WEBVTT

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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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And that is.

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The derivative of divide by DX
with respect to T.

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Divided by DX by DT.

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Now, just to recap, as YY

00:16:19.136 --> 00:16:24.167
by ZX. Was equal to three
over 2 T.

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And our DX by DT.

00:16:28.330 --> 00:16:29.820
Was equal to 2 T.

00:16:30.740 --> 00:16:36.380
So now we can do the
substitution and find D2Y by the

00:16:36.380 --> 00:16:39.159
X squared. Is equal to.

00:16:40.290 --> 00:16:44.510
The derivative of divide by DX
with respect to T.

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So that's three over 2.

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Divided by. DX by BT which is

00:16:52.640 --> 00:16:59.736
2 T? And that gives
us three over 4T.

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

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

00:17:14.470 --> 00:17:21.580
equation is X equals T
cubed plus 3T squared.

00:17:22.610 --> 00:17:29.910
And why equals T to
the Power 4 - 8

00:17:29.910 --> 00:17:33.770
T squared? So we're
going to

00:17:33.770 --> 00:17:36.470
differentiate X with
respect to T.

00:17:38.340 --> 00:17:42.265
Which gives us 3T squared

00:17:42.265 --> 00:17:45.930
plus 60. And that is why by

00:17:45.930 --> 00:17:52.204
duty? Is equal to 40
cubed minus 16 T.

00:17:52.940 --> 00:18:00.884
Using the chain rule, divide by
DX equals DY by the T

00:18:00.884 --> 00:18:04.194
divided by DX by DT.

00:18:05.030 --> 00:18:08.966
Assuming the exploited seat does
not equal 0.

00:18:09.480 --> 00:18:17.082
So we get the why by the
T is 40 cubed minus 16 T.

00:18:17.730 --> 00:18:24.780
Divided by DX by DT
which is 3 T squared

00:18:24.780 --> 00:18:28.350
plus 60. Now that let's tidy
this up a bit.

00:18:28.990 --> 00:18:31.393
And see if there's things
that we can cancel.

00:18:32.970 --> 00:18:38.052
Here at the top we've got 40
cubed takeaway 16 T so common to

00:18:38.052 --> 00:18:44.586
both parts of this is a four and
a T, so if we take four and a T

00:18:44.586 --> 00:18:45.675
outside the bracket.

00:18:46.730 --> 00:18:52.578
Inside will have left
TI squared that makes

00:18:52.578 --> 00:18:55.502
40 cubed takeaway 4.

00:18:56.680 --> 00:19:03.322
Underneath common to both these
parts is 3 T.

00:19:04.340 --> 00:19:07.466
So take 3T outside of bracket.

00:19:08.200 --> 00:19:13.426
And inside we're left with TI so
that when it's multiplied out we

00:19:13.426 --> 00:19:14.632
get 3T squared.

00:19:15.140 --> 00:19:21.560
+2 again three 2 * 2
gives us our 60.

00:19:23.240 --> 00:19:28.220
Now we can go further here
because this one here, T squared

00:19:28.220 --> 00:19:32.613
minus 4. Is actually a
difference, the minus the

00:19:32.613 --> 00:19:36.294
takeaway between 2 square
numbers? It's a difference of

00:19:36.294 --> 00:19:39.746
two squares. So we can express

00:19:39.746 --> 00:19:46.772
that. As T plus
2 multiplied by T

00:19:46.772 --> 00:19:52.482
minus 2. And that's going to
help us because we can do some

00:19:52.482 --> 00:19:56.170
more. Counseling and make it
simpler for us before we

00:19:56.170 --> 00:19:57.650
differentiate a second time.

00:19:58.440 --> 00:20:06.210
So here T goes into T once
2 + 2 goes into 2 +

00:20:06.210 --> 00:20:12.870
2 once, so we're left with four
2 - 2 over 3.

00:20:14.630 --> 00:20:17.286
Now differentiating a second

00:20:17.286 --> 00:20:23.922
time. The two
Y by the X

00:20:23.922 --> 00:20:31.234
squared. Is the differential
of DY by DX with

00:20:31.234 --> 00:20:36.738
respect to T divided by
DX by BT.

00:20:38.600 --> 00:20:45.420
Now recapping from before, let's
just note down the why by

00:20:45.420 --> 00:20:50.616
DX. What is 4 thirds of
T minus 2?

00:20:51.620 --> 00:20:54.210
And our DX by DT.

00:20:55.350 --> 00:20:58.940
Was 3T squared plus 60.

00:21:00.090 --> 00:21:04.626
So differentiating divided by DX
with respect to T.

00:21:05.420 --> 00:21:07.968
We get 4 thirds.

00:21:08.550 --> 00:21:11.518
And then we divide by DX by BT.

00:21:12.330 --> 00:21:15.767
Which is 3 T squared plus 60.

00:21:16.370 --> 00:21:23.318
So that gives us 4 over
3 lots of three 2 squared

00:21:23.318 --> 00:21:30.160
plus 60. So do 2 white
by DX squared is equal to.

00:21:30.730 --> 00:21:37.450
Full And here we can
take another three and a T

00:21:37.450 --> 00:21:41.590
outside of a bracket to tidy
this up 90.

00:21:42.350 --> 00:21:45.659
Into T +2.

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I'm not so there is to it.