### Video Transcript

In this video, weβre going to learn
how to convert between parametric and rectangular or Cartesian equations. Weβll learn how this process can
help us to sketch the curve given by a pair of parametric equations and how to
parameterize circles with centres at the origin and line segments between two
points.

Letβs begin by recalling that a
Cartesian equation or a rectangular equation is one given in the variables π₯ and
π¦. Itβs usually given as π¦ is equal
to some function of π₯, although thatβs not always the case. A pair of parametric equations are
given in terms of a third variable, usually π‘, such that π₯ is equal to some
function of π‘ and π¦ is equal to some other function of π‘. And one method we have for
converting from parametric to rectangular form is rearranging to eliminate the
variable π‘.

Letβs see what that might look
like.

Convert the parametric equations π₯
equals π‘ squared plus two and π¦ equals three π‘ minus one to rectangular form.

Here, we have a pair of parametric
equations. We have π₯ is equal to some
function of π‘ and π¦ is equal to some other function of π‘. To convert parametric equations to
rectangular form, we need to find a way to eliminate the π‘. So looking at our equations, we can
see that we can rearrange the equation in π¦ to make π‘ the subject. We begin by adding one to both
sides. And then, we divide through by
three. So we see that π‘ is equal to π¦
plus one all over three.

Now, we go back to our equation for
π₯. We replace π‘ with π¦ plus one over
three. And we find that π₯ equals π¦ plus
one over three all squared plus two. And there will be certain
circumstances where weβre required to distribute the parentheses and simplify. In this case, thatβs not
necessary. And so, weβre finished. Weβve converted the parametric
equations π₯ equals π‘ squared plus two and π¦ equals three π‘ minus one into
rectangular or Cartesian form. Itβs π₯ equals π¦ plus one over
three all squared plus two.

In our next example, weβre going to
look at how trigonometric identities can help us to eliminate the variable π‘.

Find the Cartesian equation of the
curve defined by the parametric equations π₯ equals two plus cos π‘ and π¦ equals
four cos of two π‘.

Remember, a Cartesian equation is
one which contains only the variables π₯ and π¦. So weβre going to need to find a
way to eliminate our third variable π‘ from our parametric equations. And at first glance, it doesnβt
seem to be a nice way to do so. But we can begin by recalling some
trigonometric identities. We have cos of two π‘ in our second
parametric equation. And we know that cos of two π‘ is
equal to two times cos squared π‘ minus one. This means we can rewrite our
equation for π¦ as four times two cos squared π‘ minus one.

Next, weβll look to rearrange our
equation for π₯ to make cos of π‘ the subject. Once weβve done that, weβll be able
to find an expression for cos squared π‘ in terms of π₯. We can subtract two from both
sides. And we see that π₯ minus two equals
cos of π‘. Then, by squaring both sides of
this equation, we find that cos squared π‘ is equal to π₯ minus two all squared. And so, weβre now able to replace
cos squared π‘ with π₯ minus two squared. That gives us π¦ equals four times
two times π₯ minus two all squared minus one.

We distribute this first pair of
parentheses. And we find that π₯ minus two all
squared is equal to π₯ squared minus four π₯ plus four. We distribute again by multiplying
each of these terms by two and then simplifying: eight minus one is seven. Well, finally, we distribute one
more time by multiplying each term of two π₯ squared minus eight π₯ plus seven by
four. And we find the Cartesian equation
of the curve defined by parametric equations π₯ equals two plus cos π‘ and π¦ equals
four cos of two π‘ is π¦ equals eight π₯ squared minus 32π₯ plus 28.

Now, we might be able to see that
this process can be really useful in helping us to sketch curves given in parametric
form. For example, in this case, we were
given a pair of parametric equation. It might now be instantly obvious
what the curve defined by these parametric equations looks like. But by writing it in Cartesian
form, we can see that we have a quadratic curve with a positive leading coefficient
of π₯. So we know weβre going to have that
typical parabola shape. And we can use our usual methods
for sketching quadratic curves to sketch the curve defined by these parametric
equations.

Letβs see what the whole process
might look like.

Convert the parametric equations π₯
equals three cos π‘ and π¦ equals three sin π‘ to rectangular form.

Remember, the rectangular form of
an equation is one which contains the variables π₯ and π¦ only. So weβll need to find a way to
eliminate the third variable π‘ from our parametric equations. And at first glance, it doesnβt
seem to be a nice way to do so. But letβs recall some trigonometric
identities. We know that cos squared π plus
sin squared π is equal to one. So letβs begin by simply squaring
our expressions for π₯ and π¦.

With π₯, we get π₯ squared equals
three cos π‘ all squared, which is equal to nine cos squared π‘. And so, we can say that cos squared
π‘ must be equal to π₯ squared over nine. Similarly, we can say that π¦
squared is equal to three sin π‘ all squared. We distribute the parentheses and
we find that π¦ squared is nine times sin squared π‘. And then, we divide through by
nine. And we see that sin squared π‘ is
equal to π¦ squared over nine.

Then, if we replace π with π‘ in
our identity, remember that doesnβt change the identity. We see that we can replace cos
squared π‘ with π₯ squared over nine. We can replace sin squared π‘ with
π¦ squared over nine. And then, this is all equal to
one. We next multiply through by
nine. And we find that the rectangular
form of our equation is equal to π₯ squared plus π¦ squared equals nine.

Hence sketch the curve described by
this pair of parametric equations.

Well, weβve just seen that this
pair of parametric equations π₯ equals three cos π‘ and π¦ equals three sin π‘ is π₯
squared plus π¦ squared equals nine in rectangular form. And so, we recall that the equation
of a circle whose centre is at π, π and whose radius is π is π₯ minus π all
squared plus π¦ minus π all squared equals π squared. We compare our equation β thatβs π₯
squared plus π¦ squared equals nine. And in doing so, we see that π and
π simply have to be zero. We also find that π squared is
equal to nine. Well, if we square root this, we
find that π is equal to three. And so, we see that the parametric
equations π₯ equals three cos π‘ and π¦ equals three sin π‘ give a circle centred at
zero, zero β thatβs the origin β with a radius of three. That might look a little something
like this.

We do, however, need to work out
the direction which our curve is being sketched. So letβs work out the value of our
π₯- and π¦-coordinates when π‘ is equal to zero. When π‘ is equal to zero, π₯ is
equal to three times cos of zero, which is simply three. And when π‘ is equal to zero, π¦ is
equal to three sin of zero, which is zero. So we would begin by plotting the
point with Cartesian coordinates three, zero. Letβs next choose π‘ is equal to π
by two.

Now, we could, of course, choose π‘
equals one. But since weβre working with
trigonometric expressions, π by two seems to make a little more sense. So when π‘ is equal to π by two,
π₯ is equal to three cos of π by π‘, which is zero and π¦ is equal to three times
sin of π by two, which is equal to three. So the second point we plot has
Cartesian coordinates zero, three. We can, therefore, see that we move
in a counterclockwise direction while sketching the circle.

Next, weβre going to look at how we
can go up from rectangular into parametric form.

Convert the rectangular equation π₯
squared plus π¦ squared equals 25 to parametric form.

Letβs begin by recalling what we
actually know about this equation. We know that a circle whose centre
is at the origin and whose radius is π can be given by the Cartesian equation π₯
squared plus π¦ squared equals π squared. By rewriting our rectangular
equation as π₯ squared plus π¦ squared equals five squared, we see that we have a
circle whose centre is at the origin and whose radius is five units. And so, Iβve sketched that on the
π₯π¦-plane, as shown.

Weβre looking to convert this
equation to parametric form. So we know that a pair of
parametric equations describe the π₯- and π¦-coordinates in terms of a third
parameter, π‘. So letβs pick a general point π₯,
π¦. Iβm going to choose this one in the
first quadrant. We can drop in a right triangle,
whose height is π¦ units and whose width is π₯ units. And then, we can label the included
angle π‘. Since the radius of the circle is
five units, we know that the hypotenuse of our triangle is five. Then, using standard conventions,
we label the sides of our triangle. We have the adjacent; thatβs
π₯. We have the opposite; thatβs
π¦. We have the hypotenuse; thatβs
five.

We also know that in rectangular
trigonometry, sin π is equal to the opposite divided by the hypotenuse and cos π
is equal to the adjacent over the hypotenuse. So we can say that sin π‘ equals π¦
over five and cos π‘ is equal to π₯ over five. We can multiply through by five for
both of our equations. And we find that π¦ is equal to
five sin π‘ and π₯ equals five cos π‘. So for any point in our circle, the
π¦-coordinate is given by five sin of π‘ and the π₯-coordinate is given by five cos
of π‘. Now moving in a counterclockwise
direction from the positive horizontal axis, we see that as π₯ increases from zero,
it generates our corresponding π₯- and π¦-coordinates. Then, our equations are π₯ equals
five cos π‘ and π¦ equals five sin π‘.

Now, in these previous two
examples, weβve seen that the equation of a circle whose centre is at the origin and
whose radius is π is given by the equation π₯ squared plus π¦ squared equals π
squared. But then, we also saw we can
convert this into parametric form. In this case, we had π¦ equals five
sin π‘ and π₯ equals five cos π‘, where five is the radius.

In general, we can state this as a
result. We define a circle as the locus of
the points that satisfy the equations π₯ equals π cos π‘ and π¦ equals π sin π‘,
where π₯ and π¦ are the coordinates of any point on the circle, π is the radius,
and π‘ is the parameter; itβs the angle subtended by the point at the circle
centre.

We can also use our knowledge of
simple coordinate geometry to parameterize a line segment. Letβs see what that might look
like.

Consider the points π΄ equals
negative one, one and π΅ equals four, two. Parameterize the segment π΄π΅,
where π‘ is greater than or equal to zero and less than or equal to one.

Letβs begin by simply sketching the
diagram, showing our points and our line segment π΄π΅. It looks a little something like
this. Now, we know weβre going to go from
left to right generally. So weβll begin by letting π‘ be
equal to zero at the first point, at negative one, one. And that means we need to let π‘ be
equal to one at the other end of our interval at the point four, two. Now, weβre going to use the vector
form for the equation of a line. Itβs π equals π₯ nought, π¦ nought
plus π‘ times ππ.

Now, whilst this is an equation of
a line in two dimensions, this process can be extended to working with a line in
three dimensions. Weβll next say that the vector π
is of the form π₯, π¦. Then, weβll combine the vectors on
the right to π₯ nought plus π‘π, π¦ nought plus π‘π. And we see that the only way for
the vector on the left to be equal to the vector on the right is if its component
parts are equal. That is, if π₯ is equal to π₯
nought plus π‘π and π¦ is equal to π¦ naught plus π‘π. In fact, this pair of equations is
called the parametric Form of the equation of a line. Now, we can use these along with
the information in our question to parameterize the line segment from π to π.

Letβs go back to what we said
earlier. We said π‘ is equal to zero at
negative one, one. When π‘ is equal to zero, this
gives us the values for π₯ nought, π¦ nought. So π₯ nought must be equal to
negative one and π¦ nought must be equal to one. So we see the π₯ is equal to
negative one plus π‘π and π¦ is equal to one plus π‘π. Weβll now use the fact that when π‘
is equal to one, π₯ is equal to four and π¦ is equal to two. So our first equation becomes four
equals negative one plus one π. And our second equation becomes two
equals one plus one π. By adding one to both sides of our
first equation, we find π to be equal to five. And by subtracting one from both
sides of our second, we find π is equal to one. And we have our parametric
equations to describe the line segment π΄ to π΅ for values of π‘ between zero and
one. They are π₯ equals five π‘ minus
one and π¦ equals π‘ plus one.

In this video, weβve seen that if
we want to convert from a parametric equation to a rectangular equation, we need to
find a way to eliminate π‘. We also saw that, in general, a
circle whose centre is at the origin and whose radius is π is defined by the
parametric equations π₯ equals π cos π‘ and π¦ equals π sin π‘. And finally, we saw that the
parametric equation of a line is π₯ equals π₯ nought plus π‘π and π¦ equals π¦
nought plus π‘π.