18.8: Growth Models
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1.
- P0=20.Pn=Pn−1+5
- Pn=20+5n
3.
- P1=P0+15=40+15=55.P2=55+15=70
- Pn=40+15n
- P10=40+15(10)=190 thousand dollars
- 40+15n=100 when n=4 years.
5. Grew 64 in 8 weeks: 8 per week
- Pn=3+8n
- 187=3+8n.n=23 weeks
7.
- P0=200 (thousand), Pn=(1+.09)Pn−1 where n is years after 2000
- Pn=200(1.09)n
- P16=200(1.09)16=794.061( thousand )=794,061
- 200(1.09)n=400.n=log(2)/log(1.09)=8.043. In 2008
9. Let n=0 be 1983.Pn=1700(2.9)n.2005 is n=22.P22=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. P0=300.P4=500=300(1+r)4
500/300=(1+r)4.1+r=1.136.r=0.136
- P0=300.Pn=(1.136)Pn−1
- Pn=300(1.136)n
- P24=300(1.136)24=6400 bacteria
- 300(1.136)n=900.n=log(3)/log(1.136)= about 8.62 hours
13.
- P0=100Pn=Pn−1+0.70(1−Pn−1/2000)Pn−1
- P1=100+0.70(1−100/2000)(100)=166.5
- P2=166.5+0.70(1−166.5/2000)(166.5)=273.3
15. To find the growth rate, suppose n=0 was 1968. Then P0 would be 1.60 and P8=2.30=1.60(1+r)8,r=0.0464. since we want n=0 to correspond to 1960 , then we don't know P0, but P8 would 1.60=P0(1.0464)8. P0=1.113
- Pn=1.113(1.0464)n
- P0=$1.113, or about $1.11
- 1996 would be n=36.P36=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.