by natural increase, a city whose population is 40,000 will double in 50 years. there is a net addition of 400 persons per yr because of the people leaving and moving unt the city. estimate the population in 10 yrs. hint: first find the natural growth proportionality factor. Ans: 50, 239 my answer is 49, 965. will post solution below
double time formula: \[\frac{\ln 2}{k} = t\] \[\frac{\ln 2}{t} = k\] \[\frac{\ln 2}{50} = 0.0139\] \[\ln(\frac{x}{x_o}) = kt\] \[\frac{x}{x_o} = e^{kt}\] \[x = x_o e^kt\] \[x = 40,000 e^{0.0139 \times 10}\] \[x = 45, 965\] 400 people pr year so in 10 years there are 4,000 people increase so x = 49, 965
so what did i do wrong
400 * 10 = 4000 this cannot be linear growth it grows naturally after 10 years it sums to : 400e^9k + 400e^8k +..... 400e^k + 400
so im going to solve that one by one?
geometric sequence r = e^-k n = 10
oh lol of course
remind me what the formula is?
nice thanks
why is r = e^-k?
400e^9k + 400e^8k +..... 400e^k + 400 common ratio : 2nd term/1st term
oh of course...
im getting that answer... maybe k has to be more significant.. i mean upto 5 or 6 decimal placces
\[s_{10} = \frac{a_1 (1-r^n)}{1-r} \implies \frac{400(1-e^{-0.0139 \times 10})}{1- e^{-0.0139}}\] is that right?
right
hmmm no wait
no?
a = 400e^9k ...
hmm okay \[\large S_{10} = \frac{400e^{9 \times 0.0139} (1 - e^{-0.0139 \times 10})}{1 - e^{-0.0139}}\] correct?
yh looks correct
i got 50, 226
ok.. put the exact value of k in caculator.. maybe we will get the answer given
k = ln2/50
good luck :)
i tried more exact and i got 50, 210
hmmm i think i give up... not getting the book answer :\
lol...i think it's a matter of significant figures...how many do you think i should use
nope.. i have put k = ln2/50 still not getting the book answer
\[40,000\times 2^{\frac{1}{5}}+400\times 10=49,948\] rounded
@satellite73 yes that is what i got too at first but the answer should be 50,239. any idea how that happened?
no
aww okay :(
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