Question relating to exponential growth (1 Viewer)

[physicss]

<!--[if gte vml 1]><v:shapetype id="_x0000_t75" coordsize="21600,21600" o:spt="75" oreferrelative="t" path="m@4@5l@4@11@9@11@9@5xe" filled="f" stroked="f"> <v:stroke joinstyle="miter"/> <v:formulas> <v:f eqn="if lineDrawn pixelLineWidth 0"/> <v:f eqn="sum @0 1 0"/> <v:f eqn="sum 0 0 @1"/> <v:f eqn="prod @2 1 2"/> <v:f eqn="prod @3 21600 pixelWidth"/> <v:f eqn="prod @3 21600 pixelHeight"/> <v:f eqn="sum @0 0 1"/> <v:f eqn="prod @6 1 2"/> <v:f eqn="prod @7 21600 pixelWidth"/> <v:f eqn="sum @8 21600 0"/> <v:f eqn="prod @7 21600 pixelHeight"/> <v:f eqn="sum @10 21600 0"/> </v:formulas> <vath o:extrusionok="f" gradientshapeok="t" o:connecttype="rect"/> <o:lock v:ext="edit" aspectratio="t"/> </v:shapetype><v:shape id="_x0000_i1025" type="#_x0000_t75" style='width:468pt; height:30pt'> <v:imagedata src="file:///C:\DOCUME~1\Admins\LOCALS~1\Temp\msohtmlclip1\01\clip_image001.png" o:title="" chromakey="white"/> </v:shape><![endif]--><!--[if !vml]-->A population N(t) varies with time according to the law N(t)=C(e^(kt)) where C,k are constants and t>= 0
i) Show that, if a and b are two positive numbers such that a+b=1, then:
N(at+bu)=[[N(t)]^a][[N(u)]^b] for any t>=0,u>=0
ii) Hence or otherwise, find N(13), given that N(3)=10 and N(18)=100.
I am stuck on part ii)

 

[random-1006]

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i) Show that, if a and b are two positive numbers such that a+b=1, then:
N(at+bu)=[[N(t)]^a][[N(u)]^b] for any t>=0,u>=0
ii) Hence or otherwise, find N(13), given that N(3)=10 and N(18)=100.
I am stuck on part ii)

take the time when t=0 and u=0 ( nah, i think this is wrong, i thought u was a time also, ill look at it bit more, but i think it will very similar, it will be like simultaneous eqns with three unknowns )

thus N(0)= N( a(0) +b(0) ) = [N(0)]^a [N(0) ] ^b = C^2

N(3)=10, N(18)= 100,



im thinking simulataneous eqns, ill look at it bit more

 

[random-1006]

but N( 0) also = C

therefore C^2=C
C(C-1)=0, C is not equal to zero ( or else we would have N(t)=0), therefore C=1, etc

you can find k from one of the two given conditions

think thats it

 

[random-1006]

ohhh man,

ive got it down to ln(C)=C

and you can find k from the two eqns, i will post up my working

i have a mental blank, how do u solve ln(C)=C lol


that means C=e^C, thats not right :S

ok so take time t=0

N(bu)= [N(0)]^a [N(u)]^b ( part 1)

therefore [N(0)]^a= N(bu) / (N(u))^b

a= 1/ [N(0)] ln { N(bu) / [N(u)]^b }

now N(bu) = C e^(buk)
N(u)= Ce^ku

[N(u)]^b = (C^b) e^(buk)

therefore a= 1 / C ln ( C ^ (1-b) )
a = [(1-b)/ C] ln(C) by log laws, but 1-b =a

therefore ln(C)/ C =1,

ln(C)=C { lol, googled it and you are meant to use some lambert function or something to solve it, god thats screwed up, never herd of that, i wouldnt worry bout it }.

we have two other eqns lets use them, solve them simulatenously , answer should pop out

 

[physicss]

The answer given is 15, but I'm still not getting it :S

 

[random-1006]

see editted post above

 

[random-1006]

N(18) = [N(3)]^2 is no help, because a+b should=1

 

[random-1006]

got it, from two conditions we have 10 =Ce^(3k) , 100= Ce^(18k) and ln(C) =C

rearrange first for C and sub into third

therefore

ln(c) = 10e^(-3k)
c = e^[ 10e^(-3k)]

we also have k= ln(10) / 15 = 0.1535... ( by dividing second eqn by the first)

therefore c= 544

that doesnt give the answer :|, answer must be wrong lol

i didnt do all that working to be wrong, and it all looks correct :S

 

[physicss]

Oops sorry answer is actually 10^(5/3)

EDIT: nvm possibly a wrong answer since it's given by tutor. Thanks for the working out

 

[hscishard]

I got 10^(5/3)

That was a good question. It was more algebra/function than natural growth.

 

[hscishard]

Oops sorry answer is actually 10^(5/3)

EDIT: nvm possibly a wrong answer since it's given by tutor. Thanks for the working out

:S I somehow got 10^(5/3)

 

[physicss]

how did u do it lol