18.8: Growth Models

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$P_{0}=20 . P_{n}=P_{n-1}+5$
$P_{n}=20+5 n$

$P_{1}=P_{0}+15=40+15=55 . P_{2}=55+15=70$
$P_{n}=40+15 n$
$P_{10}=40+15(10)=190$ thousand dollars
$40+15 n=100$ when $n=4$ years.

5. Grew 64 in 8 weeks: 8 per week

$P_{n}=3+8 n$
$187=3+8 n . n=23$ weeks

$P_{0}=200$ (thousand), $P_{n}=(1+.09) P_{n-1}$ where $n$ is years after 2000
$P_{n}=200(1.09)^{n}$
$P_{16}=200(1.09)^{16}=794.061(\text { thousand })=794,061$
$200(1.09)^{n}=400 . \quad n=\log (2) / \log (1.09)=8.043 .$ In 2008

9. Let $n=0$ be $1983 . \quad P_{n}=1700(2.9)^{n} . \quad 2005$ is $n=22 . \quad P_{22}=1700(2.9)^{22}=25,304,914,552,324$ people. Clearly not realistic, but mathematically accurate.

11. If n is in hours, better to start with the explicit form. $P_{0}=300 . P_{4}=500=300(1+r)^{4}$

$500 / 300=(1+r)^{4} . \quad 1+r=1.136 . \quad r=0.136$

$P_{0}=300 . \quad P_{n}=(1.136) P_{n-1}$
$P_{n}=300(1.136)^{n}$
$P_{24}=300(1.136)^{24}=6400$ bacteria
$300(1.136)^{n}=900 . n=\log (3) / \log (1.136)=$ about 8.62 hours

13.

$P_{0}=100 \quad P_{n}=P_{n-1}+0.70\left(1-P_{n-1} / 2000\right) P_{n-1}$
$P_{1}=100+0.70(1-100 / 2000)(100)=166.5$
$P_{2}=166.5+0.70(1-166.5 / 2000)(166.5)=273.3$

15. To find the growth rate, suppose $n=0$ was $1968 .$ Then $P_{0}$ would be 1.60 and $P_{8}=2.30=1.60(1+r)^{8}, r=0.0464 .$ since we want $n=0$ to correspond to 1960 , then we don't know $P_{0}$, but $P_{8}$ would $1.60=P_{0}(1.0464)^{8}$. $P_{0}=1.113$

$P_{n}=1.113(1.0464)^{n}$
$P_{0}=\$ 1.113,$ or about $\$ 1.11$
1996 would be $n=36 . \quad P_{36}=1.113(1.0464)^{36}=\$ 5.697 .$ Actual is slightly lower.

17. The population in the town was 4000 in 2005, and is growing by 4% per year.

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