In Differentiation, when we differentiate A.
Function F of X.
So there's our function F of X and this is a graph of it.
What do we actually doing? Well, when we find the derivative were
actually finding the gradient of a tangent. So when we
differentiate that's F dashed of X, this represents the gradient.
Of. The curve.
Or since we have a curve and the tangent at that point,
it is the gradient of.
The tangent so given.
A point where X is equal to
A. And therefore where?
The value of the function is F of a, then F dashed at that
point is the gradient.
Of. The Tangent.
Where? X equals
a. And that's what we're going to be using that the derivative.
The value of it at the point where X equals a gives us the
gradient of the tangent, because if we know the gradient of the
tangent and we know the point on the curve, we can then find the
equation of this tangent because it's just a straight line. So
let's have a look at some examples. The first example will
take F of X is equal to X cubed.
Minus three X squared plus X minus one, and what we want
is the equation of the Tangent.
To this curve at the point where X equals 3.
Well, at the point where X equals 3, so the first thing
we're going to find is where is this point on the curve? So we
take X equals 3 and we'll find F
of three. So that's three cubed minus 3
* 3 squared plus 3 - 1.
3 cubed is 27. Three squared is 9 and 3 nines are also 27.
So this builds the 27 - 27 + 3 - 1 altogether.
We come up with two, so the point that we're
actually interested in on the curve is the .32.
Now the next thing we need is the gradient. We've now got a
point on our line on our tangent. We now need its
gradient, so we need to differentiate F dashed of X. So
let's differentiate three X squared for the derivative of X
cubed. Now we differentiate this term and so that's minus six X.
Now this term, so the derivative of X is.
One and now the final term, the derivative of one is 0 because
it's a constant. We need the gradient when X is equal to
three, so we can take that value X equals 3 and substitute it
into our expression for the gradient. And so we have 3 *
3 squared minus six times.
3 + 1.
3 squared is 9 and 3 nines
are 27. Minus 6 * 3 is minus 18 plus one, and so
this works out to be 27 takeaway 18. That's nine an add on one
that's 10. And so we've got a point, and we've got a gradient,
so let's write those things down. We've got our point.
Which is 3 two.
And we've got our gradient.
Which is 10. What we want is the equation of the tangent at this
point on the curve with this gradient. So that's the equation
of a straight line.
The equation of a straight line that goes through a point X one
Y one is Y minus Y one over X Minus X one is equal to M
the gradient. So this is our X one, Y1. Let's just write
that in over the top X1Y one and this is our gradient M.
So now will substitute these numbers in so we have Y minus
two over X. Minus three is equal
to 10. Now will multiply it by this X minus three here, so
we have Y minus two is equal to 10 times X minus three. Now all
we need to do is multiply out the bracket Y minus two is 10 X
minus 30. Remembering to multiply everything inside this
bracket by what's outside. And now we just need to get this two
over here with this 30.
So we add two to both sides. Y equals 10 X, now
minus 30. Adding onto is minus 28 and there's the equation of
our tangent. To the curve at this point on the curve with
that gradient. OK, let's take another example. This time.
Let's say that what we want to be able to do is
looking at this curve Y equals X cubed minus six X squared.
Plus X +3 and what we want to find our what are the
tangents? What are the equations of the tangents that are
parallel to the line Y equals X
+5? So here's our curve and what we want are the equations of the
tangents parallel to this line.
Well, we've got to look at this line and we've got to extract
some information from it.
Information that we can get from it is what is its gradient,
because if the tangents have to be parallel to this line then
they have to have the same gradient. And what we can see is
that we've got One X here and the standard equation for a
straight line is Y equals MX plus. See where this M is the
gradient. So what we gain from looking at this standard
equation and comparing it with the straight line is that the
gradient of the straight line M is equal to 1.
So now we know what the gradients of the tangents have
to be. The gradients have to be 1, so how can we calculate that?
Well, we know that if we differentiate this curve its
equation we will get an expression for the gradients of
the tangents. Then we can put it equal to 1 and solve an equation
that will give us the points or at least will give us the X
coordinates of those points. So let's do that. We know what. Why
is. Let's differentiate the why by DX is.
Equal to. The derivative of X cubed is 3 X squared. The
derivative of minus six X squared is minus 12 X the
derivative of X is plus one, and the derivative of three is 0
because it's a constant and we want this expression to be equal
to 1. So we can take one away from each side and we
have three X squared minus 12 X
equals 0. This is a quadratic equation for X, which we can
solve, so let's do that.
Three X squared minus 12 X equals 0.
It's a quadratic. The first question we've got to ask
ourselves is, does it factor eyes? And if we look we can see
that we've got a common factor in the numbers of three and a
common factor in the ex terms of X, so we can take out three X as
a common factor.
3X times by something has to give us three X squared, so
that's got to be X and three X times by something has to
give us minus 12 eggs, so that's got to be minus four,
and so that equals 0.
2 numbers multiplied together. Give us 0, so one of them has
to be 0 or both of them have to be 0. So we can say three
X equals 0 or X minus 4 equals 0, so therefore X is 0 or X
equals 4. Now having got these two values of X, we want the
points. Where these tangents were, what we've got now the X
coordinates. We now need the Y coordinates and to do that we
need the equation of the curve again. So let me just bring back
the page and it's X cubed minus six X squared plus X +3.
So Y equals X cubed minus six X squared plus X plus three and
first of all we want to point X equals 0 and so Y equals
we put X equals 0.0 cubed. We put X equals 0 - 6 *
0 squared plus 0 + 3 each of these three terms is 0, so
we. End up with three.
X equals
4. Why is equal to 4 cubed?
Minus 6 * 4 squared plus 4 +
3. Equals. Well, 4 cubed is 4 times
by 4 times by 4 four 416 and four times by 16
is 64 takeaway. Now we've got 6 times by 16 and so
that's 96 + 4 + 3.
Equals 64 takeaway. 96 is minus 32, and
then we're adding on another Seven, so that's
minus 25. So our two points are 03.
And 4 - 25 and those are the two points where the gradients
of the tangent are equal to 1 and so where the tangents are
parallel to the line that we started out with. That's Y
equals X +5.
Let's take another example and this time I want to
introduce the word normal. What's a normal? Well, we
tend to think of the word normal in English as mean the
same everything's alright, each usual. But in
mathematics, the word normal has a very specific meaning.
It means perpendicular.
Or at right angles.
At right angles.
So if I have a curve.
Let's say that's my curve and I have a tangent at that point
there. Then what's the normal to the curve while the normal is at
right angles to the curve, so it's also at right angles to the
tangent, so there's the Tangent.
To the curve and this is the normal to the curve normal
because it's at right angles perpendicular to the Tangent.
OK.
If we can find the equation of a tangent, we can surely
find the equation of a normal, but there is one
little piece of information that we need. That is, if we
have two lines at right angles, what's the
relationship between their gradients?
So if I take 2 lines there, right angles to each other,
and let's say that this line has gradient.
M1 and let's say this line has gradient.
M2 And the relationship between these
two gradients, because they are at right angles, is that M1
times by M2 is equal to minus one, and that's for lines.
At right angles.
And so, since tangent normal at right angles, we can use this
relationship. We can calculate the gradient of the tangent and
use this to find the gradient of
the normal. Now let's have a look at that practice.
Let's take the curve Y equals X plus one over X.
At the point where X equals 2, let's ask ourselves what's the
equation of the tangent at the point where X equals 2. What's
the equation of the normal? So first of all, let's establish
what the point is, X equals 2, Y equals 2 plus one over 2, which
is 2 1/2.
But thinking ahead, I think I would prefer to have that as an
improper fraction, as five over 2. That's because I'm going to
have to do some algebra with it later, and I'd rather keep it is
5 over 2. Then keep it as 2 1/2.
OK, next we want the gradient at this point X equals 2.
So let me just write down.
The equation of the curve Y equals X plus and in order to
differentiate, be ready to differentiate the one over X.
I'm going to write it as X to the power minus one.
And now we can differentiate it DY by DX is equal to the
derivative of X is one plus differentiate this we multiply
by the minus one and we take one away from the minus one to give
us a power of minus two. So
that's one. Plus and A minus gives us a minus there one over
X to the minus 2 means one over
X squared. X is equal
to 2. And so my gradient D why by DX is
equal to 1 - 1 over 4, two squared is 4 and one
take away a quarter leaves me with three quarters.
So we have got a .25 over 2 and we've got a
gradient of 3/4 and so we can find the equation of the
Tangent. So let's just list what we know. We've got the
tangents at the point.
And this is 2, five over 2 and we know that the gradient at
that point is 3/4.
So the standard equation for a straight line Y minus Y one over
X Minus X one is equal to M the gradient. This is X one
and This is why one. So now we can substitute these things
into here, Y minus five over 2 over X minus two is equal to
3/4. Let me multiply both sides by X minus 2.
And then let's multiply both sides
by this form.
Now if I multiply out, the
brackets that I've got for, why?
Now 4 times by minus five over 2 four times by the five is 20
divided by the two is 10 and then the minus sign minus 10
equals 3 X minus six and it will be nice to be able to get all
these numbers together. So I'm going to add 10 to each side
will give me 4 Y equals 3X plus
4. So the equation of the curve is 4 Y
equals 3X plus 4.
Now we need to find the equation of normal to the curve.
Let's say the gradient of the normal.
Is M2 and that the gradient of the tangent?
Is M1. And let's recall that normal
and tangent are perpendicular.
And the thing that we said about 2 lines that were perpendicular
at right angles to each other was that if we multiplied their
gradients together, the answer we got was minus one.
Now we do know what the value of M1 is. We do know the gradient
of the tangent is 3/4. So if we put that into here 3/4 times by
M2 is equal to minus one.
So if we multiply it by the four and divide by the three we get
M2 is equal to minus four over three and so now we know the
gradient of the normal. We also know the point on the curve that
still hasn't changed. That still the .25 over 2.
So we now want the equation of the normal. Let's just write
down what we know. We know the point that's 25 over 2 an we
know the gradient that's minus four over 3.
So our standard equation for straight line Y minus Y one over
X Minus X one is equal to the
gradient. And this is the Point X one Y one. So
now we substitute that in Y minus five over 2 over X.
Minus two is equal to minus four over 3.
Let's multiply up by X minus two and by three.
So we have three times Y minus five over 2 is equal to minus
four times X minus 2.
Multiply out the brackets 3 Y minus 15 over 2 is equal
to minus four X +8.
Now let's get things together. It's awkward having this minus
4X here, so we'll add 4X to each side, and we'll add 15 over
2 to each side, so will have 3 Y plus 4X is equal to
8 + 15 over 2. Now eight is 16 over 2. So here I'll
have 30 one over 2.
So we'll have 3 Y plus 4X is 31 over 2 and this tools and
awkward thing. So let's multiply throughout by two in order to
get rid of it. 6 Y plus 8X equals 31 and that's the
equation of our normal.
Now when we've got tangents and normals because the normally
sort of inside the curve if you like or passing through the
curve, sometimes a normal can actually meet the curve again,
and we might be interested to know where it meets that curve.
So. Question we're going to ask is if we have the curve XY
equals 4. And we look at the point, X equals 2.
Find the equation of the normal. Where does the normal meet the
curve? Again? If it does, well, let's have a look at a picture
why equals 4 over X? What does this look like as a curve?
OK, if we have a very large value of X, say 104, / 100 is
very small, so we got a little bit of curve down here. If we
have a very small value of X, say point nor one.
Then 4 divided by Point N 1.1 is 100, so 4 / 100 is 400 very
large, so we're very small values of X. We've got a bit of
curve there and we can join it up like that.
Now if we take negative values of X, the same thing happens,
except we get negative values of Y and so the curve looks like
that. Notice we do not allow X to be 0 because we cannot divide
by zero, so there is a hole in this curve. There is a gap here
at X equals 0. Well, here's the point. Let's say X equals 2.
That's the point on the curve.
There is the tangent to the curve, and there's the normal to
the curve, and as we can see, this normal goes through there.
So the question is where is it where is that point?
So first of all, let's establish what this point is up here so we
know that X is equal to two and Y is equal to 4 over 2 equals
2. So our point is the .22.
Next, we want the gradient of the tangent in order that we can
find the gradient of the normal.
So here we have Y equals 4 over X, which is 4 times X to the
minus one. So I'll differentiate that the why by DX equals minus
4 multiplied by the minus one and then taking one of the index
that gets it to be minus two. So we have minus four over X
squared. X is equal
to 2. And so DY by the X is minus 4
over 4 is minus one.
So what have I got? I've got the .22 and I've got the
gradient of the tangent and remember tangent and normal
are right angles.
So for two lines at right angles, M1 times by M2 is minus
one. I know the value of this, it's minus one times by N 2
equals minus one. The only number M2 can be there is one.
So now I have got for the normal I want its equation. I've got
the point that it goes through which is 2 two and I've got its
gradient which is one.
So I can write down the standard equation of a straight line Y
minus Y one over X Minus X one is equal to M.
This is my X One Y1.
And I can substitute those in so I have Y minus two over X, minus
two is equal to 1.
Multiply it by X minus two, so we have Y minus two is equal
to X minus two, and so why is equal to X?
And so now we have the equation of the normal Y equals X and
we have the equation of the curve XY equals 4.
Now. Where does this line meet this curve? That's what we're
asking ourselves. Where does the normal intersect the curve?
At the points of intersection, both of these equations are true
at the same time, so that means I can take the value of Y, which
is X and substituted into here. So therefore I have X times by X
equals 4 at these points. In other words, X squared is equal
to 4. If I take the square root, that will give me the value of
X. We might think that's just
two. But remember, when you take a square root you get a plus and
A minus. So we have X equals 2 or minus 2.
And so if X equals 2, then why must be equal to X? So that
tells us Y equals 2, or and if we take minus two, Y is equal to
X. That gives us minus two and so our two points are two 2 and
minus 2 - 2.
Those are the two points where the normal meets the curve.
Notice this is the first point that we started off with.
And indeed, when we're doing this kind of question, the point
where we started off with always is going to be a part of the
solution. So we've dealt with applications to tangents and
normals. We've seen that in order to find the gradient of
the tangent you differentiate, substituting the value of X and
that gives you the gradient of the curve and hence the gradient
of the tangent and the other relationship that we found was
that a normal was perpendicular to the tangent and that the
product result. Of multiplying two gradients together, where
the two lines are perpendicular was minus one. That's an
important relationship when we're looking at 2 lines that
are perpendicular, as is the case for tangent and normal.