Hello. I'm Professor Von Schmohawk and welcome to Why U.
In the last lecture, we saw how a system of three linear equations in three variables
could be used to solve a problem with three unknown quantities
the price of a ticket
the price of a drink
and the price of popcorn.
We knew that three different groups of people had purchased different quantities of these items
and the total price that each group paid.
Given this information, we were able to create three mathematical relationships
involving these unknowns.
We then solved this system of three linear equations in three variables
to determine the price of a ticket, the price of a drink, and the price of popcorn.
Together, these three values formed the single unique solution to this system of equations.
Now let's look at example of how a system of three linear equations in three variables
can be used to solve a real-world problem represented by an equation which is not linear.
Hoping to break the Why U altitude record for student slingshot launches
A.V. Geekman has constructed a giant slingshot
which he plans to use to launch himself into the Why U campus pond.
He knows that objects in free-fall always follow a parabolic path.
He also knows that the standard form equation for any parabola with a vertical axis
is "y = A(x-squared) + Bx + C"
where x and y are the horizontal and vertical coordinates of the parabola
and A, B, and C are numbers which determine the dimensions and orientation of the parabola.
However, since this is the first test flight of this highly-technical apparatus
A.V. does not know exactly what his parabolic trajectory will be.
To determine whether he has attained sufficient altitude to break the record
A.V. has set up a camera to automatically take photos of his flight.
The camera captures three images of A.V.'s flight
one on the way up
one on the way down
and one at the point that he makes contact with terra-firma.
Unfortunately, none of these photos recorded the point that he reached his maximum altitude.
However, from these photos, A.V. can determine the x and y coordinates of his position
at three different points during his flight.
If he can somehow use this data to construct the equation for his parabolic trajectory
then he could use that equation to determine the maximum altitude attained during his flight.
Since the equation for a parabola is "y = A(x-squared) + Bx + C"
he only needs to determine the values of A, B, and C
in order to construct the equation for his path.
At this point you may be thinking, "But wait!" "The equation for a parabola is not linear."
"Linear equations do not have squared variables, and x is squared."
However, the unknown quantities which we must calculate in this equation are A, B, and C.
Therefore, in this problem A, B, and C are the variables
which of course are not squared.
Because A.V. has recorded the x and y coordinates of three different positions along his trajectory
we can substitute these values for x and y into the equation for a parabola
thereby creating three different linear equations in three variables, A, B, and C.
Now we have a system of three linear equations in three variables, A, B, and C.
Let's start by writing these equations in standard form.
We can now solve this system to determine the values of the variables A, B, and C.
We will start by eliminating the variable C
using the first and third equations.
Multiplying the bottom equation by negative one
and adding the equations
we get -140A - 10B = 5
or dividing by five
-28A - 2B = 1 .
We will now eliminate C again using the second and third equations.
Multiplying the top equation by negative one
and adding the equations
we get 44A + 2B = -5 .
This system of two equations in two variables
can now be solved to find the values of A and B.
Adding the equations
we get 16A = -4
or dividing by sixteen
A = -1/4 .
This value can now be substituted for A in either of the two equations
to find the value of B.
Substituting negative one-fourth for A in the bottom equation
we get -11 + 2B = -5 .
And adding eleven to both sides
we get 2B = 6
or B = 3 .
Now that we know the values of A and B
we can substitute those values back into any of the original equations and solve for C.
Choosing the top equation
we complete the arithmetic
to get 5 + C = 6 .
Subtracting five from both sides
gives us C = 1 .
Now that we have solved this system to find the values of A, B, and C
we can plug those values into the equation for a parabola
to produce the specific equation for A.V.'s parabolic trajectory.
Graphing this equation
we see that the graph includes the positions of all three points recorded by the photos
and that A.V. achieved a maximum altitude of ten meters
setting a new altitude record for Why U student slingshot launches.
In a future Algebra lecture we will see that it is not necessary
to physically plot the equation for a parabola in order to determine its maximum height.
It is possible to convert the standard form equation for a parabola
to what is called "vertex" form
where h and k are the coordinates of the highest
or lowest point on the parabola.
Even though this problem may seem quite different
from the problem we solved in the previous lecture
which involved calculating the prices of three items
both problems were similar in that they both had three unknowns
in this case the values of A, B, and C
and we were able to construct three linear equations involving these unknowns.
And once again, since there were at least as many independent equations as unknowns
it was possible to determine unique values for each unknown.
Together, these three values formed the single unique solution
to this system of equations.
In this last example, we saw that systems of linear equations
can be used to calculate shapes which are not linear
such as the equation for a parabolic trajectory.
In the next lecture, we will use a system of three linear equations in three variables
to calculate a different kind of non-linear shape.