What is the Growth Rate Constant for Population Growth?

[Slimsta]

Homework Statement

http://img14.imageshack.us/img14/1286/99892583.jpg

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

 

[lanedance]

how do you find p(0) without knowing k?

try dividing you 2 equations together, then use logs

 

[Slimsta]

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 let's me plug it in the main equation and get the k value..
but looks like its wrong eh?

 

[lanedance]

so dividing the 2 equations gives
e^{6k} = 10

can you solve for k?

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

the equation then becomes
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}
p(0) = p(2).10^{-2/a}

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

 

[Slimsta]

so dividing the 2 equations gives
e^{6k} = 10

can you solve for k?

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

the equation then becomes
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}
p(0) = p(2).10^{-2/a}

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%

 

[lanedance]

i would take the growth rate to mean p'(t)

 

[Slimsta]

i would take the growth rate to mean p'(t)

oh i got it! tnx man