14
Partial Differentiation
14.1 Fun tions of Several Variables
In single-variable calculus we were concerned with functions that map the real numbers R
to R, sometimes called “real functions of one variable”, meaning the “input” is a single real
number and the “output” is likewise a single real number. In the last chapter we considered
functions taking a real number to a vector, which may also be viewed as functions f : R →
R3 , that is, for each input value we get a position in space. Now we turn to functions
of several variables, meaning several input variables, functions f : Rn → R. We will deal
primarily with n = 2 and to a lesser extent n = 3; in fact many of the techniques we
discuss can be applied to larger values of n as well.
A function f : R2 → R maps a pair of values (x, y) to a single real number. The three-
dimensional coordinate system we have already used is a convenient way to visualize such
functions: above each point (x, y) in the x-y plane we graph the point (x, y, z), where of
course z = f (x, y).
EXAMPLE 14.1.1 Consider f (x, y) = 3x + 4y − 5. Writing this as z = 3x + 4y − 5 and
then 3x +4y −z = 5 we recognize the equation of a plane. In the form f (x, y) = 3x +4y −5
the emphasis has shifted: we now think of x and y as independent variables and z as a
variable dependent on them, but the geometry is unchanged.
EXAMPLE 14.1.2 We have seen that x2 + y 2 + z 2 = 4 represents a sphere of radius 2.
We cannot write this in the form f (x, y), since for each x and y in the disk x2 +y 2 < 4 there
are two corresponding points on the sphere. As with the equation of a circle, we can resolve
349
,350 Chapter 14 Partial Differentiation
p p
this equation into two functions, f (x, y) = 4 − x2 − y 2 and f (x, y) = − 4 − x2 − y 2 ,
representing the upper and lower hemispheres. Each of these is an example of a function
with a restricted domain: only certain values of x and y make sense (namely, those for
which x2 + y 2 ≤ 4) and the graphs of these functions are limited to a small region of the
plane.
√ √
EXAMPLE 14.1.3 Consider f = x + y. This function is defined only when both
√
x and y are non-negative. When y = 0 we get f (x, y) = x, the familiar square root
function in the x-z plane, and when x = 0 we get the same curve in the y-z plane.
Generally speaking, we see that starting from f (0, 0) = 0 this function gets larger in every
direction in roughly the same way that the square root function gets larger. For example,
√
if we restrict attention to the line x = y, we get f (x, y) = 2 x and along the line y = 2x
√ √ √ √
we have f (x, y) = x + 2x = (1 + 2) x.
6
5
4
3 10.0
2
7.5
1
0 5.0
0.0 y
2.5
2.5
5.0
x 7.5
0.0
10.0
√ √
Figure 14.1.1 f (x, y) = x+ y (AP)
A computer program that plots such surfaces can be very useful, as it is often difficult
to get a good idea of what they look like. Still, it is valuable to be able to visualize
relatively simple surfaces without such aids. As in the previous example, it is often a good
idea to examine the function on restricted subsets of the plane, especially lines. It can also
be useful to identify those points (x, y) that share a common z-value.
EXAMPLE 14.1.4 Consider f (x, y) = x2 + y 2 . When x = 0 this becomes f = y 2 , a
parabola in the y-z plane; when y = 0 we get the “same” parabola f = x2 in the x-z plane.
Now consider the line y = kx. If we simply replace y by kx we get f (x, y) = (1 + k 2 )x2
which is a parabola, but it does not really “represent” the cross-section along y = kx,
because the cross-section has the line y = kx where the horizontal axis should be. In
, 14.1 Functions of Several Variables 351
order to pretend that this line is the horizontal axis, we need to write the function in
p p
terms of the distance from the origin, which is x2 + y 2 = x2 + k 2 x2 . Now f (x, y) =
p
x2 + k 2 x2 = ( x2 + k 2 x2 )2 . So the cross-section is the “same” parabola as in the x-z and
y-z planes, namely, the height is always the distance from the origin squared. This means
that f (x, y) = x2 + y 2 can be formed by starting with z = x2 and rotating this curve
around the z axis.
Finally, picking a value z = k, at what points does f (x, y) = k? This means x2 +y 2 = k,
√
which we recognize as the equation of a circle of radius k. So the graph of f (x, y) has
parabolic cross-sections, and the same height everywhere on concentric circles with center
at the origin. This fits with what we have already discovered.
8
6
4
2
0 −3
−3
−2
−2
−1
−1
0
0
1 1
2 2
3 3
Figure 14.1.2 f (x, y) = x2 + y2 (AP)
As in this example, the points (x, y) such that f (x, y) = k usually form a curve, called
a level curve of the function. A graph of some level curves can give a good idea of the
shape of the surface; it looks much like a topographic map of the surface. In figure 14.1.2
both the surface and its associated level curves are shown. Note that, as with a topographic
map, the heights corresponding to the level curves are evenly spaced, so that where curves
are closer together the surface is steeper.
Functions f : Rn → R behave much like functions of two variables; we will on occasion
discuss functions of three variables. The principal difficulty with such functions is visual-
izing them, as they do not “fit” in the three dimensions we are familiar with. For three
variables there are various ways to interpret functions that make them easier to under-
stand. For example, f (x, y, z) could represent the temperature at the point (x, y, z), or the
pressure, or the strength of a magnetic field. It remains useful to consider those points at
which f (x, y, z) = k, where k is some constant value. If f (x, y, z) is temperature, the set of
points (x, y, z) such that f (x, y, z) = k is the collection of points in space with temperature
, 352 Chapter 14 Partial Differentiation
k; in general this is called a level set; for three variables, a level set is typically a surface,
called a level surface.
2 2 2
EXAMPLE 14.1.5 Suppose the temperature at (x, y, z) is T (x, y, z) = e−(x +y +z ) .
This function has a maximum value of 1 at the origin, and tends to 0 in all directions.
If k is positive and at most 1, the set of points for which T (x, y, z) = k is those points
satisfying x2 + y 2 + z 2 = − ln k, a sphere centered at the origin. The level surfaces are the
concentric spheres centered at the origin.
Exercises 14.1.
1. Let f (x, y) = (x − y)2 . Determine the equations and shapes of the cross-sections when x = 0,
y = 0, x = y, and describe the level curves. Use a three-dimensional graphing tool to graph
the surface. ⇒
2. Let f (x, y) = |x| + |y|. Determine the equations and shapes of the cross-sections when x = 0,
y = 0, x = y, and describe the level curves. Use a three-dimensional graphing tool to graph
the surface. ⇒
2 2
3. Let f (x, y) = e−(x +y ) sin(x2 +y2 ). Determine the equations and shapes of the cross-sections
when x = 0, y = 0, x = y, and describe the level curves. Use a three-dimensional graphing
tool to graph the surface. ⇒
4. Let f (x, y) = sin(x − y). Determine the equations and shapes of the cross-sections when
x = 0, y = 0, x = y, and describe the level curves. Use a three-dimensional graphing tool to
graph the surface. ⇒
5. Let f (x, y) = (x2 − y2 )2 . Determine the equations and shapes of the cross-sections when
x = 0, y = 0, x = y, and describe the level curves. Use a three-dimensional graphing tool to
graph the surface. ⇒
6. Find the domain of each of the following functions of two variables:
p p
a. 9 − x2 + y2 − 4
b. arcsin(x2 + y2 − 2)
p
c. 16 − x2 − 4y2
⇒
7. Below are two sets of level curves. One is for a cone, one is for a paraboloid. Which is which?
Explain.
