P3 Revision

Hi stuck on tackling these qu's:

dP/dt = ((10 - m)P) / 1000

If the population (t = year, P is population at start of t (year), m = deaths) is to double in 100 years find the value of m?

Explain why the population cannot double in less than 69 years?

Also:

Given P > 1/3, dP/dt = 0.5(3P^2 - P)sin(t), where P if the size of the population in thousands at time t. Given that P = 0.5 when t = 0, show that ln((3P - 1)/P) = 0.5( 1 - cos(t))? Rearrange this equation to show that P = 1/ (3 - e^0.5(1 - cos(t))?

Thus calculate the smallest possible value of t for which P = 1, and find the two values between which number of animals in the population oscillates.

Cheers

Streety

Reply 1

jaffakidds

if dP/dt = Rate of change of population per year =((100-m)P)/1000

then after a bit of algebra
1/p dP = (100-m)/1000 dt

Integrating between p and 2p (initial population and double pupulation) and 100 years and 0 years (initially) you get

ln 2p -lnp = (100-m)/1000 *(100-0)

ln2 = (100-m)/10

10ln2 = 100 - m
m = 100-10 ln 2
:smile:

I hope i havent confused ya too much... try the second bit yourself...i'l post my reply tomorrow (try putting t in as 68..hopefuly summat funny wil happen like a negative m or summat-proving that69 is the minimum)

ill finish the other bit tomorrow as well

Reply 2

streetyfatb

Anyone for the second part plz?

Reply 3

mathz

streetyfatb

Given P > 1/3, dP/dt = 0.5(3P^2 - P)sin(t), where P if the size of the population in thousands at time t. Given that P = 0.5 when t = 0, show that ln((3P - 1)/P) = 0.5( 1 - cos(t))? Rearrange this equation to show that P = 1/ (3 - e^0.5(1 - cos(t))?

using ln((3P - 1)/P) = 0.5( 1 - cos(t)) put t=0 in to get (3p-1/)p=1 and so p=1/2 so the expression satisfies p=1/2 at t=0
differentiate
(p/3p-1)(3/p-(3p-1)/p^2)dp/dt=-0.5sint
(p/3p-1)(3p-3p+1)p^2dp/dt=-0.5sint
(p/3p-1)(1/p^2)dp/dt=-0.5sint
1/p(3p-1)dp/dt=-0.5sint and so dP/dt = 0.5(3P^2 - P)sin(t) as requried

or
dp/dt=0.5(3P^2 - P)sin(t),
so int 1/p(3p-1) dp=int -0.5sint dt
1/p(3p-1)=-1/p+3/(3p-1) so int 1/p(3p-1) dp=ln(3p-1)-lnp=ln((3p-1)/p)
int -0.5sint dt=-0.5cost+Const
so ln((3p-1)/p)=-0.5cost+Const
t=0 at p=1/2 gives c=0.5
so ln((3p-1)/p=0.5( 1 - cos(t)) as required.
raise each side to the power of e
then (3p-1)/p=e^(0.5(1-cost)
p(3-e^(0.5(1-cost))=1
hence p= 1/ (3 - e^0.5(1 - cos(t))?