 1.8 l out of 1.8 is 0.5878 0.5878 and we have the same value here. So one A of t is equal to 1000 times E to the race of the 0.5 878 times, three days. So two times three. And you actually put this value into a calculator and we get Ah, so we get we multiply 1000 by e race of 5878 power and multiply the expanded by another three and we get a total of 5832.23 0.23 So after three days, after three days of colony will be about this size and weaken ground up to the nearest whole number. Things were asking for a number of mosquitoes, so you can just leave it as 5832 and were also asked to solve apart, see where were asked to find how long until they're what the population is equal to. 10,000 mosquitoes. So 10,000 is any of t and we're gonna set it equal to 1000 starting population time. See to the K east of the case, zero point five feet 78 And he So it's on for tea. Over here, we divide both sides by 1000. We got 10 is equal to E to the 0.5878 Who on now we take the hell out of both sides. So this on the right side, we're just left with the destiny. Is there a point? The power, Their 0.0.5878 t and on the right side, we have l and R left that we have Ella enough tea to take Ellen of natural log of both sides. And we can divide both sides by 0.5878 now. So l end of 10 divided by 0.5878 is approximately equal to 3.917 So this is the amount of days, and we can actually just round this up to four days until the population reaches 10,000 mosquitoes


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