/sci/ - Science & Math » Thread #13611747

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Anonymous Tue 07 Sep 03:09:46 2021 No.13611747 View Reply Original Report

Quoted By: >>13611931 >>13612018

If a city has a starting population of 10,000 and grows by 5 percent a year what function would show that?

Anonymous

Anonymous Tue 07 Sep 2021 03:21:06 No.13611778 Report

Quoted By: >>13611785 >>13611791 >>13612281

10 000(1 + 0.05)^t

Anonymous

Anonymous Tue 07 Sep 2021 03:23:46 No.13611785 Report

Quoted By: >>13611842

>>13611778
Where does the 1 come from?

Anonymous

Anonymous Tue 07 Sep 2021 03:24:45 No.13611791 Report

Quoted By: >>13611855

>>13611778
I want to meet that 1/4 person in the third year of growth.

Anonymous

Anonymous Tue 07 Sep 2021 03:40:29 No.13611842 Report

Quoted By: >>13611856

>>13611785
growth factor.

if the population grows 5% per year, the total (at the end of the year) would be 100% of the base population + 5% of it.

105% is 1.05, hence 1 + 0.05.

Anonymous

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Anonymous Tue 07 Sep 2021 03:42:59 No.13611855 Report

Quoted By: >>13611945

>>13611791

Anonymous

Anonymous Tue 07 Sep 2021 03:43:11 No.13611856 Report

Quoted By:

>>13611842
Thanks that makes sense

Anonymous

Anonymous Tue 07 Sep 2021 04:20:08 No.13611931 Report

Quoted By: >>13611937

>>13611747
10,000+(10,000 x .05y)

Anonymous

Anonymous Tue 07 Sep 2021 04:22:06 No.13611937 Report

Quoted By: >>13611941

>>13611931
this only works for the first year btw, if you want to use it more than once substitute a variable for 10k whose value = current population

Anonymous

Anonymous Tue 07 Sep 2021 04:23:27 No.13611941 Report

Quoted By:

>>13611937
z + z x .05

z = current population

Anonymous

Anonymous Tue 07 Sep 2021 04:25:10 No.13611945 Report

Quoted By:

>>13611855
scientist

Anonymous

Anonymous Tue 07 Sep 2021 04:56:58 No.13612018 Report

Quoted By: >>13612022

>>13611747
$ \\ \text{Continuous compounding} \\ \displaystyle P(t)=P_{0} \, e^{rt} \\ \text{ } P_0 \; \, \text{initial value} \\ \text{ } r \quad \text{rate of growth} \\ \text{ } t \quad \text{time} \\ ~~~~----- \\ P(t_2) = 2P_0 \Rightarrow 2P_0 = P_{0} \, e^{rt_2} \\ 2 = e^{rt_2} \\ e^{ln(2)} = e^{rt_2} \\ ln(2) = rt_2 \\ t_2 = \frac{ln(2)}{r} \approx \frac{70\%}{100r\%} \\ P(t_{10}) = 10P_0 \Rightarrow 10P_0 = P_{0} \, e^{rt_{10}} \\ 10 = e^{rt_{10}} \\ e^{ln(10)} = e^{rt_{10}} \\ ln(10) = rt_{10} \\ t_{10} = \frac{ln(10)}{r} \approx \frac{230\%}{100r\%} $

Anonymous

Anonymous Tue 07 Sep 2021 04:58:34 No.13612022 Report

Quoted By:

>>13612018
just remember 70
https://youtu.be/O133ppiVnWY?t=1m30s

Anonymous

Anonymous Tue 07 Sep 2021 07:07:26 No.13612281 Report

Quoted By:

>>13611778
No reatrd there wont be one only 0.5

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