
Some relationships between two
quantities or variables are so

complicated that we sometimes
introduce a third variable or

quantity to make things easier.

In mathematics, this third
quantity is called a parameter,

and instead of having one
equation, say relating X&Y, we

have two equations, one relating
the parameter with X and one

relating the parameter with Y.

Let's have a look at an example.

X equals Cos T.

And why it was

Scienti? It's our parametric
equations we have.

X.

And T the relationship in one
equation and Y&T related in the

other equation. Let's have a
look at what the graph looks

like and to do that, we
substitute some values for T

into both the equations and we
workout values for X&Y.

Let's take some values of tea.

Calculate X&Y.

And will take
some values Zero

Pi 2π? 35 by
2 and 2π.

And to make it a little bit

easier. Well, draw the curves.

Of cause T and sign
TA little bit more.

Call
Zetty.

He just put some labels on.

Now we have one.

Minus one. So that's
our graph of Costi.

And Scienti.

And.

This

scientist.
OK, so when T is 0.

X is cause T.

And that's one.

When she is, OY is scienti
and that zero.

20 is π by two, X is the cause
of Π by two, which is 0.

20 is π by two, Y is the sign of
Π by two, which is one.

20 is π the cause
of Pi is minus one?

Anne for why?

The sign of Π Zero.

20 is 3 Pi by two the cause of
three Pi by two is 0.

And the sign of three
Pi by two is minus one.

And at 2π. Twenty is 2π the
cause of 2π is one and the sign

of 2π is 0.

So we now have X&Y coordinates
that we can plot.

To show the curve.

How
about

X&Y?

Once he was zero,
X is one, Y

is 0, so 10.

01

Minus 10.

0  1.

And back again to 10.

Now, with those points, we've
not actually plotted enough to

be able to see what's happening
in between these points, but if

we were to take values for T
between 0 and Π by two and some

more between pie by two and Π
and so on, what we'd actually

find is that these are the

parametric equations. That
describe a circle.

Sensor. 00 and with
a radius of 1.

Now what we often want to find
out is how to variables are

changing in relationship to each
other. So when exchange is, how

is why changing what's the rate
of change? So we need to be able

to differentiate. Now what we
don't want to do is to actually.

Eliminate the parameter.

And get back to an equation
directly relating X&Y, 'cause

the whole point of having it's a
parameter is that it makes it

easier for us and simpler, So
what we need to do is to find a

way of differentiating when we
got them in the parametric form.

And that's what we do.

Thanks, right, the two equations
again X equals Cos T.

Y equals sign T. What we're
going to do to differentiate?

Is to differentiate each
equation with respect to

the parameter T.

So the X5 DT, the derivative
of Cos T is minus sign

T. 4 divided by DT.
The derivative of Scienti.

Is cause T?

Now using the chain rule.

Which says that DY by
the T is equal to DY

by DX.

Times by DX by DT.

What we have here is DX by DT&EY

by DT. What we wish to find is

divided by DX. So if we
rearrange that equation D, why

by DX is multiplied by DX by DT
so to get divided by DX on its

own, we divide by the X by DT.

We have divide by DX equals
DY by DT all divided by

DX by BT.

So if we now

substitute. Ty by DT
is cause T.

And The X by the T is minus sign

T. So what we have is
the derivative divided by DX is

Mina Scott T.

Let's look at another
example. One is a little

bit more complicated.

The parametric equations
for this example

RX equals T
cubed minus T&Y

equals 4 minus
T squared.

Again, to find the gradient

function. Of the equation, we're
going to differentiate each with

respect to the parameter, so DX
by DT is 3 T squared

minus one. Until why by duty?

Is equal to minus 2 T?

Again, using the chain rule
D, why by DX equals

DY by DT?

Divided by the X by DT. That
is assuming that DX by DT does

not equal 0.

Let's substituting do why by DT
is minus 2 T.

And EX by DT is
3 two squared minus one.

So again, we found the gradient
function of the curve.

From The parametric equations.

But it's in terms
of the parameter T.

Let's look at

another example. This
time a parametric equations are

X equals T cubed.

And why he cause?

T squared minus T.

So let's have a look at what
this curve looks like before we

differentiate and find the

gradient function. So we're
going to substitute for some

values of tea again to workout
some values of X&Y so that we

can plot the curve.

Let's take values of tea from
minus two through to two.

So when T is minus, 2X is minus
2 cubed, which is minus 8.

When T is minus two, Y is minus
2 squared, which is 4.

Takeaway minus 2?

Four takeaway minus two
gives us 6.

20 is minus One X is minus 1
cubed, which is minus one.

20 is minus one, Y is going to
be minus one squared, which is

one takeaway minus one which
gives us 2.

Went to 0, then access 0.

And why is era?

20 is One X is one.

I'm 20 is one, Y is

one. Take away one giving a

0 again. When T is 2.

The next is 8.

And when T is 2, Y is 2
squared, four takeaway, two

giving us 2.

So let's plot curve.

X axis.

And IY axis.

And we've got to go from minus 8

to +8. So would take fairly

large. Steps.

So we plot minus

eight 6.

So. Minus 1
two.

00

10

And eight
2.

So those are curve.

And here we're not. Perhaps
certain what happens.

It does look as if that is a

turning point. But let's
investigate a bit further

and actually differentiate
these parametric equations.

So as before the X5

ET. The derivative of T cubed is
3 two squared.

And if we look at the why by DT,
the derivative of T squared is 2

T. Minus one.

Again, using the chain rule
divide by DX is equal to

DY by DT divided by DX

by beauty. And again,
assuming that the X by

DT does not equal 0.

So if we substitute.

For RDY by DT.

We get 2T minus 1 divided
by R DX by DT, which is

3 T squared.

From this we can analyze the
curve further and we can see

that in fact when divided by DX

is 0. Then T must be 1/2,
so in this section here we do

have a stationary point.

Also, we can see that when.

T is 0.

DY by DX is Infinity.

So we have got the Y axis
here being a tangent to the

curve at the .00.

Sometimes it is necessary to
differentiate a second time

and we can do this with
our parametric equations.

Let's have a look at a fairly

straightforward example. X
equals T squared.

And why equals T cubed?

And what we're going to do is to
differentiate using the chain

rule, as we've done before, and
then we're going to apply the

chain rule the second time to
find the two. Why by DX squared.

So starting us before DX bite
beauty is equal to T.

And why by DT?

Is equal to three T squared.

Using the chain rule.

Dude, why by DX equals.

Divide by BT.

Divided by DX by DT.

And assuming, of course that the
X by DT does not equal 0.

So let's substitute for divide
by DT. It's 3T squared.

Divided by DX by DT, which is

2 two. And here at TI
goes into 2 squared two times.

So we've got three over 2 times

by teeth. Now applying the
chain rule for a second

time. We have the two
Y by DX squared equals D

by DX of divide by DX
'cause we need to differentiate

divided by DX again.

And that is.

The derivative of divide by DX
with respect to T.

Divided by DX by DT.

Now, just to recap, as YY

by ZX. Was equal to three
over 2 T.

And our DX by DT.

Was equal to 2 T.

So now we can do the
substitution and find D2Y by the

X squared. Is equal to.

The derivative of divide by DX
with respect to T.

So that's three over 2.

Divided by. DX by BT which is

2 T? And that gives
us three over 4T.

So do 2 white by the X squared
is 3 / 40.

Let's do one more example.
This time are parametric

equation is X equals T
cubed plus 3T squared.

And why equals T to
the Power 4  8

T squared? So we're
going to

differentiate X with
respect to T.

Which gives us 3T squared

plus 60. And that is why by

duty? Is equal to 40
cubed minus 16 T.

Using the chain rule, divide by
DX equals DY by the T

divided by DX by DT.

Assuming the exploited seat does
not equal 0.

So we get the why by the
T is 40 cubed minus 16 T.

Divided by DX by DT
which is 3 T squared

plus 60. Now that let's tidy
this up a bit.

And see if there's things
that we can cancel.

Here at the top we've got 40
cubed takeaway 16 T so common to

both parts of this is a four and
a T, so if we take four and a T

outside the bracket.

Inside will have left
TI squared that makes

40 cubed takeaway 4.

Underneath common to both these
parts is 3 T.

So take 3T outside of bracket.

And inside we're left with TI so
that when it's multiplied out we

get 3T squared.

+2 again three 2 * 2
gives us our 60.

Now we can go further here
because this one here, T squared

minus 4. Is actually a
difference, the minus the

takeaway between 2 square
numbers? It's a difference of

two squares. So we can express

that. As T plus
2 multiplied by T

minus 2. And that's going to
help us because we can do some

more. Counseling and make it
simpler for us before we

differentiate a second time.

So here T goes into T once
2 + 2 goes into 2 +

2 once, so we're left with four
2  2 over 3.

Now differentiating a second

time. The two
Y by the X

squared. Is the differential
of DY by DX with

respect to T divided by
DX by BT.

Now recapping from before, let's
just note down the why by

DX. What is 4 thirds of
T minus 2?

And our DX by DT.

Was 3T squared plus 60.

So differentiating divided by DX
with respect to T.

We get 4 thirds.

And then we divide by DX by BT.

Which is 3 T squared plus 60.

So that gives us 4 over
3 lots of three 2 squared

plus 60. So do 2 white
by DX squared is equal to.

Full And here we can
take another three and a T

outside of a bracket to tidy
this up 90.

Into T +2.

I'm not so there is to it.