MATH 355: Calculus of Several Variables1.1Functions of Several Variables 1.1.1unctionsF of two...

MATH 355: Calculus of SeveralVariables

Dr. Joseph K. Ansong

ii c©Dr Joseph K. Ansong

Contents

1 Partial Di�erentiation 1

1.1 Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Functions of two variables . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.2 Level Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.3 Functions of three or more variables . . . . . . . . . . . . . . . . . . 7

1.2 Limits & Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

iii

CONTENTS

iv c©Dr Joseph K. Ansong

1 | Partial Di�erentiation

1.1 Functions of Several Variables

1.1.1 Functions of two variables

De�nition 1. A function of two variables is a rule that assigns to each ordered pair ofreal numbers (x, y) in a set D a unique real number denoted by f(x, y). The set D is thedomain of f and its range is the set of values that f takes on, that is, f(x, y)|(x, y) ∈ D.

Figure 1.1 gives a graphical illustration of a function of two variables. It is often denotedby z = f(x, y) to show that z depends on both x and y. The variable z is called thedependent variable and x and y are called the independent variables. As shownpreviously, the graphs of functions of two variables are surfaces in three-dimensionalspace, such as quadric surfaces.

Figure 1.1: f : D → R

Illustrative Example 1. Determine the domain of the following functions

1

CHAPTER 1. PARTIAL DIFFERENTIATION

(a) f(x, y) =√x2 + y2 − 1

(b) f(x, y) = ln (1− x2 − y2)

(c) f(x, y) =√y − x

(d) f(x, y) = x ln (y2 − x)

Illustrative Solution 1. (a) f(x, y) =√x2 + y2 − 1. In this case, we expect the

number in the square root sign to be non-negative, as such

x2 + y2 − 1 ≥ 0 =⇒ x2 + y2 ≥ 1

The domain isD = {(x, y)|x2 + y2 ≥ 1}.

So the domain is the region outside the circle of radius 1 and center (0, 0) asshown in Figure 1.2(a).

(b) f(x, y) = ln (1− x2 − y2). Here, the argument of ln must be greater than 0, so thefunction is de�ned for

1− x2 − y2 > 0 =⇒ x2 + y2 < 1.

So D = {(x, y)|x2 + y2 < 1}. This is the region inside the circle of radius 1 andcenter (0, 0) as displayed in Figure 1.2(b).

(c) f(x, y) =√y − x. Similar to (a), the function is de�ned for y ≥ x, hence

D = {(x, y) : y ≥ x}, as shown in Figure 1.2(c).

(d) f(x, y) = x ln (y2 − x). Similar to (b), the function is de�ned fory2 − x > 0 =⇒ x < y2, thus D = {(x, y)|x < y2}. These are points to the left ofthe parabola x = y2 as depicted on Figure 1.2(d).

Graph

De�nition 2. Let f be a function of two variables with domain D, then the graph of fis the set of all points (x, y, z) ∈ R3 such that z = f(x, y) and (x, y) ∈ D.

The graph of a surface in R3 can be visualized as lying directly above or below itsdomain D as depicted in Figure 1.3.

Illustrative Example 2. Find the domain and range and sketch the graph of

(a) z = 9x2 + y2. (b) z =√x2 + y2

Illustrative Solution 2. (a) For z = 9x2 + y2, the domain is the set of all(x, y) ∈ R2. We note that z ≥ 0 since x2 ≥ 0 and y2 ≥ 0, hence the range of z is[0,∞). The graph of z = f(x, y) is shown in Figure 1.4(a).

(b) The domain and range of z = f(x, y) are the same as in (a), and the graph isdisplayed in Figure 1.4(b).

2 c©Dr Joseph K. Ansong

1.1. FUNCTIONS OF SEVERAL VARIABLES

(a) (b)

(c) (d)

Figure 1.2: The shaded region represent the domain of the function (a) f(x, y) =√x2 + y2 − 1 (b) f(x, y) = ln (1− x2 − y2) (c) f(x, y) =

√y − x (d) f(x, y) =

x ln (y2 − x)

1.1.2 Level Curves

De�nition 3. The level curves of a function f of two variables are the curves withequations f(x, y) = k, where k is a constant in the range of f .

The level curves (or contour curves) f(x, y) = k are a two-dimensional curves that

c©Dr Joseph K. Ansong 3


Figure 1.3: The graph of a surface and its domain D.

(a) (b)

Figure 1.4: Graphs of (a) f(x, y) = 9x2 + y2 (b) f(x, y) =√x2 + y2.



show where the graph of f has height k. Examples of level curves are topographic orelevation maps showing the elevation above sea-level of a region as displayed in Figure1.5(a-b). A painting of sea surface height is shown in Figure 1.6(a). Withoutcorresponding level curves it is impossible the determine the variations in sea surfaceheight from the �gure. Thus, for some purposes, it is more useful to have a contour mapthan graph. This is especially true if we want to estimate the value of a function at apoint. Another good example of level curves is the map of the global sea surfacetemperature displayed in Figure 1.6(b). The level curves are called isothermals and arethe lines that separate the colored regions on the map.

(a) (b)

Figure 1.5: [(a)-(b)] Example topographic maps: (a) Courtesy:�https://slideplayer.com/slide/9918701/� (b) The upper �gure shows a mountainfrom the side and the lower plot shows the level curves (topographic map); courtesy:�https://study.com/academy/lesson/topographic-maps-lesson-for-kids.html�.

Example 1. Graph the level curves of the following functions

(a) z = 9x2 + y2. (b) z = −10xye−(x2+y2).

Solution 1. (a) z = 9x2 + y2. The level curves are 9x2 + y2 = k, where k is a

constant. The equation may be written asx2

k/9+y2

k= 1, showing that the level

curves are ellipses. Figure 1.7(a) displays the graph of this surface with a few levelcurves plotted on the surface and also projected onto the xy−plane. Figure 1.7(b)shows a plot of only the level curves plotted on Figure 1.7(a).

(b) z = −10xye−(x2+y2). The level curves of this surface are shown in Figure 1.9. Thecurves for z > 0 are shown in black and those for z < 0 are in red.



(a)

(b)

Figure 1.6: (a) A schematic of sea surace height. (b) Contour map ofglobal sea surface temperature �http://www.salinityremotesensing.ifremer.fr/sea-surface-salinity/salinity-distribution-at-the-ocean-surface".

Question 1. (a) A contour map for a function is shown in Figure 1.8(a). Use it toestimate the values of f(−3, 3) and f(3,−2). What can you say about the shapeof the graph?

(b) Use the contour map in Figure 1.8(b) to make a rough sketch of the graph off(x, y).



(a) (b)

Figure 1.7: (a) Graph of z = 9x2 + y2 together with a few level curves plotted on boththe surface and on the xy−plane. (b) The level curves of the surface in (a).

(a) (b)

Figure 1.8: Contour maps of some functions.

1.1.3 Functions of three or more variables

Given the function f : D ⊃ R3 → R of three variables, we can describe f by examiningthe level surfaces of f . These are surfaces in R3 given by the equations f(x, y, z) = k,where k ∈ R. An example of a function of three variables is the salinity S at a point on



(a) (b)

Figure 1.9: (a) Graph of z = −10xye−(x2+y2) (b) Level curves of the surface in (a) withthe part of the surface above the xy−plane in black and the part below the xy−plane inred.

the ocean surface and described by S = f(x, y, t), where x is the longitude, y is thelatitude, and t is time. It is much more di�cult to visualize functions of three or morevariables.

In general, a function f of n variables may written in vectorial notation as

f(x) = c · x

where c =⟨c1, c2, · · · cn

⟩and x =

⟨x1, x2, · · ·xn

⟩.

Illustrative Example 3. Consider the function f(x, y, z) = x2 + y2 + z2. Describe thelevel surfaces of f .

Illustrative Solution 3. The level surfaces of f are given by x2 + y2 + z2 = k, wherek ≥ 0. They describe a family of concentric spheres with radius

√k. Try to sketch a few

of these concentric spheres using GeoGebra.

1.2 Limits & Continuity

Recall that in the case of a function f(x) of one variable, we say that

limx→a

f(x) = L, if and only if limx→a+

f(x) = limx→a−

f(x) = L.

That is, limx→a

f(x) = L, means that we can let f(x) approach L as close as we like by

letting x approach a from both sides without letting x = a. In this case, x can approach


1.2. LIMITS & CONTINUITY

a from only two directions; this is not case for functions of two variables as we would seeshortly. Also recall that a function of one variable f(x) is continuous at a point x = a if

limx→a

f(x) = f(a).

This property also applies in the case of functions of two variables and would be statedlater in the lesson. Below is a formal de�nition of the limit of a function of tow variables.

De�nition 4.Let f be a function of two variables whose domainD includes points arbitrarily close to(a, b). Then we say that the limit of f(x, y) as (x, y) approaches (a, b) is L and we write

lim(x,y)→(a,b)

f(x, y) = L

if for every number ε > 0 there is a corresponding number δ > 0 such that if (x, y) ∈ Dand

0 <√

(x− a)2 + (y − b)2 < δ then |f(x, y)− L| < ε.

Other standard notations for the limit of two variables is given below

limx→ay→b

f(x, y) = L.

The limit of a function of two variables lim(x,y)→(a,b)

f(x, y) = L, means that we can let

f(x, y) get as close as we like to the number L by letting (x, y) approach the point (a, b)without letting it be equal to (a, b). The limit of functions of two variables is moredi�cult to �nd than the limit of functions of a single variable. This is because, for afunction of two variables, the point (a, b) can be approached in an in�nite number ofdirections unlike single-variables functions in which only two directions are considered.This is illustrated in Figure 1.10. If lim

(x,y)→(a,b)f(x, y) = L exits, then the limit must be

the same irrespective of the direction through which (x, y) approach the point (a, b).This is stated mathematically below

PowerPoint If lim(x,y)→(a,b)

f(x, y) = L1 along a path C1 and lim(x,y)→(a,b)

f(x, y) = L2 along a path C2, where

L1 6= L2, then lim(x,y)→(a,b)

f(x, y) does not exist.

Thus, one of the ways of determining whether a limit does not exist is to take thelimit along two or three di�erent paths, if you obtain di�erent values, then the limitdoes not exist. In fact, even if the limits along these few number of paths is the same,this is still not a proof that the limit exists! However, you may try to use the ε− δde�nition of a limit to proof existence of the limit. Another handy apporach is to applya theorem on the continuity of f(x, y), this would be stated later. Some of the commonpaths to try are the x and y−axis (y = 0, x = 0), the straight line y = mx, or theparabolas y = x2 and x = y2.

Example 2. Find the following limits, if they exist.



Figure 1.10: Illustration of di�erent directions of approach to the point (a, b).

(a) lim(x,y)→(0,0)

x2

x2 + y2

(b) lim(x,y)→(0,0)

xy

2x2 + 3y2

(c) lim(x,y)→(0,0)

5x2y

x2 + y2

Solution 2. (a) Along the x−axis, y = 0, and the limit becomes

lim(x,y)→(0,0)

x2

x2 + y2= lim

(x,y)→(0,0)

x2

x2= lim

(x,y)→(0,0)1 = 1.

Along the y−axis, x = 0, and the limit is given by

lim(x,y)→(0,0)

0

0 + y2= lim

(x,y)→(0,0)0 = 0.

Thus, since the limit is di�erent for the two paths we can conclude that the limitdoes not exist.

(b) Along the line y = 0, the limit is given by

lim(x,y)→(0,0)

xy

2x2 + 3y2= lim

(x,y)→(0,0)

0

2x2 + 02= lim

(x,y)→(0,0)0 = 0.

Along the line x = 0, we get

lim(x,y)→(0,0)

xy

2x2 + 3y2= lim

(x,y)→(0,0)

0

02 + 3y2= lim

(x,y)→(0,0)0 = 0.

Thus, we get the same limit along the two paths. But this does not mean that thelimit exists or that the value of the limit is 0. What this means is that along the



two paths (the axes) the function f(x, y) approach the same value. Now let'sconsider the path y = x. The limit becomes

lim(x,y)→(0,0)

xy

2x2 + 3y2= lim

(x,y)→(0,0)

x2

5x2= lim

(x,y)→(0,0)

1

5=

1

5.

So the limit does not exist, since the limit along the path y = x is di�erent fromthe value of the limit taken along x and y axes. Note that we could have used themore general straight line path y = mx instead of y = x, where m is a constant.

(c) Along the path y = mx, the limit is given by

lim(x,y)→(0,0)

5x2y

x2 + y2= lim

(x,y)→(0,0)

5mx3

x2 +mx4= lim

x→0

5mx

1 +mx2= 0,

for all m. Note that the function that results after making the substitutiony = mx is single-variable, and so we can apply the methods from single-variablecalculus. It's straightforward to show that we get the same limit along the y−axis.Now, along the parabola y = x2, we get

lim(x,y)→(0,0)

5x2y

x2 + y2= lim

x→0

5x4

x2 + x4= lim

x→0

5x2

1 + x2= 0.

Similarly, we can show that along the parabola x = y2, the limit is 0. Asmentioned before, this does not show that the limit exists and is zeo, but becausewe are getting the same limit along these di�erent paths, we begin to suspect thatthe limit exists. We are going to apply the de�nition of the limit to try to proofthat the limit actually exists and is equal to 0.

Let ε > 0. According to the de�nition, we want to �nd δ > 0 such that

if 0 <√x2 + y2 < δ then

∣∣∣ 5x2y

x2 + y2− 0∣∣∣ < ε

That is, if 0 <√x2 + y2 < δ then

5x2|y|x2 + y2

< ε

Since y2 ≥ 0, we have x2 ≤ x2 + y2 =⇒ x2

x2 + y2≤ 1. Multiplying through by 5|y|,

we get5x2|y|x2 + y2

≤ 5|y| = 5√y2 ≤ 5

√x2 + y2 < 5δ

since√x2 + y2 < δ. Hence, if 0 <

√x2 + y2 < δ, we can choose δ = ε/5 such that∣∣∣ 5x2y

x2 + y2− 0∣∣∣ ≤ 5

√x2 + y2 < 5δ = 5

(ε5

)= ε

Question

2. Determine if lim(x,y)→(0,0)

f(x, y) exists, given that



(a) f(x, y) =x2y2

x4 + 7y4

(b) f(x, y) =x3y

x6 + y2

(c) f(x, y) =y4

x4 + 3y4

(d) f(x, y) =xy4

x2 + y8

Continuity

As in single-variable calculus, the continuity of a function of two variables is helpful in�nding the limit the function as de�ned below.

De�nition 5.A function of two variables f(x, y) is said to be continuous at a point (a, b) if

lim(x,y)→(a,b)

f(x, y) = f(a, b).

f(x, y) is said to be continous on a domain D if f is continuous at every point (a, b) ∈ D.

Thus, if we know that f(x, y) is a continuous function, then we can compute the limit ata point (a, b) by substituting the point into the function and evaluating it:

lim(x,y)→(a,b)

f(x, y) = f(a, b). For instance, all polynomials are continuous on R2 whereas all

rational functions are continuous in their domain.

Example 3. Let f(x, y) =x2 − y2

x2 + y2. Where is f(x, y) continuous?

Solution 3. The function f(x, y) is a rational function, therefore it is continous on itsdomain:

D = {(x, y)|(x, y) 6= (0, 0)}

Illustrative Example 4. Consider the function

f(x, y) =

x2−y2x2+y2

if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0)

Illustrative Solution 4. Note that lim(x,y)→(0,0)

x2 − y2

x2 + y2does not exist. This is because,

along the path x = 0, the limit is −1 whereas it is 1 along the path y = 0. Thus, eventhough f(x, y) is de�ned at (0, 0), it is not continuous there because the limit as (x, y)approach (0, 0) does not exist.

Question

3. Determine if the following function is continuous at (0, 0).

f(x, y) =

5x2yx2+y2

if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0)

Example 4. Evaluate the following limits, if the limit exists



(a) lim(x,y)→(1,1)

5x2 − 2xy + y3

(b) lim(x,y)→(2,−1)

xy2

x+ y

(c) lim(x,y)→(1,1)

2x2 − xy − y2

x2 − y2

(d) lim(x,y,z)→(2,1,0)

5x2 − 5y cos(xπ − π + z)

Solution 4. (a) lim(x,y)→(1,1)

5x2 − 2xy + y3. Here, the function is a polynomial and so

it's continuous everywhere. Thus, we can just substitute the point (1, 1) todetermine the limit.

lim(x,y)→(1,1)

5x2 − 2xy + y3 = 5(1)2 − 2(1)(1) + (1)3 = 5− 2 + 1 = 4.

(b) lim(x,y)→(2,−1)

xy2

x+ y. The function is not continuous along the line y = −x since it is

unde�ned there. But this does not mean that we cannot take the limit, since, inparticular, the point (2,−1) is not along this line. Thus

lim(x,y)→(2,−1)

xy2

x+ y=

2(−1)2

2− 1=

2

1= 2

(c) lim(x,y)→(1,1)

2x2 − xy − y2

x2 − y2. The function in this case is not continuous at the point

(1, 1) because it's not de�ned there. Again, this does not mean that we cannottake the limit. We can take the limit after observing that the function can besimpli�ed by factorizing the numerator such that

lim(x,y)→(1,1)

2x2 − xy − y2

x2 − y2= lim

(x,y)→(1,1)

2x2 − 2xy + xy − y2

x2 − y2

= lim(x,y)→(1,1)

(2x+ y)(x− y)(x− y)(x+ y)

= lim(x,y)→(1,1)

(2x+ y)

(x+ y)=

2(1) + 1

1 + 1=

3

2.

(d) lim(x,y,z)→(2,1,0)

5x2 − 5y cos(xπ − π + z). The function is continous at (2, 1, 0) so we

can substitute the values into it to get

lim(x,y,z)→(2,1,0)

5x2−5y cos(xπ−π+z) = 5(2)2−5(1) cos(2π−π+0) = 20−5 cos(π) = 20+5 = 25

Question

4. Evaluate the limit lim(x,y)→(1,1)

x4 − y4

x2 + y2

Remark 1. The limit laws from calculus I still hold in the case of functions of severalvariables. In particular

(a) lim(x,y)→(a,b)

[f(x, y) + g(x, y)] = lim(x,y)→(a,b)

f(x, y) + lim(x,y)→(a,b)

g(x, y)

(b) lim(x,y)→(a,b)

x = a; lim(x,y)→(a,b)

c = c, where c is a constant.



(c) The Squeeze Theorem also holds.

Remark 2. Consider the composite function h(x, y) = g(f(x, y)). If f(x, y) is acontinuous function and g is also continuous, then h(x, y) is continuous.

Remark 3. Polar coordinates may also be used to simplify the evaluation of limits,especially limits at the origin and for expressions that are independent of the polar angleθ.

Example 5. Given that f(x, y) = cos−1(y/x). Where is f(x, y) continuous.

Solution 5. The function g(x, y) = y/x is continuous except at x = 0. Thus, f(x, y) isalso continuous on the same domain D = {(x, y)|x 6= 0}.

Example 6. Evaluate the limit lim(x,y)→(0,0)

(x2 + y2) ln(x2 + y2).

Solution 6. We can employ polar coordinates in this case: r2 = x2 + y2, x = r cos θ,and y = r sin θ.

lim(x,y)→(0,0)

(x2 + y2) ln(x2 + y2) = limr→0

r2 ln(r2) = limr→0

2 ln r

r−2

Using L'Hopital's rule, we can write

limr→0

2 ln r

r−2= lim

r→0

(2/r)

−2r−3= lim

r→0−r2 = 0


MATH 355: Calculus of Several Variables1.1Functions of Several Variables 1.1.1unctionsF of two...

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