[Course Instructor] Since 1990 the number of fish
in Lake Beckett has been decreasing
at a rate of 1.7% per year.
In 1990 the population of fish was estimated
to be 121 million.
We want to use this information
to answer the following questions.
A, write the exponential function p of t for this scenario
where P of t is the fish population in millions,
t years after 1990.
So our exponential function is going to be in the form
of P of t equals a times b raised to the power of t.
So looking at our notes below,
we have an exponential function in this form here.
A is the initial value, or starting population,
at time, t equals zero.
And since t is the number of years after 1990,
in our case a is going to equal 121 since p of t
is in millions.
And then the base b is going to be the decay factor
since the population is decreasing at 1.7% per year.
Where b, the decay factor is equal to one minus r,
where r is the percent of decrease per year.
So notice here, r,
that a percent of decrease per year equals 1.7%,
but this must be expressed as a decimal,
which would be 0.017, and therefore the base b,
because we have exponential decay,
is equal to one minus r, or one minus 0.017,
which is equal to 0.983.
Now that we know a and b,
we know our exponential function
is p of t equals a which is 121 times the base,
0.983 raise the power of t.
Part b, we're asked to determine
the fish population in 1993.
Notice 1993 is three years after 1990.
So to find the fish population in 1993,
we want to determine the function value p of 3,
which is equal to 121 times 0.983
raised to the power of three.
If you're ever not sure what value of t to use,
we can always find t by taking the desired year,
in this case 1993,
and subtracting the base year, which is 1990.
Notice how this gives us t equals three.
Let's go and evaluate this on a calculator.
We are asked to round to two decimal places.
So from the home screen,
we could just enter 121 times 0.983, close parenthesis,
raise the power of three, press Enter.
To two decimal places, this would be approximately 114,93,
remember this is in millions.
Because we're going to solve c graphically,
I also want to show another way
that we can find this function value.
Let's go ahead and enter p of t ny one.
To do this we'll press y equals,
clear out any old functions.
So we'll press, Clear, Enter, Clear,
go back up to y one, and we'll enter 121 times 0.983,
close parenthesis, raised to the power of x.
We'll use x on the calculator, not t.
And now, because we have our function ny one,
we could go back to the home screen
by pressing Second Mode.
And then y one of three will also give us
the function value of p of three.
So if we press Vars, right arrow, Enter, Enter,
and then in parenthesis, three.
Of course we'll get the same function value.
Now let's look at part c,
determine in what year the fish population
will be half the amount it was in 1990.
We're asked to round to the nearest year.
Whenever we're determining how long it takes
for half the original amount to exist,
we're talking about half life
where if you have an exponential function
in the form of a times b raised to the power of x,
remember a is the initial value, and therefore,
a divided by two would be half of the starting amount.
So if we set our equation up in this form
and it's all for x, or in our case, t,
we can determine the half life.
So we set p of t equal to half of the starting population,
which would be 121 divided by two equals p of t,
which is 121 times 0.983 raised to power of t.
121 divided by two equals 60.5.
So we have 60.5 equals 121 times 0.983,
raised to power of t.
And since we'll be solving this graphically,
and we already have p of t, ny one, this side is ny one,
if we enter y two equals 60.5,
we can determine the value of t
that satisfies this equation,
by determining where y one intersects y two.
So let's go ahead and do that.
Again if we press y equals,
we already the writhes side ny one, so we'll press Enter.
And then for y two we'll enter 60.5.
And now before we graph this,
we do want to adjust our window.
So let's press the Window key.
We'll go ahead and leave the horizontal axis
from negative five to 25,
we can always come back and change this if we need to.
So we'll press Enter, Enter, Enter,
and now for the vertical axis,
because the starting population was 121 million,
let's set the horizontal axis to go from,
let's say, negative 40, Enter, through, let's say, 150.
Now let's press Graph and see how this looks.
So there's the exponential decay function.
And here's the constant function, 60.5.
Notice how we don't see the point of intersection,
which means we need to increase the x maxim,
or extend the horizontal axis more to the right.
So press Window,
and let's change the x maxim to 50 instead of 25.
Now press Graph.
So we're looking for the x coordinate, or t coordinate,
at this point of intersection,
which will be how long it takes for the fish population
to be half of the starting population.
So we'll press Second Trace for the calculation menu,
Option five for intersection, and then Enter three times.
Enter, Enter, Enter.
So notice how the point of intersection shows here below,
notice how it takes approximately 40.4 years
for the population to reach half of the starting population.
But our question does ask us to round to the nearest year,
so we'll round the value of t to 40,
so t is approximately 40,
but this question doesn't ask how long it will take,
it asks in what year this would occur.
So to find the year, we take the base year of 1990,
and add the approximate number of years, which is 40,
which will give us the year 2030.
So the answer is not 40, it's the year 2030.
I hope you found this helpful.