## www.mathcentre.ac.uk/.../8.10%20Tangents%20and%20Normals.mp4

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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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a tangent. So when we
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differentiate that's F dashed of
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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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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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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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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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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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Which is 10. What we want is the
equation of the tangent at this
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point on the curve with this
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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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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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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
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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
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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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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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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
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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
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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
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M1 and let's say
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M2 And
the relationship between these
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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,
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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 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
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gradient of 3/4 and so we
can find the equation of the
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Tangent. So let's just list
what we know. We've got the
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tangents at the point.
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And this is 2, five over 2 and
we know that the gradient at
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that point is 3/4.
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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
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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.
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And then let's
multiply both sides
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by this form.
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Now if I
multiply out, the
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brackets that I've
got for, why?
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Now 4 times by minus five over 2
four times by the five is 20
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divided by the two is 10 and
then the minus sign minus 10
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equals 3 X minus six and it will
be nice to be able to get all
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these numbers together. So I'm
going to add 10 to each side
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will give me 4 Y equals 3X plus
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4. So the equation of
the curve is 4 Y
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equals 3X plus 4.
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Now we need to find the
equation of normal to the curve.
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the normal.
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Is M2 and that the gradient
of the tangent?
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Is M1.
And let's recall that normal
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and tangent are perpendicular.
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And the thing that we said about
2 lines that were perpendicular
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at right angles to each other
was that if we multiplied their
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we got was minus one.
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Now we do know what the value of
M1 is. We do know the gradient
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of the tangent is 3/4. So if we
put that into here 3/4 times by
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M2 is equal to minus one.
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So if we multiply it by the four
and divide by the three we get
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M2 is equal to minus four over
three and so now we know the
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gradient of the normal. We also
know the point on the curve that
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still hasn't changed. That still
the .25 over 2.
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So we now want the equation of
the normal. Let's just write
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down what we know. We know the
point that's 25 over 2 an we
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four over 3.
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So our standard equation for
straight line Y minus Y one over
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X Minus X one is equal to the
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Point X one Y one. So
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now we substitute that in Y
minus five over 2 over X.
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Minus two is equal to minus
four over 3.
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Let's multiply up by X
minus two and by three.
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So we have three times Y minus
five over 2 is equal to minus
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four times X minus 2.
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Multiply out the brackets 3 Y
minus 15 over 2 is equal
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to minus four X +8.
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Now let's get things together.
It's awkward having this minus
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4X here, so we'll add 4X to
each side, and we'll add 15 over
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2 to each side, so will have
3 Y plus 4X is equal to
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8 + 15 over 2. Now eight
is 16 over 2. So here I'll
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have 30 one over 2.
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So we'll have 3 Y plus 4X is
31 over 2 and this tools and
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awkward thing. So let's multiply
throughout by two in order to
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get rid of it. 6 Y plus 8X
equals 31 and that's the
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equation of our normal.
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Now when we've got tangents and
normals because the normally
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sort of inside the curve if you
like or passing through the
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curve, sometimes a normal can
actually meet the curve again,
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and we might be interested to
know where it meets that curve.
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So. Question we're going to ask
is if we have the curve XY
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equals 4. And we look at
the point, X equals 2.
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Find the equation of the normal.
Where does the normal meet the
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curve? Again? If it does, well,
let's have a look at a picture
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why equals 4 over X? What does
this look like as a curve?
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OK, if we have a very large
value of X, say 104, / 100 is
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very small, so we got a little
bit of curve down here. If we
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have a very small value of X,
say point nor one.
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Then 4 divided by Point N 1.1 is
100, so 4 / 100 is 400 very
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large, so we're very small
values of X. We've got a bit of
• 25:21 - 25:25
curve there and we can join it
up like that.
• 25:25 - 25:30
Now if we take negative values
of X, the same thing happens,
• 25:30 - 25:35
except we get negative values of
Y and so the curve looks like
• 25:35 - 25:40
that. Notice we do not allow X
to be 0 because we cannot divide
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by zero, so there is a hole in
this curve. There is a gap here
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at X equals 0. Well, here's the
point. Let's say X equals 2.
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That's the point on the curve.
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There is the tangent to the
curve, and there's the normal to
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the curve, and as we can see,
this normal goes through there.
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So the question is where is it
where is that point?
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So first of all, let's establish
what this point is up here so we
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know that X is equal to two and
Y is equal to 4 over 2 equals
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2. So our point is the .22.
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Next, we want the gradient of
the tangent in order that we can
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find the gradient of the normal.
• 26:41 - 26:49
So here we have Y equals 4 over
X, which is 4 times X to the
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minus one. So I'll differentiate
that the why by DX equals minus
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4 multiplied by the minus one
and then taking one of the index
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that gets it to be minus two. So
we have minus four over X
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squared. X is equal
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to 2. And so DY
by the X is minus 4
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over 4 is minus one.
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So what have I got? I've got
the .22 and I've got the
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remember tangent and normal
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are right angles.
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So for two lines at right
angles, M1 times by M2 is minus
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one. I know the value of this,
it's minus one times by N 2
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equals minus one. The only
number M2 can be there is one.
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So now I have got for the normal
I want its equation. I've got
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the point that it goes through
which is 2 two and I've got its
• 28:13 - 28:15
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So I can write down the standard
equation of a straight line Y
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minus Y one over X Minus X one
is equal to M.
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This is my X One Y1.
• 28:30 - 28:38
And I can substitute those in so
I have Y minus two over X, minus
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two is equal to 1.
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Multiply it by X minus two, so
we have Y minus two is equal
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to X minus two, and so why
is equal to X?
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And so now we have the equation
of the normal Y equals X and
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we have the equation of the
curve XY equals 4.
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Now. Where does this line meet
this curve? That's what we're
• 29:14 - 29:18
normal intersect the curve?
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At the points of intersection,
both of these equations are true
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at the same time, so that means
I can take the value of Y, which
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is X and substituted into here.
So therefore I have X times by X
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equals 4 at these points. In
other words, X squared is equal
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to 4. If I take the square root,
that will give me the value of
• 29:48 - 29:51
X. We might think that's just
• 29:51 - 29:57
two. But remember, when you take
a square root you get a plus and
• 29:57 - 30:02
A minus. So we have X equals 2
or minus 2.
• 30:03 - 30:10
And so if X equals 2, then why
must be equal to X? So that
• 30:10 - 30:17
tells us Y equals 2, or and if
we take minus two, Y is equal to
• 30:17 - 30:23
X. That gives us minus two and
so our two points are two 2 and
• 30:23 - 30:25
minus 2 - 2.
• 30:25 - 30:29
Those are the two points where
the normal meets the curve.
• 30:29 - 30:33
Notice this is the first point
that we started off with.
• 30:33 - 30:37
And indeed, when we're doing
this kind of question, the point
• 30:37 - 30:41
where we started off with always
is going to be a part of the
• 30:41 - 30:45
solution. So we've dealt with
applications to tangents and
• 30:45 - 30:50
normals. We've seen that in
order to find the gradient of
• 30:50 - 30:55
the tangent you differentiate,
substituting the value of X and
• 30:55 - 31:01
that gives you the gradient of
the curve and hence the gradient
• 31:01 - 31:06
of the tangent and the other
relationship that we found was
• 31:06 - 31:11
that a normal was perpendicular
to the tangent and that the
• 31:11 - 31:15
product result. Of multiplying
• 31:15 - 31:21
the two lines are perpendicular
was minus one. That's an
• 31:21 - 31:26
important relationship when
we're looking at 2 lines that
• 31:26 - 31:32
are perpendicular, as is the
case for tangent and normal.
Title:
www.mathcentre.ac.uk/.../8.10%20Tangents%20and%20Normals.mp4
Video Language:
English