Use a calculator and analyze Example 6 in the book.
The population of fish in the lake stabilizes at about 1600 fish over time.
Practice makes perfect
We are asked what happens to the population of fish over time if we suppose that 75 % of the fish remain each time. A lake initially contains 5200 fish and each year the lake is restocked with 400 new fish. Let a_n be the number of fish at start of the n^(th) year. First, we will find a recursive rule for a_n.
Recursive Rule
At the start of the (n-1)^(th) year the population of fish is a_(n-1). Since 75 %=0.75 of fish remain in the lake from year to the next year, 0.75a_(n-1) of old fish will be still in the lake in the n^(th) year. Remember that the lake is restocked with 400 new fish. Here is a verbal model for the recursive equation.
ccccc
c Fish at start of the $n^(th)$ year
&
=0.75 *
&
c Fish at start of the $(n-1)^(th)$ year
&
+
&
New fish
⇓ & & ⇓ & & ⇓
a_n & =0.75 * & a_(n-1) & + & 400
Therefore, we can write a recursive rule for the given sequence.
a_1=5200, a_n=0.75a_(n-1)+400forn>1.
Now, we will describe what happens to the population of fish over time. We will use a calculator. First, enter the value of a_1, which is 5200, in a graphing calculator.
Then, enter a rule 0.75*Ans+400 to find a_2.
When you hit enter you will get the next term of the sequence. Press ENTER so many times that you see the stabilization. After around 30 hits we will get the following.
After around 45 hits we will get the following.
This tells us that after each next hit we will get the same value, 1600.000002. Therefore, the population of fish in the lake stabilizes at about 1600 fish over time.
