18.8: Growth Models
1.
- \(P_{0}=20 . P_{n}=P_{n-1}+5\)
- \(P_{n}=20+5 n\)
3.
- \(P_{1}=P_{0}+15=40+15=55 . P_{2}=55+15=70\)
- \(P_{n}=40+15 n\)
- \(P_{10}=40+15(10)=190\) 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
7.
- \(P_{0}=200\) (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.