Multivariable Functions
6.1. Functions of two variablesIn this lecture we introduce multivariable functions. These are functions that depend on mul- tiple inputs (usually 2, 3, or 4), but produce only a single output. We start our discussion with the simplest example: functions of two variables. For example, the temperatureT at any point on the surface of the Earth at this very moment depends on two pieces of information: the longitude x and the latitudeyof the point. Thus, we can think ofT as a functionT(x,y) of two variablesxand yor as a function of the ordered pair (x,y). We now give a formal definition.
Definition. A function f of two variables is a rule that assigns to each ordered pair of real numbers (x,y) in a setDa unique real number denoted f(x,y). The set Dis thedomainof f and the set of all values that f takes on is itsrange, that is, the range of f is{f(x,y) : (x,y)∈D}.
6.2. Functions of three and more variables
The example that we used above to motivate the introduction of functions of two variables is somewhat artificial. A more natural (and also more useful) quantity is the temperatureT at a point (x,y,z) in space at this very moment. Then T is a function of three variables—the coordinates (x,y,z) of the point. Note that with this function in hand we don’t have to assume that the temper- ature in a classroom and the temperature on the wing of an airplane flying 39,000 feet above that classroom are the same. An even more realistic example is that of the temperatureT at the point (x,y,z) in space at timet. This quantity is a functionT(x,y,z,t) of four variables.
Definition. A function f of n variables x1,x2, . . . ,xn is a rule that assigns to each orderedn-
tuple of real numbers (x1,x2, . . . ,xn) in a setDa unique real number denoted f(x1,x2, . . . ,xn). The
setDis thedomainof f and the set of all values that f takes on is itsrange. In particular, afunction f of three variablesis a rule that assigns to each ordered triple of real numbers (x,y,z) in a setD (inR3) a unique real number denoted f(x,y,z).
6.3. Examples of domains and ranges of multivariable functions Example6.1. The function f(x,y)=
√
1−xy is defined whenever 1− xy≥0 ⇐⇒ xy≤1.
If x> 0, the last inequality is satisfied fory ≤ x−1; ifx < 0, the inequality is satisfied fory ≥ x−1; and if x = 0, it is satisfied for ally. That is, the domain of f contains the points in I quadrant on and below the graph ofy = x−1, the points in III quadrant on and above the graph of y= x−1, and all the points in II and IV quadrants, including thex- andy-axes (see Figure 6.1).
The range of f is [0,∞). Clearly, f cannot attain negative values. On the other hand, ifk≥ 0,
p
Hence, ifk≥ 0,k ,1, f(x,y)=kfor all points (x,y) on the hyperbolaxy= 1−k2; also, f(x,y)=1 for all points (x,y) on the coordinate axes. That is, every number k ∈ [0,∞) does belong to the
range of f.
x y
Figure6.1. The domain ofz= √1−xy
y x
z
Figure6.2. z=3− x2−y2
Example6.2. The domain of the function f(x,y,z)= ln(16−x2−y2−9z2) consists of the points (x,y,z) in space whose coordinates satisfy
16−x2−y2−9z2 >0 ⇐⇒ x 2 16 + y2 16 + z2 16/9 < 1.
Since lnxis an increasing function, the range of f consists of the logarithms of the positive num- bers in the range of the function w = 16− x2− y2 −9z2. Because x2 +y2 +9z2 can attain any
non-negative value and cannot attain any negative value, it follows thatwtakes on the numbers in the interval (−∞,16]. Thus, the range of f consists of the logarithms of the numbers in (0,16], that
is, the range of f is (−∞,ln 16].
6.4. Graphs
Definition. If f is a function ofnvariables x1,x2, . . . ,xn with domainD, then thegraphof f
is the set of all points (x1,x2, . . . ,xn,z) inRn+1 with (x1,x2, . . . ,xn) ∈ Dandz = f(x1,x2, . . . ,xn),
that is, the graph of f is the set
(x1,x2, . . . ,xn, f(x1,x2, . . . ,xn)) : (x1,x2, . . . ,xn)∈D .
In particular, the graph of a function f(x,y) defined on a setDinR2is the surface given by
(x,y,
f(x,y)) : (x,y)∈D .
Example6.3. The graph of the function f(x,y)=3− x2−y2is the surface z=3−x2−y2.
6.5. Level curves and surfaces
While the graph of a function of two variables provides an excellent visualization of the func- tion, it might be quite difficult to sketch without advanced technology. What can we do using only manual (and mental) labor? One answer is well-known to cartographers. When drawing topographic or termal maps, cartographers often produce two-dimensional drawings marked with curves of constant elevation or constant temperature. We can do the same for any function f of two variables: we can graph f in the xy-plane by marking its “level curves”.
Definition. Let f be a function of two veriables, denotedx andy, and letk be a real number. Thek-level curveof f is the set of all points (x,y) in the domainDof f which satisfy the equation
f(x,y)=k.
This idea becomes even more important in the case of functions of three variables. Note that the graph w = f(x,y,z) of such a function is a four-dimensional object, which we can study analytically but cannot represent graphically in our three-dimensional universe. However, we can draw the “level surfaces” of f.
Definition. Let f be a function of three veriables, denotedx,yandz, and letkbe a real number. The k-level surface of f is the set of all points (x,y,z) in the domain D of f which satisfy the equation f(x,y,z)= k. 1 3 1 2 1 2 −13 −12 −1 −2 x y
Figure6.3. The level curves of f(x,y)= x/(x2+y2)
Example6.4. Describe the level curves of the function f(x,y)= x
x2+y2.
Solution. We want to draw the curve x
x2+y2 =k
for allk. If we exclude the origin from consideration (since the curve is undefined there), we can represent this curve by the equation
kx2+ky2−x=0.
Whenk = 0, this equation represents they-axisx = 0, so the level curve f(x,y) = 0 is they-axis with the origin removed from it. When k , 0, we can transform the equation of the level curve
further to x2− x k +y 2 =0 ⇐⇒ x− 1 2k 2 +y2= 1 4k2,
which is the equation of a circle centered at (2k1,0) of radius 21|k| (kcan be negative). Since the origin does lie on this circle, the level curve f(x,y) = kis the circle with the origin removed from it. A contour map of f showing the level curves withk= 0,±1
3,± 1
2,±1,±2 is shown on Figure 6.3.
6.6. Limits of multivariable functions
Since all the important concepts in single-variable calculus were defined in terms of limits, if we are to generalize those concepts to functions of two and more variables, we must first define the notion of a limit of a multivariable function. The following definition generalizes the formal definition of limit to functions of two variables.
Definition. Let f be a function of two variables whose domain Dincludes points arbitrarily close to (a,b). We say that thelimit of f(x,y)as(x,y)approaches(a,b)is Land write
lim
(x,y)→(a,b) f(x,y)=L or f(x,y)→L as (x,y)→(a,b),
if for any given numberε> 0, there is a corresponding numberδ =δ(ε)>0 such that 0< p(x−a)2+(y−b)2< δ =⇒ |f(x,y)−L|<ε.
This definition isveryformal. It is also very useful in proofs (something we will not be con- cerned with) and very imposing at first sight. A more intuitive definition reads as follows.
Definition. Let f be a function of two variables whose domain Dincludes points arbitrarily close to (a,b). We say that thelimit of f(x,y)as(x,y)approaches(a,b)is Land write
lim
(x,y)→(a,b) f(x,y)=L or f(x,y)→L as (x,y)→(a,b),
if f(x,y) approachesLas (x,y) approaches (a,b), independent of how (x,y) approaches (a,b). We can define limits of functions of three or more variables in a similar fashion.
Definition. Let f be a function of three variables whose domain Dincludes points arbitrarily close to (a,b,c). We say that thelimit of f(x,y,z)as(x,y,z)approaches(a,b,c)is Land write
lim
(x,y,z)→(a,b,c)f(x,y,z)=L or f(x,y,z)
→ L as (x,y,z)→ (a,b,c),
if f(x,y,z) approaches L as (x,y,z) approaches (a,b,c), independent of how (x,y,z) approaches (a,b,c). More precisely, the limit of f(x,y,z) as (x,y,z) approaches (a,b,c) is L, if for any given numberε>0, there is a corresponding numberδ=δ(ε)> 0 such that
0< p(x−a)2+(y−b)2+(z−c)2< δ =⇒ |f(x,y,z)−L|<ε.
Limits of multivariable functions can be much trickier than those of single-variable functions. The primary reason for that is the infinitude of ways in which the arguments (x,y) (or (x,y,z)) can approach the point (a,b) (or (a,b,c)). This point is best illustrated by examples of limits that do not exist.
Example6.5. The limit
lim
(x,y)→(0,0)
x x2+y2
does not exist. Indeed, we know from Example 6.4 (see also Figure 6.3) that the level curves of this function all want to pass through (0,0) (but since the point is not in the domain of f, they can’t). Thus, if we approach (0,0) along two distinct level curves (say, along f(x,y) = 1 and f(x,y)=−1), the function values will approach distinct values (1 and−1, respectively). However, if the limit existed, f(x,y) would have to approach the same value independent of the way (x,y) approaches (0,0). Therefore, the limit does not exist.
Example6.6. Show that the limit lim
(x,y)→(0,0)
xy4
x2+y8 does not exist.
Solution. First, consider the behavior of the function as (x,y) → (0,0) along the straight line y= x. We have lim x→0 f(x,x)=limx→0 x5 x2+ x8 = limx→0 x3 1+x6 =0.
On the other hand, if we let (x,y)→ (0,0) along the curvex= y4, we get lim y→0 f(y 4,y)=lim y→0 y8 y8+y8 = 1 2.
Since different approach paths lead to different limiting values, the limit does not exist. Example6.7. For any numbersa,b,c, we have
lim
(x,y)→(a,b)x=a, (x,y)lim→(a,b)y= b, (x,y)lim→(a,b)c=c.
Using these facts and the two-variable versions of the limit laws: lim
(x,y)→(a,b)[f(x,y)±g(x,y)]= (x,y)lim→(a,b) f(x,y)±(x,y)lim→(a,b)g(x,y), etc.,
we can compute the limit of any polynomial inxandyat any point (a,b) in the plane. For example, lim
(x,y)→(2,3)xy= (x,y)lim→(2,3)x lim (x,y)→(2,3)y =2·3=6.
Next, we look at some examples of evaluation of limits of functions of two and three variables. Example6.8. Evaluate lim
(x,y)→(0,0)
sin(x2+y2)
x2+y2 .
Solution. If we change the coordinates of the point (x,y) from Cartesian to polar, the condition (x,y)→ (0,0) is equivalent tor→0 and the limit takes the form
lim (x,y)→(0,0) sin(x2+y2) x2+y2 = limr→0 sin(r2) r2 = limr→0 2rcos(r2) 2r = 1,
Example6.9. Evaluate(x,y,z)lim
→(0,0,0)
(x2+y2)z
x2+y2+z2.
Solution. This time, we change the coordinates of the point (x,y,z) from Cartesian to spheri- cal, and the condition (x,y,z)→(0,0,0) becomesρ→0. Hence,
lim (x,y,z)→(0,0,0) (x2+y2)z x2+y2+z2 =limρ→0 (ρsinφ)2(ρcosφ) ρ2 =lim ρ→0ρsin 2φ cosφ =0.
Example6.10. Evaluate lim
(x,y)→(1,2)
x2y−2x2−2xy+6x+5y−12 x2+y2−2x−4y+5 .
Solution. Both the numerator and the denominator of the function vanish at the point (1,2), so the limit is an indeterminate form of type 0/0. We will try to argue similarly to the previous example, but before we can do so, we need to change the limit from one at the point (1,2) to one at the origin. Let us define new variablesuandvby
x= 1+u, y= 2+v.
The condition (x,y)→(1,2) is equivalent to (u,v)→(0,0), so we can rewrite the original limit as lim (x,y)→(1,2) x2y−2x2−2xy+6x+5y−12 x2+y2−2x−4y+5 = lim (u,v)→(0,0) (1+u)2(2+v)−2(1+u)2−2(1+u)(2+v)+6(1+u)+5(2+v)−12 (1+u)2+(2+v)2−2(1+u)−4(2+v)+5 = lim (u,v)→(0,0) u2v+2u+4v u2+v2 .
We can now change the coordinates of the point (u,v) from Cartesian to polar. We find that lim
(u,v)→(0,0)
u2v+2u+4v u2+v2 =limr→0
r3cosθsinθ+2rcosθ+4rsinθ
r2
=lim
r→0 rcosθsinθ+(2 cosθ+4 sinθ)r
−1 =lim r→0(2 cosθ+4 sinθ)r −1 =
+∞ when 2 cosθ+4 sinθ>0,
0 when 2 cosθ+4 sinθ=0,
−∞ when 2 cosθ+4 sinθ<0.
Hence, the given limit does not exist.
6.7. Continuity of multivariable functions
We can use the notion of limit of a multivariable function to define the notion of continuity. Definition. Let f be a function of two variables whose domain Dincludes (a,b) and points arbitrarily close to (a,b). We say that f(x,y) iscontinuous at(a,b), if
lim
We say that f iscontinuous in D, if it is continuous at every point (a,b)∈D.
Definition. Let f be a function of three variables whose domainDincludes (a,b,c) and points arbitrarily close to (a,b,c). We say that f iscontinuous at(a,b,c), if
lim
(x,y,z)→(a,b,c)f(x,y,z)= f(a,b,c).
We say that f iscontinuous in D, if it is continuous at every point (a,b,c)∈D.
We want also to extend the notion of piecewise continuity to multivariable functions, and in particular, to functions of two variables. Recall that a single-variable function f is called piecewise continuous on [a,b] if it is continuous everywhere in [a,b] except for a finite number of points, where it can have only jump or removable discontinuities. The two-dimensional version of this definition reads as follows.
Definition. Let f be a function of two variables with domain D. We say that f is piecewise continuous in D, if it is continuous at every point (a,b) ∈ D, except possibly for the points on a finite number of smooth curves inD, sayγ1,γ2, . . . ,γn, where f may have “jump discontinuities.”
Remark. As in the case of single-variable functions, polynomials in several variables are con- tinuous. Likewise, rational functions in several variables are continuous except for the points where the denominator is 0. Finally, if f is a continuous function (of a single variable) andgis a con- tinuous function of the variablesx,y, . . ., the composition f(g(x,y, . . .)) is continuous at all points (x,y, . . .) where it is defined.
Example6.11. The function
f(x,y)= cos
x2−y2
x2+y2
is continuous everywhere in its domain, which is all the points inR2except (0,0). Indeed, f is the
composition of the continuous functionw= costand the rational function t= x
2−y2
x2+y2.
Example6.12. Show that the function f(x,y)= sin(x2+y2) x2+y2 if (x,y), (0,0), 1 if (x,y)= (0,0), is continuous everywhere.
Solution. Recall that
lim
x→0
sinx x = 1. Hence, the function
g(x)= sinx x if x,0, 1 if x=0,
is continuous everywhere. Furthermore, the polynomial x2 + y2 is also continuous everywhere.
Exercises
Find the domain and the range of the given function. 6.1. f(x,y)=y2e−x2/2
6.2. f(x,y)=ln x+y2
6.3. f(x,y,z)= z
x2−y2−1
Sketch the domain of the given function.
6.4. f(x,y)=ln x+y2 6.5. f(x,y)= √
x+2y x2+y2−1
For each function, draw the level curves for levels−1,0,1,2, and 3; then sketch the graphz= f(x,y). 6.6. f(x,y)=1+y2 6.7. f(x,y)=2x+3y−5 6.8. f(x,y)=x2+y2
Find the given limit or show that it does not exist. 6.9. lim (x,y)→(1,3) x 4− 4xy2+2x3y 6.10. lim (x,y)→(0,0) xy x2+y2 6.11. lim (x,y,z)→(1,0,2) sin(2xy) 1−zy2 6.12. lim (x,y)→(0,0) 3x2y x4+y2 6.13. lim (x,y)→(0,0) xy p x2+y2 6.14. lim (x,y)→(0,0) xy √ 1+xy−1 Find the set of points inR2or
R3where the given function is continuous.
6.15. f(x,y)=sin x4+x2y2 6.16. f(x,y)= xy x2+y2 if (x,y),(0,0), 1 if (x,y)=(0,0). 6.17. f(x,y,z)=ln 4−x2−y2−4z2 6.18. f(x,y)= xy √ 1+xy−1 if (x,y),(0,0), 1 if (x,y)=(0,0).