Natural growth & decay question (1 Viewer)

[rsingh]

Hey guys, this question is from last year's Catholic paper.

After t years (t greater than or equal to 0), the number N of individuals in a population is given by N=A +Be^-0.5t for some constants A>0 and B>0. The initial population size is 500 individuals and the limiting population is 100 individuals.

(i) Find the values of A and B.
(ii) Find the time taken for the population size to fall within 10 of its limiting value, giving the answer correct to the nearest month.

I can do most questions like this, howver this is one question that has got me stumped?

 

[acmilan]

N=A +Be^-0.5t

(i)
when t -> infinity, N = 100

100 = A + Be^-infinity

100 = A + B.0 (e^-infinity = 1/e^infinity = 1/infinity = 0)
A = 100

N = 100 + Be^-0.5t

When t = 0, N = 500

500 = 100 + Be⁰
B = 400

(ii)

Within 10 of limiting value would be 110.

When N = 110,
110 = 100 + 400e^-0.5t
e^-0.5t = 1/40

-0.5t = ln(1/40)
t = ? (havent got calculator around)

 

[rsingh]

Thanks for your help.
I get it now!
Just how is this different from the 2u ones? I heard something about the 3u questions having a "limiting value" and I'm not quite sure what that means?

 

[FinalFantasy]

Thanks for your help.
I get it now!
Just how is this different from the 2u ones? I heard something about the 3u questions having a "limiting value" and I'm not quite sure what that means?

do u mean the t--> inifinity thing?
maybe that's wat u mean by a "limiting value"

 

[acmilan]

No the limiting value is common in 2unit also, perhaps not in the exact same context though. Put simply, the difference between this and 2 unit is the 'A' in the equation, which means as time goes on, the value of N will eventually reach A if the equation involves a negative power of the exponential term. In 2 unit you would most likely get N = Be^-kt

 

[Abtari]

but doesnt the question in part b say "fall" within 10 of its limiting value --> shouldn't that mean the value of N would be substituted as 90?

i did that and got log negative something which is obviously a maths error. i see what uve done but just wasnt sure if that was what the q asked...

 

[SaHbEeWaH]

the initial population was 500.
it then 'falls' to the limiting population of 100.
for it to fall within 10 of the limiting population, the population would be 110.

if the population were intially 2 and the limiting population were 100
then it would ask for the time taken for the population to rise within 10 of its population which would be 90...

 

[acmilan]

Here's a graph of the function N = 1 + 4e^-0.5t, which is basically the same thing except the N-axis is scaled down so 1 represents 100, 2 represents 200,... 5 represents 500 etc.

As you can see, it never goes below 100 and hence to be within 10 of 100 it has to do so from above 100, not below

 

[JamiL]

yer preety much wot acmilan said, the difference is that in 3u they have a constant added. in 2u when dealing with decay there is also a limit, and that is at zero.
2u Qu is
Q=Ae^(-kt), at t=infinity, e^-kt=0, therefore Q=0 and that is the limit... in 3u they have as seen Q=B+Ae^-kt, and hence the limit is B. they are basicly the same, except when they ask 3u ones, they dont make it has easy as in 2u, and they try and trick u a lil... u just have 2 be carful