18. How Many Deer Are There? The number of deer in a state park was 1,000 six ye
ID: 3142586 • Letter: 1
Question
18. How Many Deer Are There? The number of deer in a state park was 1,000 six years ago and today it is 2,000. Based upon data from other habitats, the park service estimates that no more than 10,000 deer can inhabit the park. (If there are more than 10,000 deer in the park, overgrazing may change the character of the park and/or large numbers of deer may starve in the winter.) Your group has been hired to assist the park service in predicting both long and short term population trends. Part 1 Your first task is to investigate a family of possible models for the population of the deer. Call P(t) the number of deer in the park t years after some starting time. Your group should decide what the starting time is and use it throughout your report. The group must also agree on each member's number since this will be needed to investigate the models Your report should include the following information for each model . A verbal description 2. A differential equation including initial conditions. 3·The deer population 1 year from today, the deer population 5 years from today and the deer population 20 years 29dco16746446151d5bo3b 1950 el from today 4. A general rule for forecasting the popu 5. The time (if ever) when the population will reach 10,000. 6. If the deer population follows this model any problems or advantages for the park service lation of the deer for any time in the future Model 1 a. Find the constant rate of change for the deer population which gives a solution that fits the two given population counts exactly. Use this model for the population from 6 years ago until today b. Using only the information that the population today is 2,000 and that the change in the population of deer is (member#) x 100 deer per year every year from now on, predict the population starting today. So you should end up with a piecewise function for the deer population starting 6 years ago. Altermative Models Each member o can be described by the differential equation dP/dt = k pa or P' = k Pa Member 1 will investigate u = .2, 1.2 Member 2 will investigated= .3.1.3 Member 3 will investigated=4.1.4 Member 4 will investigate 5.1.5 fyour group will investigate these models for selected values of a. These models The following problems as well as any other interesting facts and properties of the models: Which models( if any) keep the population under 10,000 in the next 5 years? Which models (if any) will give a model which keeps the population under 10,000 in the next n years? Which models (if any) will give a model which keeps the population under l10,000 forever? group should now have investigations of at least 10 different models. Analyze the models and discuss the The remaining parts of the project are all to be done by the group.Explanation / Answer
Frist of all, these are too many questions. Here at Chegg, we get paid on question basis and these are at least
No. of models - 14 (constant rate of change, some change fucntion and then 12 models)
No. of questions - 6
Total questions - 14 x 6 = 84
Additional questions - 1 + 6 + 4 = 11
i.e. in total 95 questions. For approximately $2, this is too much work. (You can try posting these as multiple questions or contact someone from Chegg to check how to post it.
So I wil just answer the subparts of question 1 (i.e. Model 1 answers)
Model 1 - Constant rate of change
Let's use the compound interest rate formula to find out constant rate of change
2000 = 1000 (1+r)6
r = 12.25%
1. Verbal description
Every year the population of deers increases by r% from last year. So if deer population was A in the first year, it will be A x (1+r) in the next year, and in the year next to it, population will be A x (1+r)2
Every year the population of deers increases by 12.25% from previous year.
2. Differential Equation
P = P0 (1+r)t
dP/dt = ( P0 ) x ( ln(1+r) ) x (1+r)t
dP/dt = k (1+r)t
3. Population at different time
Pt = P0 (1+r)t
Where P0 is 2000 (today's population)
P1 = 2,245
P5 = 3,564
P20 = 20,159
4. General rule of forecasting
Pt = P0 (1+r)t
Where P0 is today's population : 2000
and r is rate of change : 12.25%
5. Time to reach 10,000
14 years from now, the population will reach 10,079
6. Problem
After 14 years, the park will run into the problem due to overgrazing of 10,000 inhabitats.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.