WEBVTT

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In Differentiation, when
we differentiate A.

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Function F of X.

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So there's our function F of X
and this is a graph of it.

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What do we actually doing? Well,
when we find the derivative were

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actually finding the gradient of
a tangent. So when we

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differentiate that's F dashed of
X, this represents the gradient.

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Of. The curve.

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Or since we have a curve
and the tangent at that point,

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it is the gradient of.

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The tangent so given.

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A point where X is equal to

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A. And therefore where?

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The value of the function is F
of a, then F dashed at that

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point is the gradient.

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Of. The Tangent.

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Where? X equals

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a. And that's what we're going
to be using that the derivative.

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The value of it at the point
where X equals a gives us the

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gradient of the tangent, because
if we know the gradient of the

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tangent and we know the point on
the curve, we can then find the

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equation of this tangent because
it's just a straight line. So

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let's have a look at some
examples. The first example will

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take F of X is equal to X cubed.

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Minus three X squared plus X
minus one, and what we want

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is the equation of the Tangent.

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To this curve at the point where
X equals 3.

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Well, at the point where X
equals 3, so the first thing

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we're going to find is where is
this point on the curve? So we

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take X equals 3 and we'll find F

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of three. So that's
three cubed minus 3

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* 3 squared plus
3 - 1.

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3 cubed is 27. Three squared is
9 and 3 nines are also 27.

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So this builds the 27 -
27 + 3 - 1 altogether.

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We come up with two, so
the point that we're

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actually interested in
on the curve is the .32.

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Now the next thing we need is
the gradient. We've now got a

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point on our line on our
tangent. We now need its

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gradient, so we need to
differentiate F dashed of X. So

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let's differentiate three X
squared for the derivative of X

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cubed. Now we differentiate this
term and so that's minus six X.

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Now this term, so the derivative
of X is.

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One and now the final term, the
derivative of one is 0 because

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it's a constant. We need the
gradient when X is equal to

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three, so we can take that value
X equals 3 and substitute it

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into our expression for the
gradient. And so we have 3 *

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3 squared minus six times.

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3 + 1.

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3 squared is 9 and 3 nines

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are 27. Minus 6 * 3
is minus 18 plus one, and so

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this works out to be 27 takeaway
18. That's nine an add on one

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that's 10. And so we've got a
point, and we've got a gradient,

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so let's write those things
down. We've got our point.

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Which is 3 two.

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And we've got our gradient.

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Which is 10. What we want is the
equation of the tangent at this

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point on the curve with this
gradient. So that's the equation

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of a straight line.

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The equation of a straight line
that goes through a point X one

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Y one is Y minus Y one over
X Minus X one is equal to M

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the gradient. So this is
our X one, Y1. Let's just write

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that in over the top X1Y one
and this is our gradient M.

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So now will substitute these
numbers in so we have Y minus

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two over X. Minus three is equal

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to 10. Now will multiply it
by this X minus three here, so

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we have Y minus two is equal to
10 times X minus three. Now all

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we need to do is multiply out
the bracket Y minus two is 10 X

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minus 30. Remembering to
multiply everything inside this

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bracket by what's outside. And
now we just need to get this two

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over here with this 30.

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So we add two to both
sides. Y equals 10 X, now

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minus 30. Adding onto is minus
28 and there's the equation of

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our tangent. To the curve at
this point on the curve with

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that gradient. OK, let's take
another example. This time.

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Let's say that what we want
to be able to do is

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looking at this curve Y equals
X cubed minus six X squared.

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Plus X +3 and what we want
to find our what are the

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tangents? What are the equations
of the tangents that are

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parallel to the line Y equals X

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+5? So here's our curve and what
we want are the equations of the

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tangents parallel to this line.

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Well, we've got to look at this
line and we've got to extract

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some information from it.

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Information that we can get from
it is what is its gradient,

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because if the tangents have to
be parallel to this line then

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they have to have the same
gradient. And what we can see is

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that we've got One X here and
the standard equation for a

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straight line is Y equals MX
plus. See where this M is the

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gradient. So what we gain from
looking at this standard

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equation and comparing it with
the straight line is that the

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gradient of the straight line M
is equal to 1.

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So now we know what the
gradients of the tangents have

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to be. The gradients have to be
1, so how can we calculate that?

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Well, we know that if we
differentiate this curve its

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equation we will get an
expression for the gradients of

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the tangents. Then we can put it
equal to 1 and solve an equation

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that will give us the points or
at least will give us the X

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coordinates of those points. So
let's do that. We know what. Why

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is. Let's differentiate the why
by DX is.

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Equal to. The derivative of X
cubed is 3 X squared. The

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derivative of minus six X
squared is minus 12 X the

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derivative of X is plus one, and
the derivative of three is 0

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because it's a constant and we
want this expression to be equal

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to 1. So we can take
one away from each side and we

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have three X squared minus 12 X

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equals 0. This is a quadratic
equation for X, which we can

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solve, so let's do that.

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Three X squared minus
12 X equals 0.

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It's a quadratic. The first
question we've got to ask

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ourselves is, does it factor
eyes? And if we look we can see

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that we've got a common factor
in the numbers of three and a

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common factor in the ex terms of
X, so we can take out three X as

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a common factor.

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3X times by something has to
give us three X squared, so

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that's got to be X and three
X times by something has to

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give us minus 12 eggs, so
that's got to be minus four,

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and so that equals 0.

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2 numbers multiplied together.
Give us 0, so one of them has

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to be 0 or both of them have
to be 0. So we can say three

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X equals 0 or X minus 4 equals
0, so therefore X is 0 or X

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equals 4. Now having got these
two values of X, we want the

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points. Where these tangents
were, what we've got now the X

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coordinates. We now need the Y
coordinates and to do that we

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need the equation of the curve
again. So let me just bring back

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the page and it's X cubed minus
six X squared plus X +3.

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So Y equals X cubed minus six
X squared plus X plus three and

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first of all we want to point
X equals 0 and so Y equals

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we put X equals 0.0 cubed. We
put X equals 0 - 6 *

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0 squared plus 0 + 3 each
of these three terms is 0, so

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we. End up with three.

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X equals

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4. Why is
equal to 4 cubed?

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Minus 6 * 4
squared plus 4 +

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3. Equals.
Well, 4 cubed is 4 times

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by 4 times by 4 four
416 and four times by 16

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is 64 takeaway. Now we've got
6 times by 16 and so

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that's 96 + 4 + 3.

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Equals 64 takeaway. 96
is minus 32, and

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then we're adding on
another Seven, so that's

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minus 25. So our
two points are 03.

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And 4 - 25 and those are the
two points where the gradients

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of the tangent are equal to 1
and so where the tangents are

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parallel to the line that we
started out with. That's Y

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equals X +5.

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Let's take another example
and this time I want to

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introduce the word normal.
What's a normal? Well, we

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tend to think of the word
normal in English as mean the

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same everything's alright,
each usual. But in

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mathematics, the word normal
has a very specific meaning.

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It means perpendicular.

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Or at right angles.

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At right
angles.

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So if I have a curve.

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Let's say that's my curve and I
have a tangent at that point

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there. Then what's the normal to
the curve while the normal is at

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right angles to the curve, so
it's also at right angles to the

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tangent, so there's the Tangent.

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To the curve and this is
the normal to the curve normal

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because it's at right angles
perpendicular to the Tangent.

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OK.

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If we can find the equation
of a tangent, we can surely

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find the equation of a
normal, but there is one

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little piece of information
that we need. That is, if we

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have two lines at right
angles, what's the

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relationship between their
gradients?

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So if I take 2 lines there,
right angles to each other,

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and let's say that this line
has gradient.

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M1 and let's say
this line has gradient.

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M2 And
the relationship between these

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two gradients, because they are
at right angles, is that M1

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times by M2 is equal to
minus one, and that's for lines.

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At right angles.

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And so, since tangent normal at
right angles, we can use this

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relationship. We can calculate
the gradient of the tangent and

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use this to find the gradient of

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the normal. Now let's have a
look at that practice.

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Let's take the curve Y equals
X plus one over X.

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At the point where X equals 2,
let's ask ourselves what's the

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equation of the tangent at the
point where X equals 2. What's

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the equation of the normal? So
first of all, let's establish

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what the point is, X equals 2, Y
equals 2 plus one over 2, which

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is 2 1/2.

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But thinking ahead, I think I
would prefer to have that as an

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improper fraction, as five over
2. That's because I'm going to

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have to do some algebra with it
later, and I'd rather keep it is

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5 over 2. Then keep it as 2 1/2.

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OK, next we want the gradient at
this point X equals 2.

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So let me just write down.

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The equation of the curve Y
equals X plus and in order to

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differentiate, be ready to
differentiate the one over X.

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I'm going to write it as X to
the power minus one.

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And now we can differentiate it
DY by DX is equal to the

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derivative of X is one plus
differentiate this we multiply

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by the minus one and we take one
away from the minus one to give

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us a power of minus two. So

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that's one. Plus and A minus
gives us a minus there one over

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X to the minus 2 means one over

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X squared. X is equal

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to 2. And so my
gradient D why by DX is

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equal to 1 - 1 over 4,
two squared is 4 and one

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take away a quarter leaves
me with three quarters.

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So we have got a .25
over 2 and we've got a

00:18:18.970 --> 00:18:26.590
gradient of 3/4 and so we
can find the equation of the

00:18:26.590 --> 00:18:33.600
Tangent. So let's just list
what we know. We've got the

00:18:33.600 --> 00:18:35.908
tangents at the point.

00:18:36.480 --> 00:18:43.480
And this is 2, five over 2 and
we know that the gradient at

00:18:43.480 --> 00:18:45.480
that point is 3/4.

00:18:46.270 --> 00:18:53.030
So the standard equation for a
straight line Y minus Y one over

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X Minus X one is equal to
M the gradient. This is X one

00:19:00.310 --> 00:19:06.550
and This is why one. So now
we can substitute these things

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into here, Y minus five over 2
over X minus two is equal to

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3/4. Let me multiply both
sides by X minus 2.

00:19:20.440 --> 00:19:27.460
And then let's
multiply both sides

00:19:27.460 --> 00:19:30.970
by this form.

00:19:31.470 --> 00:19:38.304
Now if I
multiply out, the

00:19:38.304 --> 00:19:45.138
brackets that I've
got for, why?

00:19:45.680 --> 00:19:51.335
Now 4 times by minus five over 2
four times by the five is 20

00:19:51.335 --> 00:19:56.236
divided by the two is 10 and
then the minus sign minus 10

00:19:56.236 --> 00:20:02.268
equals 3 X minus six and it will
be nice to be able to get all

00:20:02.268 --> 00:20:06.792
these numbers together. So I'm
going to add 10 to each side

00:20:06.792 --> 00:20:09.808
will give me 4 Y equals 3X plus

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4. So the equation of
the curve is 4 Y

00:20:17.450 --> 00:20:20.650
equals 3X plus 4.

00:20:20.650 --> 00:20:28.354
Now we need to find the
equation of normal to the curve.

00:20:28.390 --> 00:20:33.465
Let's say the gradient of
the normal.

00:20:35.720 --> 00:20:41.489
Is M2 and that the gradient
of the tangent?

00:20:42.330 --> 00:20:49.590
Is M1.
And let's recall that normal

00:20:49.590 --> 00:20:52.454
and tangent are perpendicular.

00:20:53.070 --> 00:20:57.966
And the thing that we said about
2 lines that were perpendicular

00:20:57.966 --> 00:21:02.862
at right angles to each other
was that if we multiplied their

00:21:02.862 --> 00:21:06.534
gradients together, the answer
we got was minus one.

00:21:07.110 --> 00:21:13.635
Now we do know what the value of
M1 is. We do know the gradient

00:21:13.635 --> 00:21:20.160
of the tangent is 3/4. So if we
put that into here 3/4 times by

00:21:20.160 --> 00:21:22.770
M2 is equal to minus one.

00:21:23.570 --> 00:21:30.155
So if we multiply it by the four
and divide by the three we get

00:21:30.155 --> 00:21:36.301
M2 is equal to minus four over
three and so now we know the

00:21:36.301 --> 00:21:42.008
gradient of the normal. We also
know the point on the curve that

00:21:42.008 --> 00:21:45.959
still hasn't changed. That still
the .25 over 2.

00:21:47.100 --> 00:21:53.832
So we now want the equation of
the normal. Let's just write

00:21:53.832 --> 00:22:01.686
down what we know. We know the
point that's 25 over 2 an we

00:22:01.686 --> 00:22:06.174
know the gradient that's minus
four over 3.

00:22:06.780 --> 00:22:12.396
So our standard equation for
straight line Y minus Y one over

00:22:12.396 --> 00:22:16.140
X Minus X one is equal to the

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gradient. And this is the
Point X one Y one. So

00:22:23.300 --> 00:22:30.260
now we substitute that in Y
minus five over 2 over X.

00:22:30.260 --> 00:22:35.480
Minus two is equal to minus
four over 3.

00:22:36.140 --> 00:22:43.760
Let's multiply up by X
minus two and by three.

00:22:44.350 --> 00:22:51.840
So we have three times Y minus
five over 2 is equal to minus

00:22:51.840 --> 00:22:54.515
four times X minus 2.

00:22:55.210 --> 00:23:03.190
Multiply out the brackets 3 Y
minus 15 over 2 is equal

00:23:03.190 --> 00:23:06.515
to minus four X +8.

00:23:07.440 --> 00:23:12.700
Now let's get things together.
It's awkward having this minus

00:23:12.700 --> 00:23:20.064
4X here, so we'll add 4X to
each side, and we'll add 15 over

00:23:20.064 --> 00:23:27.428
2 to each side, so will have
3 Y plus 4X is equal to

00:23:27.428 --> 00:23:34.792
8 + 15 over 2. Now eight
is 16 over 2. So here I'll

00:23:34.792 --> 00:23:37.422
have 30 one over 2.

00:23:37.730 --> 00:23:44.795
So we'll have 3 Y plus 4X is
31 over 2 and this tools and

00:23:44.795 --> 00:23:49.976
awkward thing. So let's multiply
throughout by two in order to

00:23:49.976 --> 00:23:56.099
get rid of it. 6 Y plus 8X
equals 31 and that's the

00:23:56.099 --> 00:23:57.983
equation of our normal.

00:23:59.420 --> 00:24:03.610
Now when we've got tangents and
normals because the normally

00:24:03.610 --> 00:24:08.638
sort of inside the curve if you
like or passing through the

00:24:08.638 --> 00:24:12.828
curve, sometimes a normal can
actually meet the curve again,

00:24:12.828 --> 00:24:17.856
and we might be interested to
know where it meets that curve.

00:24:18.440 --> 00:24:24.548
So. Question we're going to ask
is if we have the curve XY

00:24:24.548 --> 00:24:30.655
equals 4. And we look at
the point, X equals 2.

00:24:32.020 --> 00:24:36.820
Find the equation of the normal.
Where does the normal meet the

00:24:36.820 --> 00:24:42.020
curve? Again? If it does, well,
let's have a look at a picture

00:24:42.020 --> 00:24:47.220
why equals 4 over X? What does
this look like as a curve?

00:24:50.630 --> 00:24:57.590
OK, if we have a very large
value of X, say 104, / 100 is

00:24:57.590 --> 00:25:04.086
very small, so we got a little
bit of curve down here. If we

00:25:04.086 --> 00:25:09.190
have a very small value of X,
say point nor one.

00:25:09.860 --> 00:25:16.036
Then 4 divided by Point N 1.1 is
100, so 4 / 100 is 400 very

00:25:16.036 --> 00:25:21.054
large, so we're very small
values of X. We've got a bit of

00:25:21.054 --> 00:25:24.914
curve there and we can join it
up like that.

00:25:25.490 --> 00:25:30.098
Now if we take negative values
of X, the same thing happens,

00:25:30.098 --> 00:25:35.090
except we get negative values of
Y and so the curve looks like

00:25:35.090 --> 00:25:40.466
that. Notice we do not allow X
to be 0 because we cannot divide

00:25:40.466 --> 00:25:46.226
by zero, so there is a hole in
this curve. There is a gap here

00:25:46.226 --> 00:25:51.218
at X equals 0. Well, here's the
point. Let's say X equals 2.

00:25:52.620 --> 00:25:55.956
That's the point on the curve.

00:25:56.600 --> 00:26:02.264
There is the tangent to the
curve, and there's the normal to

00:26:02.264 --> 00:26:07.928
the curve, and as we can see,
this normal goes through there.

00:26:07.928 --> 00:26:13.120
So the question is where is it
where is that point?

00:26:14.410 --> 00:26:21.368
So first of all, let's establish
what this point is up here so we

00:26:21.368 --> 00:26:29.320
know that X is equal to two and
Y is equal to 4 over 2 equals

00:26:29.320 --> 00:26:32.799
2. So our point is the .22.

00:26:33.370 --> 00:26:38.050
Next, we want the gradient of
the tangent in order that we can

00:26:38.050 --> 00:26:40.210
find the gradient of the normal.

00:26:41.020 --> 00:26:48.924
So here we have Y equals 4 over
X, which is 4 times X to the

00:26:48.924 --> 00:26:54.852
minus one. So I'll differentiate
that the why by DX equals minus

00:26:54.852 --> 00:27:01.274
4 multiplied by the minus one
and then taking one of the index

00:27:01.274 --> 00:27:08.190
that gets it to be minus two. So
we have minus four over X

00:27:08.190 --> 00:27:12.118
squared. X is equal

00:27:12.118 --> 00:27:19.702
to 2. And so DY
by the X is minus 4

00:27:19.702 --> 00:27:22.942
over 4 is minus one.

00:27:23.870 --> 00:27:30.669
So what have I got? I've got
the .22 and I've got the

00:27:30.669 --> 00:27:35.376
gradient of the tangent and
remember tangent and normal

00:27:35.376 --> 00:27:36.945
are right angles.

00:27:38.620 --> 00:27:45.419
So for two lines at right
angles, M1 times by M2 is minus

00:27:45.419 --> 00:27:52.741
one. I know the value of this,
it's minus one times by N 2

00:27:52.741 --> 00:27:59.017
equals minus one. The only
number M2 can be there is one.

00:27:59.800 --> 00:28:06.366
So now I have got for the normal
I want its equation. I've got

00:28:06.366 --> 00:28:12.932
the point that it goes through
which is 2 two and I've got its

00:28:12.932 --> 00:28:14.808
gradient which is one.

00:28:15.580 --> 00:28:20.871
So I can write down the standard
equation of a straight line Y

00:28:20.871 --> 00:28:25.755
minus Y one over X Minus X one
is equal to M.

00:28:26.460 --> 00:28:29.490
This is my X One Y1.

00:28:30.460 --> 00:28:37.645
And I can substitute those in so
I have Y minus two over X, minus

00:28:37.645 --> 00:28:40.040
two is equal to 1.

00:28:41.290 --> 00:28:48.416
Multiply it by X minus two, so
we have Y minus two is equal

00:28:48.416 --> 00:28:54.015
to X minus two, and so why
is equal to X?

00:28:54.020 --> 00:29:01.790
And so now we have the equation
of the normal Y equals X and

00:29:01.790 --> 00:29:07.340
we have the equation of the
curve XY equals 4.

00:29:07.980 --> 00:29:13.910
Now. Where does this line meet
this curve? That's what we're

00:29:13.910 --> 00:29:17.915
asking ourselves. Where does the
normal intersect the curve?

00:29:19.190 --> 00:29:24.074
At the points of intersection,
both of these equations are true

00:29:24.074 --> 00:29:30.734
at the same time, so that means
I can take the value of Y, which

00:29:30.734 --> 00:29:36.950
is X and substituted into here.
So therefore I have X times by X

00:29:36.950 --> 00:29:42.278
equals 4 at these points. In
other words, X squared is equal

00:29:42.278 --> 00:29:48.393
to 4. If I take the square root,
that will give me the value of

00:29:48.393 --> 00:29:50.559
X. We might think that's just

00:29:50.559 --> 00:29:56.955
two. But remember, when you take
a square root you get a plus and

00:29:56.955 --> 00:30:01.630
A minus. So we have X equals 2
or minus 2.

00:30:02.970 --> 00:30:09.540
And so if X equals 2, then why
must be equal to X? So that

00:30:09.540 --> 00:30:16.548
tells us Y equals 2, or and if
we take minus two, Y is equal to

00:30:16.548 --> 00:30:23.118
X. That gives us minus two and
so our two points are two 2 and

00:30:23.118 --> 00:30:24.870
minus 2 - 2.

00:30:24.960 --> 00:30:28.931
Those are the two points where
the normal meets the curve.

00:30:28.931 --> 00:30:32.902
Notice this is the first point
that we started off with.

00:30:33.470 --> 00:30:36.627
And indeed, when we're doing
this kind of question, the point

00:30:36.627 --> 00:30:40.645
where we started off with always
is going to be a part of the

00:30:40.645 --> 00:30:45.270
solution. So we've dealt with
applications to tangents and

00:30:45.270 --> 00:30:50.385
normals. We've seen that in
order to find the gradient of

00:30:50.385 --> 00:30:55.035
the tangent you differentiate,
substituting the value of X and

00:30:55.035 --> 00:31:00.615
that gives you the gradient of
the curve and hence the gradient

00:31:00.615 --> 00:31:05.730
of the tangent and the other
relationship that we found was

00:31:05.730 --> 00:31:10.845
that a normal was perpendicular
to the tangent and that the

00:31:10.845 --> 00:31:15.226
product result. Of multiplying
two gradients together, where

00:31:15.226 --> 00:31:20.936
the two lines are perpendicular
was minus one. That's an

00:31:20.936 --> 00:31:26.075
important relationship when
we're looking at 2 lines that

00:31:26.075 --> 00:31:31.785
are perpendicular, as is the
case for tangent and normal.