It seems that every year I read in the newspaper about the over population of deer. Why is there such an over population? Why do humans care if the deer at a park or wildlife sanctuary are over populated? What are some ways to decrease the population? What would a graph of the population of deer taken by a census of the deer population in Sandy Point State Park near Annapolis, Maryland, on January 1st of each year look like? Does it always show an increase in deer? A decrease? Does the population remain steady?
Many populations have growth patterns. Some animal populations have a cyclical growth pattern while others continuously increase or decrease. Money (consider money to be a population) can be invested to have a growth pattern. Waste in the county landfill has a growth pattern.
We will begin this chapter by looking at number patterns and complete this chapter by looking at the growth patterns of various populations.
The Fibonacci numbers were a sequence of numbers that had a pattern. There were rules for finding the next number. Look at the following sequence and determine the next three numbers.
(a) 3, 5, 7, ______, ______, ______
Did you recognize that the next three numbers are 9, 11, and 13? How did you determine the next three numbers? Add 2 to the previous number to get the next number.
Finish this sequence
(b) 17, 14, 11, ______, ______, ______
The next three numbers are 8, 5, and 2. Add -3 to the previous number to get the next number.
Complete these sequences and state what was added each time.
(c) 2.0, 2.5, 3.0, _____, ______, ______
(d) 100, 150, 200, ______, ______, ______
We will establish recursive rules for finding these sequences. The following is the general recursive rule for arithmetic sequence, also known as a linear growth model:
Go back to the sequence in (a) 3, 5, 7, ______, ______, ______.
Let P1 = 3
P2 = 3 + 2 = 5
P3 = 5 + 2 = 7
P4 = 7 + 2 = 9
P5 = 9 + 2 = 11
P6 = 11+ 2 = 13
To get the next number, we added the previous number and 2. The general formula for this sequence is
PN = PN-1 + 2
Determine the general formula for (b) 17, 14, 11, ______, ______, ______. Because we add -3 each time the general formula is
PN = PN-1 + ( - 3)
Find the general formula for each of the following by replacing the d in the Linear Growth Model (Recursive Description) by the number you added each time. The PN and PN-1 remain in the general formula, and d will be replaced by a number.
(c) 2.0, 2.5, 3.0, ______, ______, ______
(d) 100, 150, 200, ______, ______, ______
Answer to c PN = PN-1 +0.5
Answer to d PN = PN-1+50
How would you find the 50th number in (a) 3, 5, 7, ______, ______, ______.? It will take a long time if you plan to find each of the numbers by adding two each time. The following explicit formula gives the 50th value without your having to know any other numbers in the sequence except the first number.
PN = P1 + (N - 1)d
Before finding the 50th number in the sequence, we will find the 6th using the formula since we already know that the sixth number is 13.
P6 = 3 + (6 - 1) 2 where N = 6, P1 = 3 (the first number in the sequence), and d = 2 in the above formula.
P6 = 3 + 10 = 13
We will find the 50th number by letting N = 50 and using the same P1 and d.
P50 = 3 + (50 - 1)2
P50 = 3 + 98 = 101
Find the 101th number in the sequence from above c) 2.0, 2.5, 3.0, ___,_____, ______ .
P101 = 2.0 + (101 - 1) (.5) where N = 101, P1= 2.0, and d = .5 in the above formula.
P101 = 2.0 + 50.0 = 52.0
Now that you found the 101th numbers, suppose that there was a reason that we wanted to find the sum of the first 101 numbers. Obviously finding all 101 numbers and then adding them is a daunting task. Here is a short cut by using a formula.
Sum = (first term + last term) times the number of terms divided by 2
Sum = (first term + last term)*N/2
We will first use this formula with a small number of terms so that we can see how it works. For the sequence 3, 5, 7, 9, 11, 13 find the sum of the six terms. Using the formula
Sum = (3 + 13) *6 / 2
= (16) * 6 / 2
= 48
Check this answer by adding 3 + 5 + 7 + 9 + 11 + 13.
Find the sum of the first 101 terms of the sequence c) 2.0, 2.5, 3.0, ___,_____, ______ . using the formula.
Sum = (2.0 + 52.0) 101 / 2
= 54 *101 / 2
= 2727
Why was this section named the linear growth model? The word linear means line. We will look at a graph of the points generated for example a from above.
P1 = 3 (1, 3)
P2 = 5 (2, 5)
P3 = 7 (3, 7)
P4 = 9 (4, 9)
P5 = 11 (5, 11)
P6 = 13 (6, 13)
The above six points of the sequence (a) generate a line when connected on the graph thus this is an example of a linear growth model.
We will consider a different kind of sequence with different rules than the Arithmetic Sequence rules. Consider the following patterns. Try to complete the blanks based upon the pattern you see.
a) 3, 6, 12, ______, ________, ________
b) 2, 6, 18, 54, ______, ______, ______
c) 16, 8, 4, ______, ______, ______
a) To complete this sequence multiply the previous number by 2.
3, 6, 12, 24, 48, 96
b) To complete this sequence multiply the previous number by 3.
2, 6, 18, 54, 162, 486, 1458
c) To complete this sequence multiply the previous number by ˝.
16, 8, 4, 2, 1, ˝
In each sequence we found a number to multiply by. This multiplier is represented by r, the common ratio, in the following formula.
a) To find P7in the sequence in (a) 3, 6, 12, ______, ________, ________, use the above formula where
r = 2
P6= 96
P7=96 x 2 = 192.
How do you find r if it is not obvious? Divide any number in the sequence by the previous number,
, so 
. This works for any pair of numbers in the sequence.
b) Find P8 in the sequence 2, 6, 18, 54, 162, 486, 1458
Answer 
We will also consider the explicit formula so that we can find any number in the sequence.
Exponential Growth Model (explicit description)
We will use this formula to find some of the numbers that we did earlier and then to find some that would be time-consuming to find using the recursive description.
a) Find P7in the sequence 3, 6, 12, 24, 48, 96 where
P1 =3
r = 2.
This is the same answer we found using the recursive description.
b) Find P8 in 2, 6, 18, 54, 162, 486, 1458 where
P1 =2
r = 3.
a) Find P20 in 3, 6, 12, 24, 48, 96 where
P1 =3
r = 2.
Warning: Be aware that your calculator may use scientific notation for large numbers or very small numbers. P20 in 16, 8, 4, 2, 1, ˝ was 3.0517578125E-5 on my calculator which is 3.0517578125X10(-5) or .000030517578125 .
I am skipping the sum of the numbers in a geometric sequence but will instead use a summing formula that is related. This formula is used in banking.
A is the amount accumulated in the account in t years.
P is the principal, the amount initially deposited into the account.
n is the number of times each year the money is compounded.
t is the number of years the money is invested.
i is the rate of interest as a per cent.
Consider that a grandparent deposits $1000 into a savings account for a new grandchild’s future college education. The savings account pays an interest rate of 5% and is compounded quarterly. How much will the grandchild have in the account in 18 years?
P = $1000
i = .05
n = 4 (Quarterly implies four times per year.)
t = 18
This can be worked on the calculator. Some people are not careful with the order of operations when they type this information into the calculator. Here is a method to use if you have difficulties with your calculator. (If you are confident about using your calculator, go ahead and do the problem your usual (correct) way.).
1. Multiply your exponent and write down the answer. 4 x 18 = 72
2. Divide the numerator of the fraction by the denominator. (.05 )4 = .0125
3. Add 1 to this answer. 1.0125
4. Take this answer to the number you found for the exponent in (1) above. 1.012572 =2.445920268
The grandparent will save $2445.92 for the grandchild’s future college education.
When determining the n value consider the following values of n which relate to each of the following terms.
|
|
n |
|---|---|
|
Annually |
1 |
|
Semi-annually |
2 |
|
Quarterly |
4 |
|
Monthly |
12 |
|
Daily |
365 |
We will consider how different interest rates and different lengths of time for compounding one's investment influence the amount accumulated in the account. Suppose that you wanted to open a savings account at a bank. Bank A offers savings accounts with 3% interest compounded daily. Bank B offers savings accounts with 2.75% compounded quarterly. Bank C offers savings accounts with 3.025% compounded annually.
We will consider which bank really is offering the best deal by finding the annual yield (yearly interest rate).
After finding the annual yield offered by each bank, we will choose to invest our money at the bank that pays the highest yield.
|
Bank A |
Bank B |
Bank C |
|---|---|---|
|
A = 100 (1 + .03/365)365 A = 103.041 |
A = 100 (1+.0275/4) 4 A = 102.778 |
A = 100 (1+.03025/1) 1 A = 103.025 |
|
Bank A |
Bank B |
Bank C |
|---|---|---|
|
103.041 -100.00 3.041 |
102.778 -100.00 2.778 |
103.025 -100.00 3.025 |
|
Bank A |
Bank B |
Bank C |
|---|---|---|
|
3.041/100 = .0341 |
2.778/100 = .02778 |
3.025/100=.03025 |
|
Bank A |
Bank B |
Bank C |
|---|---|---|
|
.0341 =03.41% |
.02778=2.778% |
.0325=3.025% |
The above amounts are the annual yields of each bank. Note that Bank A offers the highest annual yield even though it did not advertise the highest rate.
Why was this section named the exponential growth model? The population is growing exponentially. We will look at a graph of the points generated for example (a) 3, 6, 12, 24, 48, 96, 192, from above.
P1 = 3 (1, 3)
P2 = 6 (2, 6)
P3 = 12 (3, 12)
P4 = 24 (4, 24)
P5 = 48 (5, 48)
P6 = 96 (6, 96)
P7 = 192 (7, 192)
Notice that the x and y axis are calibrated differently. The graph of the six points shows an exponential curve.
We will return to the first paragraph about population growth.
It seems that every year I read in the newspaper about the over population of deer. Why is there such an over population? Why do humans care if the deer at a park or wildlife sanctuary are over populated? What are some ways to decrease the population? What do you think that a graph of the population of deer taken by a census of the deer population in Sandy Point State Park on January 1st of each year would look like? Does it always show an increase in deer? A decrease? Does the population remain steady?
Animal populations increase due to more animals being born than dying. When the deer have an ample food supply and no predators, the population will increase. The natural predators of the deer (wolves) have been eradicated from Maryland. The deer population will continue to grow until starvation or illness or another predator kills some of them. If the deer are starving, they may leave the state park or wild life sanctuary in search of food. They may end up eating the landscape in nearby yards and many humans are upset when their yards are destroyed by deer. Humans have become predators of deer by frequently having deer hunts.
The graph of the deer population is neither linear nor exponential.
We will look at some models of animal populations to illustrate the concept of the logistic growth model. Read about population growth at http://www.otherwise.com/population/
Then follow the links to Exponential Growth, and Logistic Growth and do the experiments.
To find the population in the next generation use
PN+1 = r ( 1 - PN) PN
Where r is a growth parameter which is related to the population's growth rate. The carrying capacity is how many members of a population a habitat can support. The original population is represented by P0. PN represents the fraction of the habitat's carrying capacity that is taken up by the actual population; PN is always a value that is from 0 to 1 and is the fraction of the habitat's carrying capacity. This is derived in your text, if you are interested. The value r is the growth parameter.
Suppose a farmer stocks his lake with 200 fish, his lake can hold 1000 fish, and he knows the growth parameter is 1.5. Find the population in the second generation. The carrying capacity, PN, is
PN+1 = r ( 1 - PN) PN
In the second generation, the lake is stocked with 24% of its carrying capacity.
Future generations are
The graph of this growth pattern is shown below using the points (1, .20), (2, .24), ( 3, .27), (4, .30), (5. .32), (6, .33), (7, .33), (8, .33)
Notice that by initially stocking the lake with 20% of its capacity of fish the population appears to level off at about 33% of its capacity. If the farmer is a businessman and wants to raise a lake full of fish so that he can sell fish, having a lake that will only reach 33% of its capacity by initially beginning with 20% will not make the farmer rich.
We will consider the same stocking of a population of a different kind of fish that has a rate of growth pattern, r, is 3.0. We will find the population of several generations and then graph them.
P1 = .20 (1, .20)
P2 = 3.0 · ( 1 - .20 ) · .20 = .48 , (2, .48)
P3 = 3.0 · (1 - .48 ) · .48 = .75 , (3, .75)
P4 = 3.0 · (1 - .75 ) · .75 = .56 , (4, .56)
P5 = 3.0 · (1 - .56 ) · .56 = .74 ,(5, .74)
P6= 3.0 · (1 - .74 ) · .74 = .75 , (6, .58)
P7= 3.0 · (1 - .58 ) · .58 = .73 , (7, .73)
P8 = 3.0 · (1 - .73 ) · .73 = .59 , (8, .59)
In this situation the population in the lake eventually varies from about 59% to about 73%. By raising a fish with a different growth pattern, the farmer was able to fill the lake with more fish which he could then sell.