18.8: Modelos de Crecimiento

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

$P_{1}=P_{0}+15=40+15=55 . P_{2}=55+15=70$
$P_{n}=40+15 n$
$P_{10}=40+15(10)=190$mil dólares
$40+15 n=100$cuando$n=4$ años.

5. Creció 64 en 8 semanas: 8 por semana

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

$P_{0}=200$(mil),$P_{n}=(1+.09) P_{n-1}$ donde$n$ es años después de 2000
$P_{n}=200(1.09)^{n}$
$P_{16}=200(1.09)^{16}=794.061(\text { thousand })=794,061$
$200(1.09)^{n}=400 . \quad n=\log (2) / \log (1.09)=8.043 .$En 2008

9. Let$n=0$ be$1983 . \quad P_{n}=1700(2.9)^{n} . \quad 2005$ es$n=22 . \quad P_{22}=1700(2.9)^{22}=25,304,914,552,324$ la gente. Claramente no es realista, sino matemáticamente exacto.

11. Si n es en horas, mejor comenzar con la forma explícita. $P_{0}=300 . P_{4}=500=300(1+r)^{4}$

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

$P_{0}=300 . \quad P_{n}=(1.136) P_{n-1}$
$P_{n}=300(1.136)^{n}$
$P_{24}=300(1.136)^{24}=6400$bacterias
$300(1.136)^{n}=900 . n=\log (3) / \log (1.136)=$cerca de 8.62 horas

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. Para encontrar la tasa de crecimiento, supongamos que$n=0$ era$1968 .$ Entonces$P_{0}$ sería 1.60 y$P_{8}=2.30=1.60(1+r)^{8}, r=0.0464 .$ como queremos corresponder$n=0$ a 1960, entonces no sabemos$P_{0}$, pero lo$P_{8}$ haríamos$1.60=P_{0}(1.0464)^{8}$. $P_{0}=1.113$

$P_{n}=1.113(1.0464)^{n}$
$P_{0}=\$ 1.113,$o sobre$\$ 1.11$
1996 sería$n=36 . \quad P_{36}=1.113(1.0464)^{36}=\$ 5.697 .$ Real es ligeramente inferior.

17. La población en la localidad era de 4000 en 2005, y está creciendo 4% anual.

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