[Guest]

Here's a seemingly simple problem that can be made rather complex! Well, I did manage to make it very complex before I saw what was wrong, so while some of you would solve it in a matter or minutes, some (like me) might fight with it for some time! And then there would be some, who would think that they know the concept to solve it (but when you solve it, the concepts might go for a toss (my case again)!

So, without more delay, here it is:

At time t=0, the population of a certain town is 100 persons and the population is increasing at the rate of 5 persons per hour.

However, with each increment of the population by 1 person, the rate of increase drops by 0.05

What will be the population of the town after 21 hours?

PS: Sounds like a simple integration type of problem, doesn't it? Enjoy!

Edited September 17, 2009 by DeeGee

[HoustonHokie]

Well, I think I decreased the rate correctly - that's my only question. The first new person comes along after 20 minutes, and then the rate of increase decreases to 4.95 persons/hour, so the next one will come along in 20 minutes 12 seconds. And so on.

If that's the case, then I get 149 people after 21 hours.

Because people are discrete (can't have part of a person), I assumed the rate of change was also discrete. So I might have gotten a different answer than you might have gotten with an integration using a continuously variable rate of change.

[superprismatic]

136. I don't see this as an integration problem because of the discrete language you used in "...with each increment of the population by 1 person, the rate of increase drops by 0.05." So, I take it to mean that after 1/5 hr, the population is 101; after 1/(0.95*5) hr, the population is 102; etc. The pieces of time add up to slightly less then 21 hours after 36 increases. It's just a matter of summing a series from 0 to N and determining the N which gives you a bit under 21 hours and where N+1 would be over 21 hours.

[Guest]

I used a spreadsheet and came up with 167

Edited September 17, 2009 by Carl

[Guest]

Well, both of you are right in saying that this is NOT an integration problem. That is the key to this problem.

Well, I banged my head on this with integration for about an hour and then realised the mistake! But you guys got it right at the start.

However, your answers are not correct!

HoustonHokie, you are probably not at 21 hours yet, and superprismatic, your approach of taking (0.95*5) as the new rate of change is not right.

[Guest]

166

Very Close! Don't round off people

[Guest]

Very Close! Don't round off people

I noticed that! I just changed my answer.

[Guest]

171

Edited September 17, 2009 by fredcadet

[superprismatic]

Well, both of you are right in saying that this is NOT an integration problem. That is the key to this problem.

Well, I banged my head on this with integration for about an hour and then realised the mistake! But you guys got it right at the start.

However, your answers are not correct!

HoustonHokie, you are probably not at 21 hours yet, and superprismatic, your approach of taking (0.95*5) as the new rate of change is not right.

It's 165!

[Guest]

I used integration, and it all checks out, I think.

First, dP/dt = 5-.05(P-100)

I know I am bootstrapping with the initial conditions, and if I made a mistake, its here, but on we go.

dP/dt = 5 - .05P + 4

dP/dt = 9 - .05P

dP/dt + .05P = 9

Use of integration variable

U = e^(.05t)

dU/dt = .05e^(.05t) = .05*U

Multiply everything in the previous equation by U

U*dP/dt + .05*U*P = 9*U

U*dP/dt + P*dU/dt = 9*U

Using chain rule in reverse

d/dx(U*P) = 9*U

Integrate

U*P = 9*int(e^(.05t))+C

U*P = 9*20e^(.05t) + C

U*P = 180e^(.05t) + C (divide by U)

P = 180 + Ce^(-.05t) (use initial conditions)

100 = 180 + C

C = -80

P = 180 - 80e^(-.05t)

This checks out with the initial conditions and rate equation, and when I plug in 21, I get

152.005, or 152 people.

Edited September 17, 2009 by brescher

[Guest]

171

hour #people #people to add

0 100

1 105

2 110

3 115

4 120 20x.05 =1 so rate now add 4

5 124 .075

6 128 1.4

7 132 1.6

8 136 1.8

9 140 2 now add 3

10 143 2.15

11 146 2.3

12 149 2.45

13 152 2.6

14 155 2.75

15 158 2.9

16 161 3.05 so now only add 2

17 163 3.15

18 165 3.25

19 167 3.35

20 169 3.45

21 171 3.55

Edited September 17, 2009 by fredcadet

[Guest]

165 people, after 20.812 hours

200 people, after 103.75 hours, beyond which the population doesn't grow.

Edited September 17, 2009 by ogden_tbsa

[Guest]

165 people at 1248 minutes

[Guest]

I took it to mean that the rate of increase decreased by 5% each time a person was born, which if that was the case I agree with Superprismatic's first guess of 136:

0.20/(0.95^0) + 0.20/(0.95^1) + 0.20/(0.95^2) + 0.20/(0.95^3) + ... + 0.20/(0.95^[x-1]) = 21 (where x is the number of people born in those 21 hours)

Which becomes:

0.20/(0.95^[x-1+1]) - 0.20 = 21/0.95 - 21

0.95^x = 0.95/6.2

x = ln(0.95/6.2)/ln(0.95)

x = 36.57... (Round down to 36)

so 136 people.

But evidently the OP meant this:

1/[5-0(0.05)] + 1/[5-1(0.05)] + 1/[5-2(0.05)] + 1/[5-3(0.05)] + ... + 1/[5-(x-1)(0.05)] = 21 (where x is the number of people born in those 21 hours)

My question is, is there a way to simplify and solve for x here like above or do we have to use a spreadsheet or something similar and brute force it?

[Guest]

167 people.

Hour 1 - new person born every 12 minutes

Hour 2 - new person born every 12.6 minutes

Hour 3 - new person born every 13.23 minutes

etc...

Hour 21 - new person born every 31.83 minutes

At hour 21 67.31 people have been added but you can't have 0.31 of a person so 67 perople have been added. 167 is the answer.

[Guest]

165 it is!

[Guest]

The population at 21 hours is 165 persons since the last one arrived at 20h48m43s. The next will not arrive for 23 minute more. The final population of this progression will be 200 persons at 4d7h44m51s at which the rate will become zero and as such no more new arrivals, therefore the rate will not change from zero.