Welcome to Converting Parametric Equations
to Rectangular Form.
Converting parametric equations to a rectangular equation
is called eliminating the parameter,
which means we need to eliminate the variable T,
and here are the general guidelines.
Number one, we're going to solve one of the equations for T.
Step two, we'll perform substitution
into the second equation.
And then step three, we'll simplify
and then normally solve for Y.
A couple extra considerations,
if the equations do involve trig functions,
we often have to use a trig identity
in order to eliminate T.
And then lastly, the domain may need to be adjusted
or restricted based upon the original parametric equation.
Let's go ahead and take a look at some examples.
Here we have X equals 4T plus four,
and Y equals T plus two.
So we'll first solve one of the equations for T.
It's going to be easier if we solve this second equation for T.
So we would have Y minus two equals T.
So now we'll use this to perform substitution
into the first equation
where we have X equals 4T plus four,
we'll then replace T with Y minus two.
So we'll have X equals four
times the quantity Y minus two plus four.
Now we'll simplify this and solve for Y.
So we'll distribute.
We'll go ahead and combine our like terms.
Now we'll add four to both sides,
so we have X plus four equals 4Y,
and the last step to solve for Y
we'll divide everything by four.
So we would have Y equals this would be 1/4X plus one.
So these parametric equations represent a line,
and in rectangular form that line is Y equals 1/4X plus one.
Again, notice there were no restrictions on T,
so T could be any real number,
and therefore there's no restrictions
on the domain of this line.
So you can go ahead and just graph the line
Y equal 1/4X plus one,
where the Y intercept is positive one and our slope is 1/4,
so we'll go up one, right four.
So our line passes through these points.
Let's go and take a look at another example.
Here we have X equals the square root of T,
and Y equals T minus five.
Now even though the restriction is not given,
we know that T can never be negative,
because we have X equals the square root of T.
So T has to be greater than equal to zero
for these parametric equations.
Next, let's go ahead and solve one of these equations for T.
Let's use the first equation this time.
If X equals the square root of T,
we could square both sides of this equation,
so we'd have T equals X squared.
And now we'll perform substitution into the second equation
where we have Y equals T minus five.
So we'll have Y equals, and then T is X squared minus five.
So the rectangular equation
is Y equals X squared minus five,
but since T has to be greater than or equal to zero,
that means X will also
always be greater than or equal to zero.
So the domain for this function must be restricted
where X is always greater than or equal to zero.
So this is a parabola shifted down five units,
but we're only going to graph the part
when X is greater than or equal to zero.
So let's go ahead and make a quick T table here.
When X is zero, Y would be negative five.
When X is one, we'd have Y equals negative four.
When X is two, we'd have four minus five.
That would be negative one.
Then lastly, when X is three,
nine minus five, that would give us Y equals four.
Let's go ahead and plot these points.
Here's zero, negative five,
one, negative four,
two, negative one,
and three, four.
This would be the graph of our curve.
Let's take a look at two more examples.
Here we have X equals E to the negative T,
and Y equals E to the T minus one.
These are going to be a little bit more challenging
to solve for T, but let's go ahead
and try to solve this first equation for T.
The way we're going to do this
is rewrite this exponential equation as a log equation.
So let's go ahead and review this for a minute.
Remember, if we have E raised to the power B equals A,
it's also true that natural log is equal to B.
So in this case, we'd have the exponent negative T
must equal natural log X.
So if we were to solve this equation for T,
we'd multiply both sides by negative one,
we'd have T equals negative one times natural log X.
And now we'll take advantage of the power property of logs,
which means we can take this coefficient
and move it to the exponent on the X.
So we have T equals natural log X
raised to the power of negative one.
And now we'll take this
and perform substitution into this second equation
when Y is equal to E to the T minus one.
So we have Y equals E to the natural log X
to the power of negative one minus one.
And now we'll take advantage
of another advantage of logarithms.
Remember, if this base and the base of the log are equal,
this is just going to give us X to the negative one.
So we'll have Y equals X to the negative one minus one,
which is the same as Y equals one over X minus one.
Now let's talk about the domain for a minute.
Notice there were no restrictions on T
in the original parametric equations,
meaning T could be any real number.
But, notice that T is also the exponent on E.
So in both cases here,
if we take E and raise it
to the power of negative T or positive T,
it's always going to be greater than zero.
So if E to the negative T
is always greater than zero,
X is also always going to be greater than zero.
Therefore, we do have to restrict the domain
to X being greater than zero.
So let's go ahead and make a T table for this graph,
but only select values when X is greater than zero.
So if X is equal to one, we'd have one minus one,
that would be zero.
When X is equal to two, we'll have 1/2 minus 1,
that would be negative 1/2.
Let's go ahead and select X equals one half.
Well, one divided by one half,
that's going to give us two, minus one, so Y is equal to one.
Let's go ahead and plot these points.
Here's one, zero,
two, negative 1/2 would be here,
and 1/2, one would be here.
Notice as X approaches infiniti,
Y is going to approach negative one.
So we do have a horizontal a-sin-toch here,
and we also have a vertical a-sin-toch, X equals zero.
So we have a graph that looks something like this
with our restricted domain.
Let's go ahead and take a look at one more.
Notice on this last example
we have X equals three cosin[t],
and Y equals three sin[t].
Both of these involve trick functions.
So in this case, we'll use the identity
sin squared data plus cosin squared data equals one
to eliminate the variable T.
The way we'll do that
is we'll solve this first equation for cosin[t],
and then we'll solve this second equation for sin[t].
So cosin[t] would be equal to X divided by three,
and sin[t] would be equal to Y divided by three.
Now using this trick identity,
we can replace sin squared theta
with Y over three squared,
and we can replace cosin squared data
with X over three squared.
And the sum must also equal one based upon this identity.
So now we'll go ahead and simplify.
We'd have Y squared over nine
plus X squared over nine equals one.
Now, multiply everything by nine.
So, we'll have Y squared plus X squared equals nine.
Now we should begin to recognize this as a circle,
where the center is at zero, zero,
and the radius is equal to three.
So here's the center,
and since the radius is equal to three,
we'll plot four points out three units from the center,
and now we can sketch the circle.
And there it is.
Okay, that's going to do it for this video.
I hope you found this helpful.