Population Growth question

  • Thread starter Slimsta
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  • #1
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Homework Statement


http://img14.imageshack.us/img14/1286/99892583.jpg [Broken]


Homework Equations


in the picture


The Attempt at a Solution


so i know that
p(0)e^2k = 950
p(0)e^8k = 9500

how do i find k?
for p(0) i get 440.950939 is that right?

btw. ignore the answers in the picture
 
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Answers and Replies

  • #2
lanedance
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how do you find p(0) without knowing k?

try dividing you 2 equations together, then use logs
 
  • #3
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how do you find p(0) without knowing k?

try dividing you 2 equations together, then use logs

what i did is made both equations equal to k
so
k1 = [ln(950/p(0))] / 2
k2 = [ln(9500/p(0))] / 8

k1 = k2
[ln(950/p(0))] / 2 = [ln(9500/p(0))] / 8
solving for p(0) i get 440.9509 which lets me plug it in the main equation and get the k value..
but looks like its wrong eh?
 
  • #4
lanedance
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so dividing the 2 equations gives
[tex] e^{6k} = 10 [/tex]

can you solve for k?

you should get some constant a such that
[tex]k = \frac{1}{a} ln(10) [/tex]

the equation then becomes
[tex]p(t) = p(0).e^{kt} = p(0).e^{(t/a) ln(10)} = p(0).e^{ln(10^{t/a}} = p(0) .10^{t/a}[/tex]
[tex]p(0) = p(2).10^{-2/a}[/tex]

which gives me the same p(0) value as you, so your way was fine
 
  • #5
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so dividing the 2 equations gives
[tex] e^{6k} = 10 [/tex]

can you solve for k?

you should get some constant a such that
[tex]k = \frac{1}{a} ln(10) [/tex]

the equation then becomes
[tex]p(t) = p(0).e^{kt} = p(0).e^{(t/a) ln(10)} = p(0).e^{ln(10^{t/a}} = p(0) .10^{t/a}[/tex]
[tex]p(0) = p(2).10^{-2/a}[/tex]

which gives me the same p(0) value as you, so your way was fine

oh so i just did it the long way... kk that makes sense.
now how do i find the growth rate after 5 hours?
k = 0.38376418 so growth rate = 38.376%
 
  • #6
lanedance
Homework Helper
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i would take the growth rate to mean p'(t)
 
  • #7
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i would take the growth rate to mean p'(t)

oh i got it!! tnx man
 
Last edited:

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