7 | Multivariable Functions

This chapter of Multivariable Calculus by Dr JH Klopper is licensed under an Attribution-NonCommercial-NoDerivatives 4.0 International Licence available at THE CREATIVE COMMONS .

7.1 Introduction

In this chapter we introduce multivariable functions. Such functions are defined for more than just a single variable.
In the chapter we consider the same concepts as with a single variable function. These concepts include the domain and range of a a multivariable functions, limits, differentiation, and integration.

7.2 Domain and range

Multivariable functions map values for more than one independent variable to a dependent variable,
f:
n


. Let
D
be a subset of
n

. A function
f
of
n
variables is a rule that assigns each
n
-tuple
(
x
1
,
x
2
,…,
x
n
)
in
D
a value
f(
x
i
)
(where
i={1,2,…,n}∈
) in

.
D
is the domain of
f
, the set of all outputs of
f
is the range. It is represented in
n+1
-space where
n
is the number of independent variables.
In (1) we have an example of a 4-space function.
w=f(x,y,z)
(
1
)
Here, the domain is in
4
-space.
​
Problem 7.2.1
What is the domain and range of the function in (2).
f(x,y)=3
xy
x-y
(
2
)
The domain,
D
, is shown in (3).
D={x,y∈|x≠y}
(
3
)
The range,
R
, is shown in (4).
R={z|z∈}
(
4
)
​
​
Problem 7.2.2
What is the domain and range of the function in (5).
f(x,y)=
16-(
2
x
+
2
y
)
(
5
)
The domain,
D
, is shown in (6).
16-
2
x
-
2
y
≥0​​
2
x
+
2
y
≤
2
4
​​D={x,y∈|0≤
2
x
+
2
y
≤
2
4
(
6
)
The range,
R
, is shown in (7).
2
x
+
2
y
=16
16-16
=0​​
2
x
+
2
y
=0
16
=4​​R={z∈|0≤z≤4
(
7
)
​
​
Problem 7.2.3
What is the domain of the function in (8).
f(x,y)=
log(y-x)
x-y+1
(
8
)
Here we have two constraints for the domain, shown in (9).
y-x>0∧x-y+1>0​​y>x∧y<x+1​​D={x,y∈0|y>x∧y<x+1}
(
9
)
Figure 7.2.1 visualizes the inequalities. The domain is the intersection of the two shaded areas.
In[]:=
Show[Plot[x,{x,-3,3},FillingTop,PlotLabel->"Figure 7.2.1",GridLines->Automatic,ImageSize->Large],Plot[x+1,{x,-3,3},FillingBottom]]
Out[]=
We can also use the RegionPlot function, shown in Figure 7.2.2.
In[]:=
RegionPlot[y>x∧y<x+1,{x,-3,3},{y,-3,3},PlotLabel->"Figure 7.2.2",GridLines->Automatic,ImageSize->Large]
Out[]=
​
​
Problem 7.2.4
What is the domain of the function in (10).
f(x,y,z)=
9-
2
x
-
2
y
-
2
z
(
10
)
We have the constraint in (11), where all the variable values are squared and their sum will always be equal to or greater then
0
.
9-(
2
x
+
2
y
+
2
z
)≥0​​
2
x
+
2
y
+
2
z
≤
2
3
(
11
)
The domain is a closed sphere with a radius of
3
centered at the origin.
​
​
Problem 7.2.5
What is the domain of the function in (12).
f(x,y,z)=
4-
2
x
-
2
y
z-3
(
12
)
We have the constraint in (13).
2
x
+
2
y
≤
2
2
∧z≠3
(
13
)
This is a cylinder with radius
2
extending along the
z
axis except at the plane
z=3
(parallel to the
xy
plane).
We can see the domain below in orange. The blue plane is excluded from the domain. Note that the actual function is in 4-space. Figure 7.2.3 represents the domain over which the functions is defined.
In[]:=
ContourPlot3D[{
2
x
+
2
y
4,z3},{x,-3,3},{y,-3,3},{z,-1,4},MeshNone,ContourStyleOpacity[0.8],BoundaryStyleNone,PlotLabel->"Figure 7.2.3",ImageSize->Large]
Out[]=
​

7.3 Level curves

Level curves are intersections along a surface, parallel to a plane, projected onto that plane. A series of such curves at different levels, projected onto the plane creates a contour plot.
​
Problem 7.3.1
Graph the contour plot of the function in (14).
​
​
Problem 7.3.2
Graph the contour plot of the function in (15).
The solution is shown in Figure 7.3.2.
​

7.4 Limits and continuity

As with single variable function, we need to know that a limit exists and that we have continuity at a point.

7.4.1 Limits

Limits for single variable functions are constrained in their approach to the point under consideration. The point can only be approached from either sides of the point (16).
On a surface in 3-space, we can approach a point from infinitely many directions. It is usually much easier to show that a limit does not exist.
We follow the same reasoning as with the previous problem, where we apply L’Hopital’s Rule, shown in (19).
Now we can state that the limit does not exist.
​
Problem 7.4.1.1
Show whether a limit exists for the function in ().
We begin with our usual strategy ().
We can state that the limit does not exist.
​
​
Problem 7.4.1.2
Show whether a limit exists for the function in (26).
Directly substituting the limit point gives us (27).
This limit exist.
​
​
Problem 7.4.1.3
Show whether a limit exists for the function in (28).
The limit does not exist.
​
​
Problem 7.4.1.4
Show whether a limit exists for the function in (31).
The limit does not exist.
​
​
Problem 7.4.1.5
Show whether a limit exists for the function in (34).
We start by choosing an axis again shown in (35).
Now we create a parametric curve again, taking cognisance of the fact that we want the same power of the parameter in the numerator and denominator shown in (36).
The limit does not exist.
​
​
Problem 7.4.1.6
Show whether a limit exists for the function in (37).
We can make use of a polar transform (38).
​
​
Problem 7.4.1.7
Show whether a limit exists for the function in (39).
We use polar transformation again (40).
The limit exists.
​
​
Problem 7.4.1.8
Show whether a limit exists for the function in (41).
We can use the Squeeze Theorem to show that the limit exists (42).
​

7.4.2 Continuity

If a function is defined in a region, it is continuous on that region since a path in the region can be found to the point, in the neighborhood of the point.
Points on the boundary can still be included as long as we follow paths contained in the region.
​
Problem 7.4.2.1
Where is the function in (46) continuous?
We look at the two variables in (47).
​
​
Problem 7.4.2.2
Where is the function in (48) continuous?
This is only defined when () holds.
​
​
Problem 7.4.2.3
We look at the composition in (51).
We note the following constraints (52).
We visualize this region in Figure 7.4.2.1.
​

7.5 Derivatives of multivariable functions

A derivative of a single variable function is a tangent line. The derivative to a surface in 3-space is a tangent plane.
We can calculate this tangent by intersection the surface at a point with a plane parallel to one of the axis planes. The parallel planes keep op of the independent variables constant and we can take the derivative, with that variable as a constant. This gives rise to partial derivatives.

7.5.1 Partial derivatives

A partial derivative with respect to one of the independent variables is denoted in (53).
For a partial derivative with respect to two variables, we do the following in (54).
​
Problem 7.5.1.1
Calculate the partial derivatives of both independent variables in the function in (55).
We can check this with the D function that takes the derivative.
​
​
Problem 7.5.1.2
Calculate the partial derivatives of both independent variables in the function in (60).
We do both partial derivatives in (61).
​
​
Problem 7.5.1.3
Calculate the partial derivatives of both independent variables in the function in (62).
We use different notation for the partial derivatives in (63).
​
​
Problem 7.5.1.4
Calculate the partial derivatives of both independent variables in the function in (64).
The partial derivatives, calculated using the product and chain rules, are shown in (65).
​
​
Problem 7.5.1.5
Calculate the partial derivatives of both independent variables in the function in (66).
The partial derivatives are shown in (67).
​
​
Problem 7.5.1.6
What is the change in BSA with respect to height and weight? Which changes the body surface area more per unit change? We take the two partial derivatives in (69).
A change in weight will lead to a bigger change in BSA.
​

7.6 Implicit partial derivatives

Implicit derivatives follow the same arguments in single variable calculus.
​
Problem 7.6.1
​
​
Problem 7.6.2
​

7.7 Higher-order partial derivatives

Consider a function with two independent variable. We can take the partial derivative of both of these. For each of them in turn, we can take a second derivative, one for each of the variables. This leaves us with four partial derivatives shown in (76).
​
Problem 7.7.1
Calculate the second derivatives of the function in (77).
Notice that the two second partial derivatives with respect to both variables, called the mixed partials, are equal.
​
​
Problem 7.7.2
Calculate the second derivatives of the function in (80).
Once again, notice how the mixed partial derivatives are equal.
​
​
Problem 7.7.3
Calculate the second partial derivatives of the function in (84).
All of this is easily checked using the D function in the Wolfram Language.
​

7.8 Tangent planes and linear approximations

We have mentioned that the derivative to a function with two independent variables (a surface in 3-space) is a plane in 3-space, tangent to the surface at a point.
The slope of the tangent line is the first derivative in (87).
From parametric curves we know that we can make use of an intermediate variable shown in (89).
Figure 7.8.3 visualizes the curve.
The tangent plane at this point is calculated from the two partial derivatives shown in (91).
The partial derivatives of the point on the surface are along the two planes in the Figure 7.8.4.
We now have the two tangent lines along the two planes at the point of tangency on the surface. Another point on each of the two lines, gives us two vectors shown in (93).
The cross product of these two vectors is the normal to the point on the surface.
We can now plot this tangent plane, shown in Figure 7.8.5.
We can derive a simple equation from this problem. Notice the scalar multiple of the tangent vector in (99).
The equation of the plane of tangency is therefor shown in (101).
This also gives us the parametric equations for the normal line at the point of tangency shown in (102).
​
Problem 7.8.1
We can now calculate the equation of the plane shown in (105).
We plot the surface, point, and tangent plane in Figure 7.8.6.
​

7.9 Differentials

​
Problem 7.9.1
We solve for the differentials shown in (112).
​
​
Problem 7.9.2
We calculate the differential in (115).
​
​
Problem 7.9.3
We use the equation for the approximation of the differential in (117).
We do this simple calculation using the Wolfram Language.
​

7.10 Chain rule for multivariable function differentiation

From the composition of functions we have the following derivation in (120).
Now consider the equation in (121).
​
Problem 7.10.1
We apply the chain rule in (127).
​
​
Problem 7.10.2
We can calculate all of the components separately with substitution in (130).
​

7.11 Implicit partial differentiation

​
Problem 7.11.1
We calculate the derivative in (134).
We get the result using the Wolfram Language.
​

7.12 Directional derivatives

We can now write the directional derivative as in (139).
​
​
Problem 7.12.1
We create the unit vector in (141).
The partial derivatives are calculated in (142).
Now we can calculate the slope of a line tangent to the surface in the direction specified in (143).
​
​
Problem 7.12.2
The partial derivatives are shown in (145).
The directional derivative is shown in (146).
​

7.13 Gradient

​
Problem 7.13.1
Calculate the gradient of the function in (148).
The gradient is shown in (149).
We use the Grad function to calculate the gradient.
​