CHAPTER 1
Curves and surfaces
Recall that a function is a rule 𝑓 that assigns to each element 𝑥 of a set 𝐴 an element
𝑦 of a set 𝐵. The set 𝐴 is the domain of the function, while 𝐵 is the codomain. For each
𝑥 ∈ 𝐴, the value of the function at 𝑥 is 𝑦 = 𝑓 (𝑥). Notice that the domain 𝐴 consists of
all the 𝑥’s for which 𝑓 (𝑥) is defined. Sometimes, the set 𝐴 is not stated explicitly, but
has to be deduced from the context. The set of all 𝑦’s which are actually values of the
function 𝑓 is the range of 𝑓 ; this is the set
{ 𝑓 (𝑥) : 𝑥 ∈ 𝐴} .
There is a subtle difference between the range and the codomain; the two are the same
if and only if the function is onto.
In MAM1020, you worked mainly with functions whose domains and codomains were
subsets of R. Such a function has a graph
{(𝑥, 𝑦) : 𝑥 ∈ 𝐴, 𝑦 = 𝑓 (𝑥)} ,
which is a subset of the cartesian plane. We can often deduce the properties of a
function from its graph; which is why there is so much emphasis in first year on
"sketching the graph of the function".
Example 1.1.
𝑦
𝑦 = 𝑓 (𝑥)
𝑎
𝑐
𝑑
𝑒
This function is continuous at 𝑥 = 𝑐 because
lim 𝑓 (𝑥) = 𝑓 (𝑐).
𝑥→𝑐
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𝑏
𝑥
It is differentiable at 𝑥 = 𝑒. At the point (𝑒, 𝑓 (𝑒)), there is a tangent line with slope 𝑓 ′(𝑒).
This function is Riemann integrable on the interval [𝑎, 𝑏], with
∫ 𝑏
𝑎
𝑓 (𝑥) d𝑥 =
∫ 𝑑
𝑎
𝑓 (𝑥) d𝑥 +
∫ 𝑏
𝑑
𝑓 (𝑥) d𝑥.
There is a discontinuity at 𝑥 = 𝑑, since
lim 𝑓 (𝑥) ≠ lim+ 𝑓 (𝑥).
𝑥→𝑑 −
𝑥→𝑑
Of course, if a function is too "wild", for example, if its domain can’t be broken into a
few intervals in R, or if it has lots of discontinuities (like Dirichlet’s function), then we
can’t draw its graph. Our preference is to work with functions such as the one above
whose graphs are made up of some nice smooth curves in R2 , but we don’t always get
what we want.
In MAM2083, we shall study functions whose domains are subsets of R𝑛 and whose
codomains are subsets of R𝑚 . If 𝑛 > 1, then we are speaking about a "function of
several variables". If 𝑚 > 1, then the function is "vector-valued".
1.1. Curves and surfaces in R3
1.1.1. Functions of two variables. Let’s start with functions of the form
𝑧 = 𝑓 (𝑥, 𝑦).
This is a function of two variables. We call 𝑥 and 𝑦 the independent variables, while 𝑧
is the dependent variable. The domain of this function consists of all (𝑥, 𝑦) for which
𝑓 (𝑥, 𝑦) is defined. Again, this could be a "weird" subset of R2 , but our preference is to
work with functions whose domains are made up of nice simple things like polygons,
or sectors of circles, and so on. Sometimes it is useful to sketch the domain as a subset
of the cartesian plane:
𝑦
(𝑎, 𝑏)
𝑥
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At other times, we prefer to draw the domain as if it is lying in the plane 𝑧 = 0 in R3 .
𝑧
𝑦
(𝑎, 𝑏, 0)
𝑥
Since the values of this function are numbers, its graph {(𝑥, 𝑦, 𝑧) : (𝑥, 𝑦) ∈ 𝐴, 𝑧 = 𝑓 (𝑥, 𝑦)}
is a subset of R3 . As long as 𝑓 (𝑥, 𝑦) is a "nice" function, its graph is some kind of a
surface in R3 .
𝑧
𝑧 = 𝑓 (𝑥, 𝑦)
(𝑎, 𝑏, 𝑓 (𝑎, 𝑏))
𝑦
(𝑎, 𝑏, 0)
𝑥
There are many graphics programs that enable you to plot the graph of a function
𝑧 = 𝑓 (𝑥, 𝑦) and it is often convenient to use them. However, there is still some merit
to learning how to sketch the graphs of some simple functions, so that you get used to
working with them and discover what to look out for. Obviously, if a function is too
complicated, you won’t manage to sketch its graph. There is always a stage when one
runs up against the limitations of one’s artistic ability.
Definition 1.2. Given any point (𝑎, 𝑏) in the domain of a function 𝑧 = 𝑓 (𝑥, 𝑦), we can
consider the two partial functions 𝑧 = 𝑓 (𝑥, 𝑏) and 𝑧 = 𝑓 (𝑎, 𝑦).
For the first partial function, we are fixing the value of 𝑦 and treating 𝑧 as if it were a
function of 𝑥 only. We get the second partial function by fixing 𝑥 and pretending that 𝑧
is a function of 𝑦 only. The two partial functions are each functions of a single variable.
Notice that the plane 𝑦 = 𝑏 is a vertical plane parallel to the 𝑥𝑧-plane. Its intersection
with the 𝑥𝑦-plane is the line parallel to the 𝑥-axis that passes through the point (𝑎, 𝑏, 0).
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The intersection of this plane with the graph of 𝑧 = 𝑓 (𝑥, 𝑦) is a curve.
𝑧
trace
𝑧 = 𝑓 (𝑥, 𝑦)
(𝑎, 𝑏, 𝑓 (𝑎, 𝑏))
(0, 𝑏, 0)
𝑦
(𝑎, 𝑏, 0)
𝑥
Definition 1.3. We call this curve the trace of 𝑧 = 𝑓 (𝑥, 𝑦) in the plane 𝑦 = 𝑏.
If you think it, you will realise that the trace of 𝑧 = 𝑓 (𝑥, 𝑦) in the plane 𝑦 = 𝑏 is identical
to the graph of the partial function 𝑧 = 𝑓 (𝑥, 𝑏). You can see this by looking at the
surface from the perspective of a point very far away on the 𝑦-axis:
𝑧
𝑧 = 𝑓 (𝑥, 𝑏)
(𝑎, 𝑏, 𝑓 (𝑎, 𝑏))
(𝑎, 𝑏, 0)
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𝑥
In the same way, you can consider the trace of 𝑧 = 𝑓 (𝑥, 𝑦) in the plane 𝑥 = 𝑎. This is
the intersection of the surface with a vertical plane parallel to the 𝑦𝑧-plane.
𝑧
trace
𝑧 = 𝑓 (𝑥, 𝑦)
(𝑎, 𝑏, 𝑓 (𝑎, 𝑏))
𝑦
(𝑎, 0, 0)
(𝑎, 𝑏, 0)
𝑥
This curve is a copy of the graph of the partial function 𝑧 = 𝑓 (𝑎, 𝑦), except that it lies
in the plane 𝑥 = 𝑎. You can see this by looking at the surface from the perspective of a
point very far away on the 𝑥-axis:
𝑧
𝑧 = 𝑓 (𝑎, 𝑦)
(𝑎, 𝑏, 𝑓 (𝑎, 𝑏))
(𝑎, 𝑏, 0)
𝑦
By sketching the traces of 𝑧 = 𝑓 (𝑥, 𝑦) in sufficiently many planes, you can get a pretty
good idea what the surface looks like. This is the approach which many graphics
programs use, when they produce "wireframes" of the surface. Of course, you aren’t
restricted to using planes that are of the form 𝑥 = 𝑎 or 𝑦 = 𝑏. Sometimes, you get
very useful information by considering the intersection of 𝑧 = 𝑓 (𝑥, 𝑦) with a horizontal
plane of the form 𝑧 = 𝑐, where 𝑐 is chosen to be a number in the range of the function
𝑓 . The difference now is that we aren’t just looking at the graph of a partial function;
in fact, the curve 𝑓 (𝑥, 𝑦) = 𝑐 might not be the graph of a function, and it might be
difficult to solve this equation. However, if you can work out what these "horizontal
cross-sections" look like, they help you to sketch the surface.
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Closely related to these traces are the level curves or contours of the function 𝑧 = 𝑓 (𝑥, 𝑦).
These are curves of the form 𝑓 (𝑥, 𝑦) = 𝑐, but in the cartesian plane. In other words,
they live in 2D, while the traces 𝑓 (𝑥, 𝑦) = 𝑐 lie on the surface 𝑧 = 𝑓 (𝑥, 𝑦), and therefore
live in 3D. Remember that the domain of the function 𝑓 is a subset of R2 . Each point
(𝑎, 𝑏) in this domain lies on a level curve 𝑓 (𝑥, 𝑦) = 𝑓 (𝑎, 𝑏). The corresponding point
(𝑎, 𝑏, 𝑓 (𝑎, 𝑏)) on the graph of 𝑧 = 𝑓 (𝑥, 𝑦) lies on the trace of 𝑧 = 𝑓 (𝑥, 𝑦) in the horizontal
plane 𝑧 = 𝑓 (𝑎, 𝑏).
𝑧
(0, 0, 𝑓 (𝑎, 𝑏))
trace
(𝑎, 𝑏, 𝑓 (𝑎, 𝑏))
𝑦
(𝑎, 𝑏, 0)
𝑥
Very often, one can deduce the behaviour of the function 𝑧 = 𝑓 (𝑥, 𝑦) by looking at its
level curves. If you sketch enough of them, you get a "contour map" of the surface,
just like you used to do in Geography: the value of 𝑧 = 𝑓 (𝑥, 𝑦) represents height
above or below the 𝑥𝑦-plane. Using such a map, you can work out in which directions
the function is increasing or decreasing, and see where it is likely to have maxima or
minima.
A very simple but important idea is that on any contour 𝑓 (𝑥, 𝑦) = 𝑐 in the cartesian
plane, the "height" 𝑓 (𝑥, 𝑦) stays constant. This is why we call the contours "level
curves".
Example 1.4. Consider the function 𝑧 = 𝑥 2 + 𝑦 2 − 6.
The first thing to note is that this function is defined for all (𝑥, 𝑦) ∈ R2 . Since its domain
is unbounded, we can’t sketch the entire graph 𝑧 = 𝑥 2 + 𝑦 2 − 6, but only a portion of
this surface.
The trace of this function in the plane 𝑦 = 3 is a parabola, 𝑧 = 𝑥 2 + 3. In the plane
𝑥 = 1, its trace is another parabola, 𝑧 = 𝑦 2 − 5. In the plane 𝑧 = 4, the trace is a circle,
𝑥 2 + 𝑦 2 = 10.
More generally, the trace of this function in any vertical plane of the form 𝑥 = 𝑎 or
𝑦 = 𝑏 is a parabola lying in that plane. All these parabolas have essentially the same
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shape, as they are either of the form 𝑧 = 𝑦 2 + 𝑘 or of the form 𝑧 = 𝑥 2 + 𝑘 for suitable
values of 𝑘. On the other hand, if we choose a number 𝑐 > −6, then the trace of
𝑧 = 𝑥 2 + 𝑦 2 − 6 in the plane 𝑧 = 𝑐 is a circle in that plane, centred at the point (0, 0, 𝑐)
√
and with radius 𝑐 + 6. If 𝑐 < −6, then the equation 𝑥 2 + 𝑦 2 − 6 = 𝑐 doesn’t have any
solutions. If 𝑐 = −6, then the trace consists of precisely one point, (0, 0, −6). The range
of the function 𝑧 = 𝑥 2 + 𝑦 2 − 6 is the interval [−6, ∞).
Since the vertical traces are all parabolas, while the horizontal traces are circles, we call
the graph of 𝑧 = 𝑥 2 + 𝑦 2 − 6 a circular paraboloid. Less formally, you could describe it as
being a "bowl". Its vertex is the point (0, 0, −6).
𝑧
𝑧 = 𝑥2 + 𝑦2 − 6
𝑦
𝑥
(0, 0, −6)
Remember that the level curves for this function lie in the cartesian plane (i.e., R2 ).
They make up a family of concentric circles, centred at (0, 0).
𝑦
𝑧 = −6
𝑧 = −5
𝑧 = −4
𝑧 = −3
𝑥
𝑧 = −2
𝑧 = −1
𝑧=0
The fact that the level curves are a family of concentric circles shows that the graph of
this function is a surface of revolution, generated in this case by rotating a parabola
𝑧 = 𝑥 2 − 6 about the 𝑧-axis. The fact that these circles get closer together as one moves
way from their centre means that in R3 the surface gets steeper and steeper as one
moves way from the 𝑧-axis.
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Example 1.5. Consider the function 𝑧 = 1 − 𝑥 2 − 4𝑦 2 .
Again, this is defined for all (𝑥, 𝑦) ∈ R2 . In this case, the range of the function
is the interval (−∞, 1]. For each (𝑎, 𝑏) ∈ R2 , the partial functions are of the form
𝑧 = (1 − 4𝑏) − 𝑥 2 and 𝑧 = 1 − 𝑥 2 − 4𝑦 2 , so the traces of this function in the vertical
planes 𝑦 = 𝑏 and 𝑥 = 𝑎 are parabolas. This tells us that the graph is again some kind
of paraboloid. However, this time the parabolas all open downwards, so the "bowl" is
upside down.
Notice also that the level curves of this function are of the form
1 − 𝑥 2 − 4𝑦 2 = 𝑐,
where 𝑐 ≤ 1. If 𝑐 < 1, then this equation can be written as
𝑦2
𝑥2
+
= 1,
1 − 𝑐 (1 − 𝑐) /4
so the level curves are a family of ellipses, centred at (0, 0), with their major axes along
the 𝑥-axis and their minor axes along the 𝑦-axis.
𝑧=1
𝑧=0
𝑧 = −1
𝑦
𝑧 = −2
𝑧 = −3
𝑥
𝑧 = −4
𝑧 = −5
𝑧 = −6
𝑧 = −7
𝑧 = −8
The fact that these curves aren’t circles tells us that this graph isn’t a surface of revolution. Since its horizontal cross-sections are ellipses, it is an elliptical paraboloid. Its
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vertex is at (0, 0, 1).
𝑧
(0, 0, 1)
𝑧 = 1 − 𝑥 2 − 4𝑦 2
𝑦
𝑥
Example 1.6. Consider the function 𝑓 (𝑥, 𝑦) =
p
𝑥2 + 𝑦2 .
This is defined for all (𝑥, 𝑦) ∈ R2 . An important point is that this function only takes on
non-negative values. Let’s start with the level curves. For each 𝑐 ≥ 0, the level curve
𝑓 (𝑥, 𝑦) = 𝑐 is a circle of the form 𝑥 2 + 𝑦 2 = 𝑐 2 . Once again, these curves form a family
of concentric circles, centre (0, 0).
𝑦
𝑧=0
𝑧=1
𝑧=2
𝑥
𝑧=3
𝑧=4
𝑧=5
However, the difference now is that if we increment 𝑐 by some fixed amount, then the
radii of the corresponding circles will increase by that same amount. In other words,
the steepness of the surface doesn’t change.
The fact that the level curves are concentric circles means that the graph of this function
√
is a surface of revolution. Its trace in the plane 𝑦 = 0 is 𝑧 = 𝑥 2 = |𝑥|. If you imagine
rotating this curve around the 𝑧-axis, you will realise that the graph of this function is
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a cone:
𝑧
𝑧=
p
𝑥2 + 𝑦2
𝑦
𝑥
(Strictly speaking, this is actually a "half-cone". A cone is a surface like 𝑧 2 = 𝑥 2 + 𝑦 2 ,
which consists of the graph shown above together with its reflection in the 𝑥𝑦-plane.)
Example 1.7. Consider the function 𝑧 =
p
9 − 𝑥2 − 𝑦2.
The first thing to notice is that this function isn’t defined for all values of (𝑥, 𝑦) ∈ R2 .
In fact, its domain is the circular disc 𝒟 = (𝑥, 𝑦) ∈ R2 : 𝑥 2 + 𝑦 2 ≤ 9 . Its range is the
closed interval [0, 3]. For each 𝑐 in this interval, the level curve corresponding to 𝑧 = 𝑐
is a circle 𝑥 2 + 𝑦 2 = 9 − 𝑐 2 . Thus the level curves form a family of concentric circles
fitting inside the disc 𝒟. Once again, we have a surface of revolution. The trace of this
√
surface in the plane 𝑦 = 0 is the semicircle 𝑧 = 9 − 𝑥 2 . Rotating this semicircle about
the 𝑧-axis generates the graph, which is thus a hemisphere.
Example 1.8. Consider the function 𝑧 = 𝑦 2 − 𝑥 2 .
The domain of this function is R2 . It’s not difficult to see that 𝑧 can take on any real
value, so the range of this function is R. For each (𝑎, 𝑏) ∈ R2 , the partial functions are
𝑧 = 𝑏 2 − 𝑥 2 and 𝑧 = 𝑦 2 − 𝑎 2 , so the trace of 𝑧 = 𝑦 2 − 𝑥 2 in the plane 𝑦 = 𝑏 is a parabola
opening downwards, while the trace in the plane 𝑥 = 𝑎 is a parabola opening upwards.
This is a tricky surface to draw. I find the best approach is to use the traces we’ve just
described:
𝑧
𝑧 = 𝑦2 − 𝑥2
𝑦
𝑥
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The level curves of this function are hyperbolas of the form 𝑦 2 − 𝑥 2 = 𝑐, where 𝑐 ∈ R.
You could classify the graph as being a hyperbolic paraboloid.
It is possible to consider more elaborate functions, such as 𝑧 = sin(𝑥 + 𝑦) cos(𝑥 − 𝑦) or
2
2
𝑧 = 𝑒 −(𝑥 +𝑦 ) using the same methods. I’ve chosen the examples above simply because
they are relatively easy to imagine and illustrate most of the points I would like to
discuss. Stewart has some pretty pictures of more complicated functions.
1.1.2. Functions of three variables. A function of the form 𝑤 = 𝑓 (𝑥, 𝑦, 𝑧), where
𝑥, 𝑦, 𝑧 ∈ R, is a function of three variables. Its domain is a subset of R3 . If the values of
the function are real numbers, then it is scalar-valued, although later we will consider
functions of three variables that are vector-valued.
Even if 𝑓 is scalar-valued, it is difficult to imagine its graph, since that would have
to be a subset of R4 . However, it is possible to extend some of the ideas which we
used in the previous section to functions of three variables. Thus, one can consider the
partial functions 𝑤 = 𝑓 (𝑥, 𝑦, 𝑐), 𝑤 = 𝑓 (𝑥, 𝑏, 𝑧) and 𝑤 = 𝑓 (𝑎, 𝑦, 𝑧), or even 𝑤 = 𝑓 (𝑥, 𝑏, 𝑐),
𝑤 = 𝑓 (𝑎, 𝑦, 𝑐) and 𝑤 = 𝑓 (𝑎, 𝑏, 𝑧). Also useful are the level surfaces 𝑓 (𝑥, 𝑦, 𝑧) = constant.
These are surfaces in R3 .
Example 1.9. Consider the function
𝑓 (𝑥, 𝑦, 𝑧) =
27
(𝑥 − 1)2 + (𝑦 − 3)2 + (𝑧 − 4)2 + 1
.
It’s not difficult to see that the level surfaces of this function form a family of concentric
spheres, centred at my favourite point, 𝑃(1, 3, 4). Although this is a function of three
variables, its value actually only depends on the distance between (𝑥, 𝑦, 𝑧) and the
point 𝑃. As one moves away from 𝑃, the value of 𝑓 (𝑥, 𝑦, 𝑧) decreases asymptotically to
0. The function therefore has its maximum value at the point 𝑃. Since 𝑓 (1, 3, 4) = 27,
we have that 0 < 𝑓 (𝑥, 𝑦, 𝑧) ≤ 27 for all (𝑥, 𝑦, 𝑧) ∈ R3 .
Level surfaces are a useful tool even in cases where one isn’t explicitly studying the
behaviour of a function. For example, a sphere such as 𝑥 2 + 𝑦 2 + 𝑧 2 = 16 is not the graph
of a function of two variables. (It doesn’t satisfy the "vertical line test".) However, this
sphere is a level surface of the function 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥 2 + 𝑦 2 + 𝑧 2 . We’ll see later how
this gives us some insight into its properties.
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