Confusion about this question (1 Viewer)

[Interdice]

This question is replacement question 14 2012 hsc from the excel 2022, if anyone wants to see.

Question 14(4 marks)
(a) The population of a gold mining town is initially 8000 people. This population would increase at a rate of 1.5% per year, except that there is a steady flow of people leaving the town.

The population P after t years may be modelled by the differential
equation dP/dt = (3P)/(200) - k, where k is the number pf people leaving per year minus the number of people arriving per year.
(i) did that with ease
(ii) For k = 600, determine the number of years until the population is zero.

My solution
(3P)/(200) = 600

Their solution
(3P)/(200) = 0

The constant is stated to be K = 600. It kinda just disappears. Can someone tell me what im missing.

Also it's amazing how much more difficult yr 11 math extension is than yr 12 math advanced lol

 

[Trebla]

Not sure what you're trying to do, but the goal is to find the value of t such that P = 0.

dP/dt is the rate of change of the population. The population itself is represented by P. You need to find the solution to the differential equation to P in terms of t. After that you can them solve for t when P = 0.

 

[Interdice]

Not sure what you're trying to do, but the goal is to find the value of t such that P = 0.

dP/dt is the rate of change of the population. The population itself is represented by P. You need to find the solution to the differential equation to P in terms of t. After that you can them solve for t when P = 0.

Yes. I am trying to find when (3P)/(200) - K = 0

We are finding when (3P)/(200) - K = 0
K = 600
SO (3P)/(200) = -600.
This answer isn't correct.

However the answers completely omit the K value

 

[cossine]

Yes. I am trying to find when (3P)/(200) - K = 0

We are finding when (3P)/(200) - K = 0
K = 600
SO (3P)/(200) = -600.
This answer isn't correct.

However the answers completely omit the K value

Your post is irrelevant. Have you understood the question and trebla's post

 

[chilli 412]

Yes. I am trying to find when (3P)/(200) - K = 0

We are finding when (3P)/(200) - K = 0
K = 600
SO (3P)/(200) = -600.
This answer isn't correct.

However the answers completely omit the K value

if you're finding when (3P)/(200) - K = 0, that means you are finding when dP/dt = 0 (a stationary point on the graph). when the derivative equals 0, it does not necessarily mean that the population is 0.

 

[Interdice]

if you're finding when (3P)/(200) - K = 0, that means you are finding when dP/dt = 0 (a stationary point on the graph). when the derivative equals 0, it does not necessarily mean that the population is 0.

Ahh. I get it now. WE are finding when P = 0 not when DP/DT = 0. I read the question wrong. Thanks

 

[Interdice]

Your post is irrelevant. Have you understood the question and trebla's post

I must have bene half asleep when I asked this question XD. Now I get it