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## www.mathcentre.ac.uk/.../8.7%20Parametric%20Differentiation.mp4

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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
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by ZX. Was equal to three
over 2 T.
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And our DX by DT.
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Was equal to 2 T.
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So now we can do the
substitution and find D2Y by the
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X squared. Is equal to.
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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
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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
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equation is X equals T
cubed plus 3T squared.
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And why equals T to
the Power 4 - 8
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T squared? So we're
going to
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differentiate X with
respect to T.
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Which gives us 3T squared
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plus 60. And that is why by
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duty? Is equal to 40
cubed minus 16 T.
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Using the chain rule, divide by
DX equals DY by the T
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divided by DX by DT.
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Assuming the exploited seat does
not equal 0.
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So we get the why by the
T is 40 cubed minus 16 T.
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Divided by DX by DT
which is 3 T squared
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plus 60. Now that let's tidy
this up a bit.
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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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Underneath common to both these
parts is 3 T.
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So take 3T outside of bracket.
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And inside we're left with TI so
that when it's multiplied out we
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get 3T squared.
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+2 again three 2 * 2
gives us our 60.
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Now we can go further here
because this one here, T squared
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minus 4. Is actually a
difference, the minus the
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takeaway between 2 square
numbers? It's a difference of
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two squares. So we can express
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that. As T plus
2 multiplied by T
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minus 2. And that's going to
help us because we can do some
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more. Counseling and make it
simpler for us before we
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differentiate a second time.
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So here T goes into T once
2 + 2 goes into 2 +
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2 once, so we're left with four
2 - 2 over 3.
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Now differentiating a second
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time. The two
Y by the X
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squared. Is the differential
of DY by DX with
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respect to T divided by
DX by BT.
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Now recapping from before, let's
just note down the why by
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DX. What is 4 thirds of
T minus 2?
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And our DX by DT.
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Was 3T squared plus 60.
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So differentiating divided by DX
with respect to T.
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We get 4 thirds.
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And then we divide by DX by BT.
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Which is 3 T squared plus 60.
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So that gives us 4 over
3 lots of three 2 squared
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plus 60. So do 2 white
by DX squared is equal to.
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Full And here we can
take another three and a T
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outside of a bracket to tidy
this up 90.
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Into T +2.
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I'm not so there is to it.
Title:
www.mathcentre.ac.uk/.../8.7%20Parametric%20Differentiation.mp4
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