### Video Transcript

In this video, we will see how to
apply differentiation to find an equation of a tangent to a curve at a given
point. We’ll also discuss what is meant by
the normal to a curve at a given point and see examples of how to find the equations
of both tangents and normals to curves.

We’ll recall, first of all, that a
tangent to a curve at a particular point is a straight line, which touches the curve
at that point but does not cross the curve. We will recall also that the
gradient or slope of a curve at a given point is defined to be the slope of the
tangent to the curve at that point. Therefore, it follows that if we
can use differentiation to find the gradient function of a curve d𝑦 by d𝑥, then we
can evaluate the slope of the curve and hence the slope of the tangent to the curve
at a given point by substituting the 𝑥-value at that point into our gradient
function d𝑦 by d𝑥.

We’ll recall also that the general
equation of a straight line with slope 𝑚 passing through the point 𝑥 one, 𝑦 one
is 𝑦 minus 𝑦 one equals 𝑚𝑥 minus 𝑥 one. So we can use the slope, which
we’ve calculated, and the coordinates of the point at which we’re looking to find
the tangent in order to find the equation of a tangent.

Let’s see how this works in an
example.

Determine the equation of the line
tangent to the curve 𝑦 equals four 𝑥 cubed minus two 𝑥 squared plus four at the
point negative one, negative two.

So we’ve been given the equation of
a curve. And we need to determine the
equation of the line that is tangent to this curve at a particular point. We’re going to use the formula for
the general equation of a straight line 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥
one. We already know the coordinates 𝑥
one, 𝑦 one. It’s the point negative one,
negative two. But what about 𝑚, the slope of
this line? Well, we recall that the slope of a
curve is equal to the slope of the tangent to the curve at that point. So in order to find the slope of
this tangent, we’re first going to find the gradient function of the curve d𝑦 by
d𝑥.

We can do this by applying the
power rule of differentiation, giving d𝑦 by d𝑥 equals four multiplied by three 𝑥
squared minus two multiplied by two 𝑥. Remember, a constant differentiates
to zero. So that plus four just
differentiates to zero in our derivative, which simplifies to 12𝑥 squared minus
four 𝑥. Now, that’s the general gradient
function of this curve. But we want to know the gradient at
a particular point. So we need to evaluate d𝑦 by d𝑥
when 𝑥 is equal to negative one because that’s our 𝑥-coordinate at this point. This gives 12 multiplied by
negative one squared minus four multiplied by negative one, which simplifies to
16.

We now know that the slope of this
tangent is 16 and the coordinates of a point that it passes through are negative
one, negative two. So we have all the information we
need in order to use the formula for the general equation of a straight line. Substituting the values of 𝑚, 𝑥
one, and 𝑦 one gives 𝑦 minus negative two equals 16 𝑥 minus negative one. That’s 𝑦 plus two equals 16𝑥 plus
16. And, then, subtracting two from
each side in order to collect the constants gives 𝑦 equals 16𝑥 plus 14. So this is the equation of the line
tangent to the given curve at the point negative one, negative two. It passes through the point
negative one, negative two and it has the same gradient as the curve at that
point.

Now, let’s consider a second
example.

Find the point on the curve 𝑦
equals negative 40𝑥 squared plus 40 at which the tangent to the curve is parallel
to the 𝑥-axis.

Now, let’s think about what it
means for a line to be parallel to the 𝑥-axis. The 𝑥-axis is a horizontal
line. So if another line is parallel to
the 𝑥-axis, then it must also be a horizontal line. And what do we know about the
slopes of horizontal lines? Well, they’re equal to zero. So we know that the slope of the
tangent we’re looking to find must be equal to zero. Remember also that the slope of a
tangent is equal to the slope of the curve at that point. So we also know that the slope of
the curve at this point must also be equal to zero.

The slope of the curve is its
gradient function d𝑦 by d𝑥. So what we’re going to do is find
the gradient function d𝑦 by d𝑥 by differentiating 𝑦 with respect to 𝑥 and then
set it equal to zero. We’ll be able to solve the
resulting equation to find the 𝑥-coordinates of the point on the curve where the
gradient is equal to zero. Step one then is to find d𝑦 by
d𝑥, which we can do by applying the power rule of differentiation. It gives negative 40 multiplied by
two 𝑥, which is negative 80𝑥. And remember, the derivative of a
constant is zero. So our gradient function d𝑦 by d𝑥
is just negative 80𝑥.

Then, we set d𝑦 by d𝑥 equal to
zero and solve the resulting equation. We have negative 80𝑥 equals
zero. And by dividing both sides of this
equation by negative 80, we find that 𝑥 is equal to zero. So we know that the 𝑥-coordinate
of the point on this curve at which the tangent is parallel to the 𝑥-axis is
zero. We also need to find the
𝑦-coordinate, which we can do by substituting 𝑥 equals zero back into the equation
of the curve. When 𝑥 is equal to zero, 𝑦 is
equal to negative 40 multiplied by zero squared plus 40 which is equal to 40. So we find that the point on this
curve at which the tangent to the curve is parallel to the 𝑥-axis is the point with
coordinates zero, 40.

Now, we could also have seen this
by considering what the graph of 𝑦 equals negative 40𝑥 squared plus 40 actually
looks like. It’s a negative parabola as the
coefficient of 𝑥 squared is negative 40 and it has a 𝑦-intercept of positive
40. We can see from our sketch that
this function has a critical point at the point with coordinates zero, 40. In fact, it’s a local maximum. At the critical points for a
function, the gradient of the curve and tangent are equal to zero. And so, we see that at the zero, 40
— the critical point of this curve — the tangent at this point will be parallel to
the 𝑥-axis.

Let’s now consider another
example.

The line 𝑥 minus 𝑦 minus three
equals zero touches the curve 𝑦 equals 𝑎𝑥 cubed plus 𝑏𝑥 squared at one,
negative two. Find 𝑎 and 𝑏.

The key information given in this
question is that the line and the curve touch at this point with coordinates one,
negative two. But the line does not cross the
curve, which means that the line 𝑥 minus 𝑦 minus three equals zero is a tangent to
the given curve at this point. We know that the gradient of a
curve is equal to the gradient of the tangent to the curve at that point. The equation of our line is 𝑥
minus 𝑦 minus three equals zero. And when rearranging, we see that
this is equivalent to 𝑦 equals 𝑥 minus three. Comparing with 𝑦 equals 𝑚𝑥 plus
𝑐, that’s the general form of the equation of a straight line in slope-intercept
form, we see that the slope of our tangent is one. Can we find an expression for the
slope of the curve? Well, we can do this by
differentiation. By applying the power rule, we see
that d𝑦 by d𝑥 is equal to three 𝑎𝑥 squared plus two 𝑏𝑥.

Next, we evaluate this gradient
function at the point one, negative, two. So we substitute 𝑥 equals one into
our gradient function, giving three 𝑎 plus two 𝑏. We can then equate the gradient of
the curve at this point with the gradient of the tangent to the curve at this
point. And it gives an equation involving
𝑎 and 𝑏: three 𝑎 plus two 𝑏 is equal to one. We can’t solve this equation
because we have only one equation and two unknowns. So we’re going to need to find a
second equation.

The point one, negative two lies on
both the curve and the tangent. So if we substitute the values of
one and negative two into the equation of the curve, we’ll get a second equation
connecting 𝑎 and 𝑏. We have 𝑎 multiplied by one cubed
plus 𝑏 multiplied by one squared is equal to negative two, simplifying to 𝑎 and 𝑏
equals negative two. We now have two linear equations in
𝑎 and 𝑏, which we need to solve simultaneously. We can multiply equation two by
two, as this will make the coefficient of 𝑏 the same as it is in equation one.

We’ll then subtract the second
equation from the first to eliminate the 𝑏 terms, giving 𝑎 equals five. Substituting this value for 𝑎 into
our original equation two that’s 𝑎 and 𝑏 equals negative two gives five plus 𝑏
equals negative two. And by subtracting five, we see
that 𝑏 is equal to negative seven. So we found the values of 𝑎 and
𝑏. 𝑎 is equal to five and 𝑏 is equal
to negative seven.

A reminder is that the key point
that we used in this question is that the slope of a curve is equal to the slope of
the tangent to the curve at that point.

Let’s consider another type of
example.

Find the equation of the tangent to
the curve 𝑦 equals 𝑥 cubed plus nine 𝑥 squared plus 26𝑥 that makes an angle of
135 degrees with the positive 𝑥-axis.

So we’ve been asked to find the
equation of a tangent to a particular curve, which we know we can do using
differentiation and the general equation of a straight line. But what does it mean when it says
this tangent makes an angle of 135 degrees with the positive 𝑥-axis? Let’s consider a sketch. Well, it will look something like
this. The tangent here is shown in
pink. And we can see that when intersects
with the 𝑥-axis, the angle between the positive 𝑥-axis and the tangent is 135
degrees.

In order to apply the general
equation of a straight line 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one, we either
need to know the slope 𝑚 of our line or the coordinates of a point 𝑥 one, 𝑦 one
which lies on the line. So how does knowing that our
tangent makes an angle of 135 degrees with the positive 𝑥-axis help with
determining either of those? Well, the angle on the other side
of this line will be 45 degrees because we know that angles on a straight line sum
to 180 degrees. We can sketch in a right-angled
triangle below this line and recall that the slope of a line is change in 𝑦 over
change in 𝑥. That’s the vertical height of this
triangle divided by the horizontal distance. But in that right triangle, those
sides are the opposite and adjacent in relation to the angle of 45 degrees. So we’re dividing the length of the
opposite by the length of the adjacent.

As the line is sloping downwards
though, that vertical change is actually the negative of the value of the
opposite. So we have that the slope is equal
to negative opposite over adjacent. Opposite divided by adjacent
defines the tangent ratio. So in fact, this is equal to
negative tan of 45 degrees. And tan of 45 degrees is just
one. So by considering this right-angled
triangle, we found that the slope of this line is negative one. So, we found the slope of our
tangent. But we don’t yet know the
coordinates of the point on the curve where this tangent is being drawn. To find this, we need to find the
point on the curve where the gradient or slope is equal to negative one.

We begin by differentiating the
equation of the curve with respect to 𝑥 and applying the power rule of
differentiation, giving d𝑦 by d𝑥 equals three 𝑥 squared plus 18𝑥 plus 26. We then set this expression equal
to negative one to find the 𝑥-coordinate of the point on the curve, where the
gradient is negative one. This simplifies to three 𝑥 squared
plus 18𝑥 plus 27 equals zero. And then dividing through by three
gives 𝑥 squared plus six 𝑥 plus nine equals zero. We should notice that this is, in
fact, a perfect square. We can write it as 𝑥 plus three
all squared. Solving this equation then, this
means that 𝑥 plus three must be equal to zero. And so, 𝑥 is equal to negative
three.

Next, we need to find the value of
𝑦 when 𝑥 is equal to negative three, which we do by substituting negative three
into the equation of the curve. And it gives negative 24. We now know that this tangent has a
slope of negative one at the point negative three, negative 24. All that’s left is to substitute
into the general equation of the straight line. 𝑦 minus negative 24 equals
negative one multiplied by 𝑥 minus negative three. That all simplifies to 𝑦 plus 𝑥
plus 27 equals zero.

The key steps in this question then
were to use some trigonometric reasoning to identify that if a line makes an angle
of 135 degrees with the positive 𝑥-axis. Then, its gradient or slope is
equal to negative tan 45 degrees, which is equal to negative one. We then use the gradient function
of the curve to identify the 𝑥-value at which the gradient was equal to negative
one. We found the corresponding 𝑦-value
by substituting into the equation of the curve and then finally used the general
equation of a straight line 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one to find the
equation of this tangent.

Finally, in this video, we’re going
to discuss what is meant by a normal to a curve. And we’ll do this in the context of
an example.

List the equations of the normals
to 𝑦 equals 𝑥 squared plus two 𝑥 at the points where the curve meets the line 𝑦
minus four 𝑥 equals zero.

What is meant by the term normal in
this context? Well, we recall first of all that
the tangent to a curve has the same gradient as the curve at that point. The normal, however, passes through
that same point, but it is perpendicular to the tangent at that point. We can use properties of
perpendicular lines to deduce the relationship that exists between the gradient of
the tangent and the gradient of the normal to a curve at a given point. The product of the two gradients
will be equal to negative one and they will be negative reciprocals of one
another.

We must make sure that we’re clear
whether we’ve been asked to find the equation of a tangent or a normal when we’re
answering questions like this. So now that we know what normals
are, let’s see how we can answer this question. We’ve been asked to list the
equations of the normals to a given curve at the point where this curve meets
another line. So our first step is going to be to
find these points of intersection.

We can rearrange the equation of
the line to give 𝑦 equals four 𝑥 and then set the two expressions for 𝑦 equal to
one another to give an equation in 𝑥 only. We can subtract four 𝑥 from each
side and then factor the resulting quadratic to give 𝑥 multiplied by 𝑥 minus two
is equal to zero. The two roots of this equation are
𝑥 equals zero or 𝑥 equals two. So we know the 𝑥-coordinates of
our points of intersection. To find the corresponding
𝑦-coordinates, we substitute each 𝑥-value back into the equation of the curve to
give 𝑦 equals zero when 𝑥 equals zero and 𝑦 equals eight when 𝑥 equals two.

So we now know the two points of
intersection. And we, therefore, know the
coordinates of one point that lies on each normal. But we need to determine the
gradient or slope of each normal. First, we can find the slope of
each tangent by differentiating 𝑦 with respect to 𝑥, giving d𝑦 by d𝑥 equals two
𝑥 plus two. When 𝑥 equals zero, the slope will
be two. And when 𝑥 equals two, the slope
will be six. But remember, this is the slope of
the tangent, not the slope of the normal. To find the slope of each normal,
we need to take the negative reciprocal of the slope of each tangent. So the slope of our first normal is
negative a half and the slope of our second is negative one-sixth.

Finally, we can apply the formula
for the general equation of a straight line. For the first normal with a slope
of negative a half passing through the point zero, zero, we get the equation two 𝑦
plus 𝑥 equals zero. And for the second with a slope of
negative one-sixth passing through the point two, eight, we get the equation six 𝑦
plus 𝑥 minus 50 equals zero. So we found the equations of the
two normals. We must be really careful on
questions like this. Remember, the slope of the normal
is not the same as the slope of the tangent. It’s equal to the negative
reciprocal of the slope of the tangent because the two lines are perpendicular to
one another.

Let’s summarize what we’ve seen in
this video. Firstly, we reminded ourselves that
the gradient of a curve is equal to the gradient of the tangent to the curve at that
point. So by differentiating and then
substituting the 𝑥-value at that point, we can find the slope of the tangent to a
curve at any given point. We can then substitute the slope
and the coordinates of the point into the general equation of a straight line 𝑦
minus 𝑦 one equals 𝑚𝑥 minus 𝑥 one in order to find the equation of the tangent
to the curve at that point.

We also saw that the normal to a
curve is perpendicular to the tangent to the curve at that point. And therefore, the product of their
slopes is equal to negative one. We can apply all of these key
results in order to find the equations of tangents and normals to a variety of
different curves.