### Video Transcript

By this stage, you should feel
confident in sketching curves to find by the equation π¦ equals π of π₯. You might look for asymptotes,
intercepts, critical points, and intervals of increase and decrease when doing
so.

In this lesson, weβre going to
learn how to graph curves defined using a third parameter π‘. We see that these curves are
defined parametrically by the equations π₯ equals π of π‘ and π¦ equals π of
π‘. Imagine that a particle moves
through the π₯π¦-plane along a curve π, as shown. It would be impossible to define
this curve in the usual way, π¦ as some function of π₯, as it fails the vertical
line test. That is to say, each unique input
can have more than one output. In other words, for a value of π₯
equals π, we can have more than one output π¦.

So instead, we look to find
alternative ways to describe curves of this form. And we do so by introducing a new
parameter for time π‘. And our π₯- and π¦-coordinates are
functions of time such that π₯ is π of π‘ β π₯ is a function in time β and π¦ is π
of π‘ β π¦ is a different function in time. Each point on our curve will now be
described by an ordered pair, π of π‘, π of π‘. And as π‘ varies, we trace our
curve π, which we call a parametric curve. A key skill is being able to sketch
such curves. And we do so by plotting key
points. There are, of course, a few curves
whose shape we can learn. And weβll consider those too.

Consider the parametric equations
π₯ of π‘ equals π‘ squared plus two and π¦ of π‘ equals three π‘ minus one, which is
greater than negative two and less than one. Which of the following is the
sketch of the given equation?

Here weβve been given a pair of
parametric equations and asked to sketch the curve over the open interval for π‘
from negative two to one. Weβre going to need to begin by
finding some coordinate pairs that satisfy our parametric equations. And whilst we donβt want to include
π‘ equals negative two and π‘ equals one, we know that π‘ approaches both of these
values. So weβll use direct substitution to
find the coordinate pair that our curve approaches at the end points of our open
interval.

Now weβre just looking to identify
which of the graphs is the sketch of our equation. So Iβve chosen seven values of
π‘. If we were looking to sketch the
graph from scratch, we may choose more values of π‘, perhaps choosing to go up in
intervals of 0.25 as supposed to 0.5.

To find our π₯- and π¦-coordinates,
weβre going to substitute each value of π‘ into the parametric equations. For example, when π‘ is negative
two, π₯ is equal to negative two squared plus two, which is six. And when π‘ is equal to negative
two, π¦ is equal to three times negative two minus one, which is negative seven. When π‘ is equal to negative 1.5,
π₯ is equal to negative 1.5 squared plus two, which is 4.25. And when π‘ is equal to negative
1.5, π¦ is equal to three times negative 1.5 minus one, which gives us a π¦-value of
negative 5.5.

We continue this process,
substituting π‘ equals negative one into π₯ to get three and into π¦ to get negative
four. When we substitute π‘ equals
negative 0.5, we get π₯ equals 2.25 and π¦ equals negative 2.5. When π‘ is zero, our coordinate
pair is two, negative one. And the final two values of π‘ give
us coordinate pairs of 2.25, 0.5 and three, two.

We could now plot each ordered pair
on a pair of axes. But of course, weβre looking to
find which of the given sketches represents our pair of parametric equations. The first coordinate that weβre
going to look to find is six, negative seven. Thatβs here. We then look to find 4.25, negative
5.5, which is around here. And is in fact, as we would expect,
the remaining five coordinates do indeed lie on this line sketched.

Notice how the arrows describe the
direction in which the graph is sketched. So our curve is sketched through
increasing values of π‘ from negative two to π‘ equals one. The sketch representing the
parametric equations π₯ of π‘ equals π‘ squared plus two and π¦ of π‘ equals three
π‘ minus one is C.

Now, in fact, it looks like we have
part of a parabola. We could use simultaneous equations
to eliminate π‘. And when we do, we do indeed obtain
the equation for our parabola, or be it a parabola on its side. There are a number of special
parametric equations for which we can learn the shape.

Consider the parametric equations
π₯ of π‘ equals two sin π‘ and π¦ of π‘ equals three cos π‘, where π‘ is greater
than zero and less than three π. Which of the following is the
sketch of the given equations?

Here weβve been given a pair of
parametric equations and asked to find the sketch of the curve over the open
interval for π‘ from zero to three π. Weβre going to need to find some
coordinate pairs that satisfy our parametric equations. Now whilst we donβt technically
want to include π‘ equals zero and π‘ equals three π, we know that π‘ approaches
both of these values. So weβll use direct substitution to
find the coordinate pair that our curve approaches at the end points on our open
interval.

Letβs choose intervals of π by two
radians. Now if we were actually looking to
sketch the curve, we may wish to choose, say, π by four as our subinterval. But weβre just simply looking to
compare our coordinates to the given graphs. To find the first ordered pair, we
substitute π‘ equals zero into each of our parametric equations. That gives us π₯ equals two sin
zero and π¦ equals three cos zero. Our first ordered pair is zero,
three.

We then substitute π‘ equals π by
two. And we get π₯ equals two sin of π
by two and π¦ equals three cos of π by two. That gives us π₯ equals two and π¦
equals zero. We continue the pattern by
substituting π‘ equals π to get two sin π for π₯ and three cos π for π¦, giving
us an ordered pair of zero, negative three. Substituting π‘ equals three π by
two into each equation, and we get π₯ equals negative two and π¦ equals zero. For π‘ equals two π, we get the
ordered pair zero, three. And our last two ordered pairs
occur when π‘ equals five π by two and three π. And they are two, zero and zero,
negative three, respectively.

Letβs compare these to each of our
sketches. And we need to ensure that we move
in increasing values of π‘. That is, we begin at π‘ equals zero
and move all the way through to π‘ equals three π. That leaves us with either B or
C. And in fact, we move in a clockwise
direction. So actually, weβre interested in
C. And you might have noticed that the
values themselves repeat. It should be quite clear now that,
due to the nature of the shape of our curve, this pattern will continue forever. The sketch of the equations π₯ of
π‘ equals two sin π‘ and π¦ of π‘ equals three cos π‘ is C.

Now can you predict what our graph
would look like if our equation for π₯ was in fact three sin π‘? The values for π‘ is π by two, π‘
is three π by two, and π‘ is five π by two would change to three, negative three,
and three, respectively. Comparing these to our curves, we
see that that gives us a circle. Now we can confirm this by
eliminating π‘ and using the identity sin squared π₯ plus cos squared π₯ equals
one. We square both our equations for π₯
and π¦. And we see that π₯ squared plus π¦
squared is equal to nine sin squared π‘ plus nine cos squared π‘. By factoring nine, we can rewrite
this right-hand side as nine sin squared π‘ plus cos squared π‘. And since sin squared π‘ plus cos
squared π‘ will be equal to one, this becomes nine times one, which is nine.

This corresponds to the general
equation for a circle with the center at the origin and a radius π. And we see that, comparing our
graph to our equation, we do indeed have a circle with a center at the origin and a
radius of three. In general, we can say that the
parametric equations of a circle centered at the origin with the radius π are π₯
equals π cos π‘ and π¦ equals π sin π‘. π‘ takes values from zero to two
π. And after this, the pattern
repeats.

This can be further
generalized. And we see that the parametric
equations of a circle with the center at π₯ nought, π¦ nought and a radius π are π₯
equals π₯ nought plus π cos π‘ and π¦ equals π¦ nought plus π sin π‘, for π‘ is
greater than or equal to zero and less than or equal to two π. In our final example, weβre going
to consider a very special curve represented by parametric equations. Itβs called a cardioid.

Sketch the curve defined by the
parametric equations π₯ equals two cos π‘ minus cos two π‘ and π¦ equals two sin π‘
minus sin two π‘, where π‘ is greater than or equal to zero and less than or equal
to two π.

Here weβve been given a pair of
parametric equations and asked to sketch the curve over the closed interval for π‘
from zero to two π. Weβll begin by finding some
coordinate pairs which satisfy our parametric equations. And since weβre sketching the curve
this time rather than just identifying the graph, Iβve chosen values of π‘ at
subintervals of π by four radians. So thatβs zero, π by four, π by
two, three π by four, all the way through to two π.

We begin by substituting π‘ equals
zero into π₯. Thatβs two cos of zero minus cos of
two times zero, which is equal to one. We then substitute zero into the
equation for π¦. Thatβs two sin of zero minus sin of
two times zero, which is zero. By substituting π‘ equals π by
four into π₯, we get two cos of π by four minus cos of two times π by four. And two times π by four is π by
two. This is simply root two.

Repeating this process for π¦, we
get two sin of π by four minus sin of π by two, which is minus one plus root
two. In fact, it will be easier to plot
these if theyβre in decimal form. So rounded to three decimal places,
we have 1.414 and 0.414. The remaining coordinate pairs that
weβre interested in are one, two; negative 1.414, 2.414; negative three, zero; and
so on.

Plotting these on a pair of
coordinate axes gives us something that looks a little like this. We then join the points as shown
and then add arrows to show the direction in which the curve is traced. Remember, we do this by following
the values of π‘ from smallest to largest. And we sketched the curve defined
by the parametric equations π₯ equals two cos π‘ minus cos two π‘ and π¦ equals two
sin π‘ minus sin two π‘. This curve has a special name. Itβs called a cardioid.

You might notice that that name is
similar to the word βcardiac,β meaning heart. And itβs a fascinating shape that
can be created by following the locus of a point on a circle as that circle rolls
around another circle of equal radius.

Now we generally use polar form to
represent these. But in general, the equation of
what we call a horizontal cardioid, like this one, in parametric form is π₯ equals
π times two cos π‘ minus cos two π‘, π¦ equals π two sin π‘ minus sin two π‘. Reversing the π₯ and π¦ so that π₯
was equal to π two sin π‘ minus sin two π‘ and π¦ was equal to π two cos π‘ minus
cos two π‘ would give us a vertical cardioid. Thatβs a curve that looks a little
bit like a heart in the usual orientation we would expect.

In this video, weβve learned that
we can sketch a parametric curve by making a table of values and drawing the
direction in which the curve is traced. We saw that the parametric
equations of a circle with center π₯ nought, π¦ nought and a radius π are π₯ equals
π₯ nought plus π cos π‘ and π¦ equals π¦ nought plus π sin π‘ for values of π‘ in
the closed interval zero to two π. And finally, we saw how a
fascinating shape called a cardioid occurs with equations of the form given. And that if we reverse the
equations for π₯ and π¦, we end up with a cardioid with a different orientation.