Hello. I'm Professor Von Schmohawk and welcome to Why U.
In the last several lectures, we introduced several forms of linear equations
whose graphs are lines in the xy-plane.
In this chapter, we will use linear equations to solve some real-world problems.
Why Hello Hulk.
I'm so happy to see that you are gainfully employed this summer.
I was looking for someone to drain my pool tomorrow.
Uh, sure Professor Ethylene
but I'm supposed to clean Mr. Squeegeeopolis' pool tomorrow.
Well I'm sorry Hulk, but I need it drained by eight p.m. tomorrow.
I'm having some very important guests over
and the pool must be completely empty by the time they arrive.
Okay Professor, I'll figure something out.
Oh Hulk, you're such a resourceful young man.
I knew I could count on you!
Ahem. What's the matter Hulk?
Have you perhaps encountered a small resource scheduling problem?
Okay Geekman, how much is this going to cost me?
Well my normal short-notice subcontracting fee is fifty dollars an hour
but I'm running a special today … only forty-nine buckeroonies an hour!
Yeah? Well I'm running a special today on knuckle sandwiches.
Perhaps I didn't mention that you were eligible for the
double-extra-special-very-best-buddy rate of three dollars an hour?
Well, just make sure you have that pool empty by eight p.m. tomorrow.
Yes sir, Mr. Moosemasher!
Rats! It's almost noon.
I'd better get crackin!
Start pool draining at twelve noon.
Hmm, It's one thirty and the water level is ten feet.
Hmm, three o'clock and the water level is eight feet.
Yikes! This pool better be empty by eight p.m.
Well, it looks like Ace Algebra student A.V. Geekman may be in hot water.
Will the pool be empty by eight p.m.?
If we graph the water depth versus time
we might be able to figure out how long it will take for the pool to completely drain.
We will plot the amount of time since the drain was first opened at noon
on the horizontal axis
and water depth on the vertical axis.
We know that at one-thirty, one and one-half hours after the drain was opened
the water level was ten feet
and at three o'clock after three hours of draining
the water level was eight feet.
However, there is one problem.
With only two data points it is impossible to determine
whether the water level is dropping by the same amount every hour
or if the water level drops more quickly as time goes on
or perhaps if the water level drops more slowly with increasing time.
Taking one additional measurement a third time
would allow us to determine which scenario more accurately represents the water level
as a function of time.
Dang! Its six o'clock already
and we still have four feet to go.
We now can add a third data point to our graph of water depth versus elapsed time
since A.V. has determined that after six hours the water depth is four feet.
It looks like these three points all fall on a line
so maybe the function we are graphing is linear.
We can verify this mathematically
by calculating the slopes of the line segments connecting the first and second points
and the second and third points.
If the slopes of both segments are the same then all three points must lie on the same line.
As we saw in the chapter on slope
we can calculate the slope of a line which passes through two points
as the difference in the point's y-coordinates, delta-y
divided by the difference in their x-coordinates, delta-x.
Dividing delta-y by delta-x
we get negative four
over three
or a slope of negative four-thirds.
Now let's calculate the slope of the other line segment.
Once again, we take the difference in the y-coordinates, delta-y
and the difference in the x-coordinates, delta-x.
Dividing delta-y by delta-x
we get negative two
over three-halves.
We can simplify this complex fraction by dividing negative two by three-halves.
This is the same as multiplying negative two times the reciprocal of three-halves, or two-thirds.
Negative two times two-thirds is negative four-thirds.
So the slope of this line segment is also negative four-thirds.
Since the slopes of these two connected line segments are the same
all three points must lie on the same line.
We can therefore assume that the water in the pool is draining at a constant rate
and that the water depth as a function of time is linear.
Since we know this line's slope, and at least one point which falls on the line
we can write the equation for this line using the point-slope form.
If we choose the bottom point which has coordinates (6,4)
to use as the known point in the point-slope formula
then x-one is six
y-one is four
and the slope m, is negative four-thirds.
Now that we have an equation which represents water level, y
as a function of elapsed time, x
we can plug in different values for x
to see what the water level will be at those points in time.
For example, an elapsed time, x, of zero
corresponds to the time when the drain was first opened.
If we set x to zero
then the resulting value of y will tell us the initial depth of the pool.
Completing the arithmetic, zero minus six
is negative six.
And negative four-thirds times negative six
is positive twenty-four thirds
or eight.
Adding four to both sides of the equation
allows us to cancel negative four on the left
leaving us with "y equals eight plus four"
or twelve.
So the initial depth of the pool when the elapsed time x, was zero, was twelve feet.
If we would like to find out how long it will take for the pool to be empty
we can set the water level, y, to zero
and solve the equation for x, to see what the elapsed time is at that point.
We start by setting the value of y to zero
and we can then write "zero minus four"
as negative four.
Since we are trying to find the value of x
we want to rearrange the equation so that x is alone on one side.
To do this, we must first eliminate the negative four-thirds
by multiplying both sides of the equation by the reciprocal of negative four-thirds
which is negative three-fourths.
Since anything multiplied by its reciprocal is one
we can replace negative three-fourths times negative four-thirds with one.
Now, since everything in the parentheses is multiplied by one
we can eliminate the one and the parentheses.
Completing the arithmetic on the left side
negative three-fourths times negative four
is twelve-fourths
or three
and adding six to both sides
allows us to cancel the negative six on the right
leaving us with the equation "three plus six equals x".
Adding three and six
we have "nine equals x"
or equivalently, "x equals nine".
So when y is zero, x is nine.
Since x is the elapsed time
and y is the water level
this result tells us that when the water level y, is zero
the elapsed time x, will be nine hours.
Since the drain was opened at noon the pool will not be empty until nine o'clock.
Excuse me Professor Von Schmohawk.
Did I just hear you say that my pool will not be empty until nine o'clock?
Uh, yes Professor Ethylene.
According to our calculations, that is correct.
But I distinctly remember telling Mr. Moosemasher
that my pool must absolutely be empty by the time my guests arrive at eight.
Uh, excuse me. If I may interject here for a moment
I believe that it is possible to calculate the exact depth of water in your pool
at the time your guests will arrive.
If we start with the equation for water depth versus elapsed time
and set the elapsed time, x, to eight hours
we can then solve this equation for y which represents the water depth
and thereby determine the depth at exactly eight o'clock post meridiem.
As we all know, eight minus six is two
and negative four-thirds times two is negative eight-thirds.
Then adding four to both sides
the fours on the left side cancel.
We can then write four as twelve-thirds
and negative eight-thirds plus twelve-thirds is four-thirds.
So according to my calculations
the water depth at eight o'clock p.m. will be exactly four-thirds feet ...
... in other words, one and one-third feet.
Now, since there are twelve inches in a foot
one third of a foot is twelve inches divided by three
or four inches.
So, one and one-third feet
is one foot, four inches.
We can therefore deduce that when your guests arrive
your pool will contain exactly one foot, four inches of water!
Gulp!
Uh, Professor Ethylene?
How much longer do I have to entertain your guests?
Why you're doing a splendid job A.V.
I'll let you know when it's nine o'clock.
Jeesh!
Thank you A.V.
It's exactly nine o'clock
and just as calculated
the pool is empty!
We have seen how linear equations can be applied to real-world applications.
In the next lecture we will see how real-world quantities can be calculated
using formulas, also referred to as "literal equations"
which relate two or more quantities.