BAM!!!
Mr. Tarrou.
In my second video about parametric equations we are going to take most of the examples
that we just used in the previous video where we graphed them using t-tables and put those
parametric equations together and eliminate the parameter.
Basically convert them into just regular rectangular form or function form.
So we are going to try and get these...well y in terms of x, or in the case of a couple
of our final examples we are going to have conic sections.
We will put those conic sections into Standard from.
This is just basic algebra.
I don't have any trig functions.
So I am going to do substitutions to get rid of the parameter 't' and do in a way that
I get y in terms of x.
So we are going to take this equation, we are going to subtract both sides by 3 and
get x-3 equals t.
Now that we know what t is equal to, we are going to take what it is equal to and plug
into the other equation.
We get y is equal to negative... or... yeah negative parenthesis x-3 squared.
Ok, that is y in terms of x.
I could take this binomial and FOIL it together or distribute it out.
I hate the word FOIL, sorry about that.
I am taking this binomial and distribute it or square it, and just expand it and have
it in General Form y=ax^2+bx+c.
But we did just study parabolas in Standard Form.
So I can take this equation, divide both sides by negative one and get negative y is equal
to (x-3)^2.
If I really wanted to push the idea of getting this into Standard form I can turn this around
and get (x-3)^2=-y.
Well this is a parabola with x being squared, which means the parabola opens either up or
down....up or down.
My coefficient of my single degree term is negative.
So my 4p value is equal to negative one.
So this is a parabola opening down.
This is a parabola with the vertex at (3,0) which opens down which is exactly what we
had in the...
I believe the first example in the last video.
Here is our equation.
I am sure you teacher, as well as myself, would accept either one of these answers for
these...
where you would say here are some parametric equations, eliminate the parameter.
Our next example is going to be x equals the square root of t+1 and then over here we have
y equals t-2.
Ok.
So we want to get this into....
I keep wanting to say get into rectangular form.
We want to eliminate the parameter.
Again you can do the substitution in any order that you like.
I am kind of getting stuck on the idea of having y in terms of x.
I am going to take this equation and I am going to square both sides to get rid of the
square root.
Bam.
We get x squared equals t+1, and x squared minus 1 equals t.
Now that we know what t is equal to, we are going to take again what it is equal to and
plug it into the other equation.
You don't always have to do this with the x equation and plugging it into y.
That is just what I am doing.
We have y is equal to whatever t is, which is x^2-1 and then minus 2.
So we have y equals x squared, and negative one minus two make negative three.
I use the parenthesis because it is habit for substitution.
This t had no leading coefficient, it had no divisor, it had no power so really those
parenthesis are not doing much except for reminding me that is where I did my substitution.
We have an upward opening parabola with a vertical shift down of three.
This would be more like your general form of a parabola.
Our next example...well I don't know.
Let's see what happens:D
There we go.
SO!
We have x equals 4sin(t) and y equals -3cos(t).
Ok.
To substitute these equations in together and only get one function and eliminate the
parameter of t, if my only goal was to get rid of t and... to do this problem like I
did my pure algebra questions where there is just squares, square roots, plus and minuses,
no trig functions I would have to do some kind of weird inverse trig function.
Have maybe something like x equals four sine of some kind of inverse cosine function.
It is going to get ugly.
So what we are going to do instead is...
I have sine and cosine in these equations.
I am going to just remind you, because we are going to use this in a minute, that sine
squared t plus cosine squared t equals one.
We would have said theta when you were studying all of your trig identities.
So we are going to make this show up in these two parametric equations after we have combined
them.
So I want to use that pare... that... excuse me... that Pythagorean identity to help me
clean up these parametric functions and get a rectangular function or in rectangular form.
I am going to square both sides of these equations.
I am going to square both sides and get x squared is equal to 4^2 which is 16 and sine
squared which is... well sine squared t.
To be honest, I just paused there for a second because I have been in class with my students
getting rid of the leading coefficient first.
It really is better to get rid of, or isolate the trig function first before you square
both sides.
I got away with it in this one because there was only one term.
So in this example...or in the parametric equation I am going to divide both sided by
-3, get y over -3 equals cosine of t.
Over here now I am going to divide both sides by 16.
I don't know why I am trying to take of both equations at the same time.
But x^2/16 equals sine squared t.
Now I need... you know... sine squared plus cosine squared equals one.
I need my cosine to be squared as well.
So I am going to square both sides of that equation.
When you square a fraction, you have to square the top and bottom.
So y squared over 9 equals cosine squared t.
Now I want sine squared plus cosine squared so I can let the entire expression equal one.
Sine is on the right, or sine squared is on the right.
Cosine squared is on the right.
I am going to add these equations.
Now I don't know how you feel about.
You go like, "How is he able to do that?"
By the way the answer is going to be x^2/16 + y^2/9 = sin^2(t) + cos^2(t).
I added the two left sides of the equations and I added the two right sides.
I probably should have had it stacked so the left sides were on top of each other and the
right sides were on top of each other.
Then sine squared plus cosine squared is going to be equal to one, so I can just erase that
sine squared plus cosine squared and just let it equal one.
Now you can see why I knew that example I graphed in my last video was going to form
an ellipse.
It is because I have already done the problem and also if you know how to take the parametric
equations and put them into a regular rectangular form (this is not a function) you can see
that we have two squared terms, both of my terms have a same sign.. in this case they
happen to be positive...and I have different denominators.
My a value is going to be 4 and my b value is going to be the square root of 9 or 3.
Getting back to this sort of idea where I was able to just add the equations, you do
realize that is not a foreign idea to you.
You have been solving linear equations, linear systems of equations since algebra 1.
They said maybe just as a little side example, i am not going to finish this, x+3y=7 and
let's say 2x-3y=4.
And they gave you this linear system of equations and asked you to solve this system, find out
where these two lines intersect.
And, you know, what did you do?
You did it with graphing which was not very good when decimals are involved.
You did it with substitution.
You also did something called Linear Combination which just said, you know, create opposite
numbers in front of one of your variables and then add the equations.
You get 1 plus 2 is 3x.
These are going to cancel out.
You get 7 plus 4 is 11 and dot, dot, dot.
You find x, you back substitute, and you find y.
So adding equations is not a new idea to your, or it should not be.
So me adding these two equations and getting x^2/16 + y^2/9 = 1, that really is ok.
You have done that before.
Ok, one last example.
Because I think this video might be running a bit long.
Of course, if you watch my videos you know I always talk a lot.
I probably over explain things a little bit, but hopefully for some of you that is a good
thing.
Let's take a look at this one.
That hyperbola one that I did at the end of my last example...or my last video.
We have got x equals 3 secant t minus 2 and y equals 2 tangent t plus 1.
Ok.
I am going to do the exact same process I just did with the last one.
And you might want to just pause and see if you can do it.
We are going to square both sides, well first you are going to isolate the trig functions.
You are then going to square both sides and you are going to use one of your three Pythagorean
Identities that relates secant squared with tangent squared.
Hmmm.
Nana..nana...
Ok, we are going to add 2 to both sides and get x+2=3sec(t).
Divide both sides by 3.
We get X+2 divided by 3 equals the secant of t.
Over here we are going to subtract both sides by 1 to undo that addition.
We are going to undo that multiplication of 2 by dividing both sides by 2.
We are going to square both of these equations and add them.
What Pythagorean Identity are we interested in using or matching up with?
This one:)
1 plus... that looks like an H. 1 plus tangent squared x equals secant squared x...or t.
So if I take this and I subtract tangent squared from both sides I get 1 equals secant squared
x minus tangent squared x.
It is bothering me that I keep writing x when I am just for this example...it does not really
matter what variable I use but currently it is the parameter of t.
So.
I want to square these.
Instead of adding these two equations, I want to subtract them in such a way that the secant
function is first.
So I am going to square both sides.
I am going to subtract them and I am going to go ahead and just kind of clean this up
a little bit.
Secant squared minus tangent squared is...
I am going to subtract these equations.
So secant squared t minus tangent squared t.
I could distribute out the top, but we are trying to get this...
We already know from the last video that this is a hyperbola.
So I don't want to expand this and get it out of Standard Form.
So (x+2)^2, then we are going to square the bottom which is 9 minus (y-1)^2 with a minus
sign in the middle.
I don't know why I wrote equals! (y-1)^2.
Then 2 squared is equal to 4.
So we have (x+2)^2/9 -(y-1)^2/4 and yes secant squared t minus tangent squared t, matching
angles, is equal to 1.
And just to really drill home that this is a hyperbola and it is the same one that we
graphed in the last video, this is a hyperbola because of the two squared terms with different
signs, one positive fraction and one negative.
The center is at (-2,1).
'a' comes from square rooting the denominator of the positive fraction, so 'a' is equal
to 3.
And 'b' is equal to two.
So there you go.
I am Mr. Tarrou.
BAM!!!
Go Do Your Homework:D