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Some relationships between two
quantities or variables are so

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complicated that we sometimes
introduce a third variable or

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quantity to make things easier.

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In mathematics, this third
quantity is called a parameter,

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

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

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relating the parameter with Y.

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Let's have a look at an example.

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X equals Cos T.

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And why it was

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Scienti? It's our parametric
equations we have.

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X.

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

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

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

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

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Let's take some values of tea.

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Calculate X&Y.

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And will take
some values Zero

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Pi 2π? 35 by
2 and 2π.

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And to make it a little bit

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easier. Well, draw the curves.

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Of cause T and sign
TA little bit more.

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Call
Zetty.

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He just put some labels on.

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Now we have one.

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

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And Scienti.

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And.

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This

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scientist.
OK, so when T is 0.

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X is cause T.

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And that's one.

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When she is, OY is scienti
and that zero.

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

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

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20 is π the cause
of Pi is minus one?

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Anne for why?

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The sign of Π Zero.

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

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

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

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of 2π is 0.

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

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To show the curve.

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How
about

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X&Y?

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Once he was zero,
X is one, Y

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is 0, so 10.

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01

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Minus 10.

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0 - 1.

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And back again to 10.

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

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

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

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

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find is that these are the

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parametric equations. That
describe a circle.

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Sensor. 00 and with
a radius of 1.

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

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

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

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

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Eliminate the parameter.

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

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

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

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

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And that's what we do.

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

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

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Is to differentiate each
equation with respect to

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the parameter T.

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

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

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Is cause T?

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Now using the chain rule.

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Which says that DY by
the T is equal to DY

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by DX.

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Times by DX by DT.

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What we have here is DX by DT&EY

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by DT. What we wish to find is

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

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

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own, we divide by the X by DT.

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

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DX by BT.

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So if we now

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substitute. Ty by DT
is cause T.

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And The X by the T is minus sign

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

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Mina Scott T.

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

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bit more complicated.

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The parametric equations
for this example

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RX equals T
cubed minus T&Y

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equals 4 minus
T squared.

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Again, to find the gradient

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

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

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minus one. Until why by duty?

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Is equal to minus 2 T?

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

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DY by DT?

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

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not equal 0.

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

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And EX by DT is
3 two squared minus one.

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

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From The parametric equations.

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But it's in terms
of the parameter T.

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Let's look at

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another example. This
time a parametric equations are

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X equals T cubed.

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And why he cause?

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T squared minus T.

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

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differentiate and find the

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

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

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can plot the curve.

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

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

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

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Takeaway minus 2?

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Four takeaway minus two
gives us 6.

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

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

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one takeaway minus one which
gives us 2.

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Went to 0, then access 0.

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And why is era?

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20 is One X is one.

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I'm 20 is one, Y is

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one. Take away one giving a

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0 again. When T is 2.

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The next is 8.

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

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giving us 2.

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So let's plot curve.

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X axis.

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And IY axis.

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And we've got to go from minus 8

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to +8. So would take fairly

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large. Steps.

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So we plot minus

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eight 6.

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So. Minus 1
two.

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00

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10

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And eight
2.

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So those are curve.

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

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It does look as if that is a

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turning point. But let's
investigate a bit further

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and actually differentiate
these parametric equations.

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So as before the X5

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

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

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T. Minus one.

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

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DY by DT divided by DX

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by beauty. And again,
assuming that the X by

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DT does not equal 0.

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So if we substitute.

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For RDY by DT.

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

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3 T squared.

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

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that in fact when divided by DX

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

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have a stationary point.

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Also, we can see that when.

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T is 0.

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DY by DX is Infinity.

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

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curve at the .00.

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Sometimes it is necessary to
differentiate a second time

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and we can do this with
our parametric equations.

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Let's have a look at a fairly

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straightforward example. X
equals T squared.

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And why equals T cubed?

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

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

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

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

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And why by DT?

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Is equal to three T squared.

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Using the chain rule.

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Dude, why by DX equals.

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Divide by BT.

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Divided by DX by DT.

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

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

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Divided by DX by DT, which is

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

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So we've got three over 2 times

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by teeth. Now applying the
chain rule for a second

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

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

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divided by DX again.

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00:16:03,320 --> 00:16:04,628
And that is.

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00:16:05,130 --> 00:16:10,700
The derivative of divide by DX
with respect to T.

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Divided by DX by DT.

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00:16:15,410 --> 00:16:19,136
Now, just to recap, as YY

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00:16:19,136 --> 00:16:24,167
by ZX. Was equal to three
over 2 T.

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00:16:24,830 --> 00:16:27,350
And our DX by DT.

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00:16:28,330 --> 00:16:29,820
Was equal to 2 T.

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00:16:30,740 --> 00:16:36,380
So now we can do the
substitution and find D2Y by the

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00:16:36,380 --> 00:16:39,159
X squared. Is equal to.

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00:16:40,290 --> 00:16:44,510
The derivative of divide by DX
with respect to T.

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00:16:45,810 --> 00:16:47,620
So that's three over 2.

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00:16:48,640 --> 00:16:52,640
Divided by. DX by BT which is

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00:16:52,640 --> 00:16:59,736
2 T? And that gives
us three over 4T.

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00:17:00,510 --> 00:17:06,402
So do 2 white by the X squared
is 3 / 40.

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00:17:07,360 --> 00:17:14,470
Let's do one more example.
This time are parametric

217
00:17:14,470 --> 00:17:21,580
equation is X equals T
cubed plus 3T squared.

218
00:17:22,610 --> 00:17:29,910
And why equals T to
the Power 4 - 8

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00:17:29,910 --> 00:17:33,770
T squared? So we're
going to

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00:17:33,770 --> 00:17:36,470
differentiate X with
respect to T.

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00:17:38,340 --> 00:17:42,265
Which gives us 3T squared

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00:17:42,265 --> 00:17:45,930
plus 60. And that is why by

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00:17:45,930 --> 00:17:52,204
duty? Is equal to 40
cubed minus 16 T.

224
00:17:52,940 --> 00:18:00,884
Using the chain rule, divide by
DX equals DY by the T

225
00:18:00,884 --> 00:18:04,194
divided by DX by DT.

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00:18:05,030 --> 00:18:08,966
Assuming the exploited seat does
not equal 0.

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00:18:09,480 --> 00:18:17,082
So we get the why by the
T is 40 cubed minus 16 T.

228
00:18:17,730 --> 00:18:24,780
Divided by DX by DT
which is 3 T squared

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00:18:24,780 --> 00:18:28,350
plus 60. Now that let's tidy
this up a bit.

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00:18:28,990 --> 00:18:31,393
And see if there's things
that we can cancel.

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

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

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outside the bracket.

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Inside will have left
TI squared that makes

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40 cubed takeaway 4.

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00:18:56,680 --> 00:19:03,322
Underneath common to both these
parts is 3 T.

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00:19:04,340 --> 00:19:07,466
So take 3T outside of bracket.

238
00:19:08,200 --> 00:19:13,426
And inside we're left with TI so
that when it's multiplied out we

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00:19:13,426 --> 00:19:14,632
get 3T squared.

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00:19:15,140 --> 00:19:21,560
+2 again three 2 * 2
gives us our 60.

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00:19:23,240 --> 00:19:28,220
Now we can go further here
because this one here, T squared

242
00:19:28,220 --> 00:19:32,613
minus 4. Is actually a
difference, the minus the

243
00:19:32,613 --> 00:19:36,294
takeaway between 2 square
numbers? It's a difference of

244
00:19:36,294 --> 00:19:39,746
two squares. So we can express

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00:19:39,746 --> 00:19:46,772
that. As T plus
2 multiplied by T

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00:19:46,772 --> 00:19:52,482
minus 2. And that's going to
help us because we can do some

247
00:19:52,482 --> 00:19:56,170
more. Counseling and make it
simpler for us before we

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00:19:56,170 --> 00:19:57,650
differentiate a second time.

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00:19:58,440 --> 00:20:06,210
So here T goes into T once
2 + 2 goes into 2 +

250
00:20:06,210 --> 00:20:12,870
2 once, so we're left with four
2 - 2 over 3.

251
00:20:14,630 --> 00:20:17,286
Now differentiating a second

252
00:20:17,286 --> 00:20:23,922
time. The two
Y by the X

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00:20:23,922 --> 00:20:31,234
squared. Is the differential
of DY by DX with

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00:20:31,234 --> 00:20:36,738
respect to T divided by
DX by BT.

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00:20:38,600 --> 00:20:45,420
Now recapping from before, let's
just note down the why by

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00:20:45,420 --> 00:20:50,616
DX. What is 4 thirds of
T minus 2?

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00:20:51,620 --> 00:20:54,210
And our DX by DT.

258
00:20:55,350 --> 00:20:58,940
Was 3T squared plus 60.

259
00:21:00,090 --> 00:21:04,626
So differentiating divided by DX
with respect to T.

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00:21:05,420 --> 00:21:07,968
We get 4 thirds.

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00:21:08,550 --> 00:21:11,518
And then we divide by DX by BT.

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00:21:12,330 --> 00:21:15,767
Which is 3 T squared plus 60.

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00:21:16,370 --> 00:21:23,318
So that gives us 4 over
3 lots of three 2 squared

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00:21:23,318 --> 00:21:30,160
plus 60. So do 2 white
by DX squared is equal to.

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00:21:30,730 --> 00:21:37,450
Full And here we can
take another three and a T

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00:21:37,450 --> 00:21:41,590
outside of a bracket to tidy
this up 90.

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00:21:42,350 --> 00:21:45,659
Into T +2.

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00:21:46,710 --> 00:21:49,944
I'm not so there is to it.