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.