Advanced calculus-I
MT3501-as per St. Xavier’s College, Ahmedabad
MT-301-as per Gujarat University
Published by:
Shanti-Peace for Mathematics
SP
M
Prepared by:
Dr. Prashantkumar Patel
For Subject introduction watch the video https://youtu.be/-94ebeidIHw
Contents
1 Introduction of Function of several variables
1
2 Neighborhoods of point in R2
2.1 Rectangular δ-Neighborhood of point in R2 . . . . .
2.2 Deleted rectangular δ-Neighborhood of point in R2
2.3 Spherical δ-Neighborhood of point in R2 . . . . . .
2.4 deleted spherical δ-Neighborhood of point in R2 . .
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3
4
4
4
5
3 Limit of a function in two variables
6
3.1 Path Dependent of the limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.2 Repeated or Iterated limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
15
5 Partial Derivatives
19
6 HIGHER-ORDER PARTIAL DERIVATIVES
23
7 PARTIAL DERIVATIVES AND CONTINUITY
27
8 Differentiability
29
9 Sufficient condition for differentiability
34
10 EQUALITY OF MIXED PARTIALS
11 Directional Derivative
12 Differential
13 THE CHAIN RULE
SP
14 IMPLICIT DIFFERENTIATION
M
4 Continuity of function of Two variables
37
46
50
53
65
15 Which variable is to be treated as constant
67
16 Jacobian
69
17 Euler’s Theorem for Homogeneous functions
69
18 MAXIMA AND MINIMA OF FUNCTIONS OF TWO VARIABLES
77
19 Lagrange’s method of undermined multiplies
86
20 Taylor’s Theorem of two variables
90
21 Multiple points
95
22 Curvature
97
1
Introduction of Function of several variables
To understand this topic, please watch the video https://youtu.be/g5-KYx7wYdU.
In this course, we shall develop concepts for understanding functions of several variables: limit, continuity, differentiability, problem of maxima or minima. We shall see how these concept resemble and are different than that of
single variable.
Since most of theorem that can also be proved for functions of two variables can be extended to functions of three
or more variables, we confine our analysis to two/three variables only.
Definition 1.1 A real valued function of n variables is a function that takes as input n real numbers, represented
by the variables x1 , x2 , . . . , xn for producing another real number, the value of the function, commonly denoted
f (x1 , x2 , x3 , . . . , xn ).
A real valued function of n real variable is a function
f :X→R
such that its domain X is a subset of Rn that contains an open set.
A such real valued function are alsoo know as Scalar field. The set X is called domain and the set f (X) = {f (x̄)|x̄ ∈ X}
is called the range.
Definition 1.2 Let f : X → R be function of two variables. Let
G(f ) = (x, y, z) ∈ R3 : z = f (x, y), (x, y) ∈ X ⊂ R3
The set G(f ) is called the graph of the function f . It is also called a surface in R3 .
Example 1.3 Let X = (a, b) ∈ R2 |a > 0, b > 0 . Define f : X → R such that
f (a, b) = ab,
M
which is the area V of a rectangle with sides a and b. The domain restricts all variables to be positive since side must
be positive.
Example 1.4 Let X = (A, h) ∈ R2 |A > 0, h > 0 . Define V : X → R such that
V (A, h) =
1
Ah,
3
which is the volume V of a cone with base area A and height h measured perpendicularly from the base. The domain
restricts all variables to be positive since lengths and area must be positive.
Example 1.5 Let X = (a, b, c) ∈ R3 |a > 0, b > 0, c > 0 . Define g : X → R such that
SP
g(a, b, c) = abc,
which is the volume V of a cuboid with sides a, b and c. The domain restricts all variables to be positive since side
must be positive.
Example 1.6 The cost of renting a bike depends on how may days you keep it and how far you drive. Represent this
using function.
Let d be the number of days you rent the bike, and m be the number of miles you drive. Then the cost of the car rental
C(d, m) is function of two variables.
Example 1.7 The demand of for Maska bun depends on the price for the bun and also on the Maska (butter). Represent
this is as a function.
The demand z = f (B, M ) is function of two variables. Here, B is price of bun and M is the prize of Maska.
Example 1.8 The relation
z=
p
1 − x2 − y 2
(1)
2
2
between x, y, z determines a value of z corresponding to every pair of number x, y which are such that x + y ≤ 1.
Denoting a pair of number x, y geometrically by a point on a XY -plane, we see that the points (x, y) for which x2 +y 2 ≤ 1
lie on or within circle whose centre is at the origin and radius 1.
The region determine by the point (x, y) is called the domain of the point (x, y).
We say that the relation (1) associated a real number z to each pair (x, y) such that x2 + y 2 ≤ 1. We say that we have
a function whose domain is the set
(x, y) : x2 + y 2 ≤ 1 .
We may write
f : (x, y) → z =
p
1 − x2 − y 2 ,
where (x, y) ∈ (x, y) ∈ R2 |x2 + y 2 ≤ 1 . The graph of this surface is plotted in figure 1.
p
1 − x2 − y 2
M
Figure 1: Graph of the function z =
p
(1 − x)(x − 2) +
SP
Figure 2: Graph of the function z =
p
(3 − y)(y − 4).
Example 1.9 Consider the relation
z=
p
p
(1 − x)(x − 2) + (3 − y)(y − 4),
(2)
Now, (1 − x)(x − 2) is nonnegative if 1 ≤ x ≤ 2 and (3 − y)(4 − y) is nonnegative if 3 ≤ y ≤ 4.
The points (x, y) for which 1 ≤ x ≤ 2, 3 ≤ y ≤ 4 determine a rectangular domain bounded by the line
x = 1; x = 2; y = 3; y = 4.
Hence, the domain of this function is
D = {(x, y) : 1 ≤ x ≤ 2; 3 ≤ y ≤ 4} .
The graph of this function plotted in figure 2.
2
2
Example 1.10 The relation z = e−x −y determine a function of two variable (x, y); the domain of the function being
whole plane, that means the set of all the ordered pairs of real numbers.
Exercise 1.11 Give the domain of the following functions. Also, draw them in the online 3Dplot.
1. z =
2
p
16 − x2 − y 2
2. z =
x+y
x−y
3. z =
1
log(x) + log(y)
4. z =
1
log(x + y)
Neighborhoods of point in R2
In this section, we learn about two type of neighborhood of point in R2 . Both these neighborhood are equivalent.
2.1
Rectangular δ-Neighborhood of point in R2
To understand this topic, please watch the video https://youtu.be/xiRGj31yraA.
Let (a, b) ∈ R2 and δ > 0 be given, then we defined the square δ-neighborhood of the point (a, b) as the set
N ((a, b), δ)
(a − δ, a + δ) × (b − δ, b + δ)
(x, y) ∈ R2 /x ∈ (a − δ, a + δ) & y ∈ (b − δ, b + δ)
(x, y) ∈ R2 /|x − a| < δ & |y − b| < δ
(x, y) ∈ R2 /a − δ < x < a + δ & b − δ < y < b + δ
=
=
=
=
It is also denoted by Nδ ((a, b)).
2.2
Deleted rectangular δ-Neighborhood of point in R2
Let (a, b) ∈ R2 and δ > 0 be given. The deleted Rectangular δ-Neighborhood of point (a, b) is defined as
N ∗ ((a, b), δ)
=
=
N ((a, b), δ) − {(a, b)}
(x, y) ∈ R2 /0 < |x − a| < δ & 0 < |y − b| < δ .
It is also denoted by Nδ∗ ((a, b)).
Solution: Consider
N ((0, 0), 1)
=
=
(0 − 1, 0 + 1) × (0 − 1, 0 + 1)
(x, y) ∈ R2 /|x − 0| < 1 & |y − 0| < 1
(x, y) ∈ R2 / − 1 < x < 1 & − 1 < y < 1
SP
=
M
Example 2.1 Draw N ((0, 0), 1) in R2 .
Figure 3: N ((0, 0), 1) in R2 .
2.3
Spherical δ-Neighborhood of point in R2
Let (a, b) ∈ R2 and δ > 0 be given, then we defined the spherical δ-neighborhood of the point (a, b) with radius δ is
denoted by S((a, b), δ) or Sδ ((a, b)) and it is defined as
S((a, b), δ) =
(x, y) ∈ R2 /(x − a)2 + (y − δ)2 < δ 2
n
o
p
=
(x, y) ∈ R2 / (x − a)2 + (y − δ)2 < δ
Example 2.2 Draw S((0, 0), 1) in R2 .
Figure 4: S((0, 0), 1) in R2 .
Solution: Consider
S((0, 0), 1)
=
Remark 2.3 We note that
or
M
=
(x, y) ∈ R2 /(x − 0)2 + (y − 0)2 < 1
o
n
p
(x, y) ∈ R2 / x2 + y 2 < 1
√
S((a, b), δ) ⊂ N ((a, b), δ) ⊂ S((a, b), 2δ)
Sδ ((a, b)) ⊂ Nδ ((a, b)) ⊂ S√2δ ((a, b)).
SP
From these property, we say that the spherical neighborhood or rectangular neighborhood are equivalent. The above
relation for the particular case is plotted in figure 5.
√
Figure 5: The relation S((0, 0), 1) ⊂ S((0, 0), 1) ⊂ S((0, 0), 2).
2.4
deleted spherical δ-Neighborhood of point in R2
Let (a, b) ∈ R2 and δ > 0 be given, then we defined the deleted spherical δ-neighborhood of the point (a, b) with radius
δ is denoted by Sδ∗ ((a, b), δ) or Sδ∗ ((a, b)) and it is defined as
S ∗ ((a, b), δ) =
(x, y) ∈ R2 /(x − a)2 + (y − δ)2 < δ 2 − {(a, b)} .
3
Limit of a function in two variables
To understand this topic, please watch the video https://youtu.be/E-VC51cdJOE.
Recall that for a function f of one variables, the fundamental concept, which is used again and again to define
various other concepts, is that of limit of f . Immediately, as for a function of one variables, we say f (x, y) approaches
a limit L ∈ R as (x, y) approaches (a, b), if f (x, y) lies ’arbitrarily close’ to L for all points ’sufficiently close’ to (a, b).
To describe this concept of ’closeness’ in R2 , we have the notion of distance between points. So, it is natural to say
that a function f of two variables has a limit L as (x, y) ∈ R2 ’approaches’ (a, b) if the distance between f (x, y) and L
is ’arbitrarily’ small for all point (x, y) sufficiently close to (a, b).
To make this definition meaningful, let us assume that for the time being that f is defined at all points ’sufficiently
closed’ to (a, b), possibly not at (a, b). That is, we assume that there exists some r > 0 such that N ∗ ((a, b), r) ⊂ D.
Definition 3.1 For every points (a, b) in D ⊂ R2 , there exist δ > 0 such that N ((a, b), δ) ⊂ D. Then D is called open set.
Definition 3.2 Let D be a open subset of R2 and (a, b) ∈ D. Let f : D → R. We say f (x, y) approaches a limit L ∈ R as
(x, y) approaches (a, b), and write
lim
f (x, y) = L, if every > 0 there exist δ > 0 such that
(x,y)→(a,b)
0 < |x − a| < δ & 0 < |y − b| < δ ⇒ |f (x, y) − L| < .
That can be write as
(x, y) ∈ N ∗ ((a, b), δ) ⇒ |f (x, y) − L| < ,
which is also equivalent as
(x, y) ∈ S ∗ ((a, b), δ) ⇒ |f (x, y) − L| < ,
OR
p
(x − a)2 + (y − b)2 < δ ⇒ |f (x, y) − L| < ,
M
0<
Remark 3.3 The above limit in the definition is also known as simultaneous limit or double limit.
Example 3.4 Find
lim
(x,y)→(1,2)
f (x, y), where f (x, y) = x.
Solution: We first to guess whether this limit can exist or nor, and if it exists, what is the possible value. Suppose
it exists and is L. Then to prove
lim
f (x, y) = L, we have to find δ > 0 for given > 0 such that
SP
(x,y)→(1,2)
0 < |x − 1| < δ & 0 < |y − 2| < δ ⇒ |f (x, y) − L| < Since f (x, y) = x and |x − 1| < δ, is obvious that if we take L = a and choose δ = then for a given > 0, choose δ = ,
0 < |x − 1| < δ & 0 < |y − 2| < δ ⇒ |f (x, y) − L| = |x − 1| < .
Hence,
lim
(x,y)→(1,2)
x = 1.
Example 3.5 Prove that lim(x,y)→(1,2) 2x + 3y = 8.
Solution: Here, f (x, y) = 2x + 3y and (a, b) = (1, 2).
Let > 0 be given, we want to find δ > 0 such that
0 < |x − 1| < δ & 0 < |y − 2| < δ ⇒ |(2x + 3y) − 8| < .
Consider
|f (x, y) − 8| = |(2x + 3y) − 8|
= |2(x − 1 + 1) + 3(y − 2 + 2) − 8|
= |2(x − 1) + 2 + 3(y − 2) + 6 − 8|
= |2(x − 1) + 3(y − 2)|
≤ 2|x − 1| + 3|y − 2|
< 2δ + 3δ = 5δ, if |x − 1| < δ & |y − 2| < δ
< , if δ < .
5
such that
5
|(2x + 3y) − 8| < , when 0 < |x − 1| < δ & 0 < |y − 2| < δ.
Thus for > 0 there exists δ <
Hence,
lim
(x,y)→(1,2)
2x + 3y = 8.
Example 3.6 Prove that lim(x,y)→(1,2) 3xy = 6.
solution: Here f (x, y) = 3xy and (a, b) = (1, 2)
Let > 0 be given, we want to find δ > 0 such that
0 < |x − 1| < δ & 0 < |y − 2| < δ ⇒ |3xy − 6| < .
Consider
|f (x, y) − 6| = |3xy − 6|
= |3(x − 1 + 1)(y − 2 + 2) − 6|
= |3(x − 1)(y − 2) + 6(x − 1) + 3(y − 2) + 6 − 6|
= |3(x − 1)(y − 2) + 6(x − 1) + 3(y − 2)|
≤ 3|x − 1|y − 2| + 6|x − 1| + 3|y − 2|
< 3δ 2 + 6δ + 3δ, if |x − 1| < δ & |y − 2| < δ
< 3δ + 6δ + 3δ = 12δ, if 0 < δ < 1
< if δ <
.
12
that means δ < 1 and δ <
.
12
n12 o
Therefore For given > 0, there exist δ = min 1,
> 0 such that
12
Thus, δ = min 1,
Hence,
lim
3xy = 6.
(x,y)→(1,2)
Example 3.7 Prove that lim(x,y)→(2,1) x2 y + xy 2 = 6.
solution: Here f (x, y) = x2 y + xy 2 and (a, b) = (2, 1)
Let > 0 be given, we want to find δ > 0 such that
M
0 < |x − 1| < δ & 0 < |y − 2| < δ ⇒ |3xy − 6| < .
SP
0 < |x − 2| < δ & 0 < |y − 1| < δ ⇒ |x2 y + yx2 − 6| < .
Consider
|f (x, y) − 6| = |x2 y + xy 2 − 6|
= |(x − 2 + 2)2 (y − 1 + 1) + (x − 2 + 2)(y − 1 + 1)2 − 6|
= | (x − 2)2 + 4(x − 2) + 4 (y − 1 + 1) + (y − 1)2 + 2(y − 1) + 1 (x − 2 + 2) − 6|
= |(x − 2)2 (y − 1) + 4(x − 2)(y − 1) + 4(y − 1) + (x − 2)2 + 4(x − 2) + 4
+(y − 1)2 (x − 2) + 2(y − 1)(x − 2) + (x − 2) + 2(y − 1)2 + 4(y − 1) + 2 − 6|
= |(x − 2)2 (y − 1) + 6(x − 2)(y − 1) + 8(y − 1) + (x − 2)2 + 5(x − 2)
+(y − 1)2 (x − 2) + 2(y − 1)2 |
≤
|x − 2|2 |y − 1| + 6|x − 2||y − 1| + 8|y − 1| + |x − 2|2 + 5|x − 2|
+|y − 1|2 |x − 2| + 2|y − 1|2
<
δ 3 + 6δ 2 + 8δ + δ 2 + 5δ + δ 3 + 2δ, if |x − 2| < δ & |y − 1| < δ
< 24δ, if 0 < δ < 1
.
< if δ <
24
n o
Thus, δ = min 1,
that means δ < 1 and δ <
.
24
n 24
o
Therefore For given > 0, there exist δ = min 1,
> 0 such that
24
0 < |x − 2| < δ & 0 < |y − 1| < δ ⇒ |x2 y + xy 2 − 6| < .
Hence,
lim
x2 y + xy 2 = 6.
(x,y)→(2,1)
Example 3.8 Prove that
lim
(x,y)→→(−1,1)
x2 + 3y = 4.
solution: Here f (x, y) = x2 + 3y and (a, b) = (−1, 1)
Let > 0 be given, we want to find δ > 0 such that
0 < |x + 1| < δ & 0 < |y − 1| < δ ⇒ |x2 + 3y − 4| < .
Consider
|f (x, y) − 6| = |x2 + 3y − 4|
= |(x + 1 − 1)2 + 3(y − 1 + 1) − 4|
= |(x + 1)2 − 2(x + 1) + 1 + 3(y − 1) + 3 − 4|
= |(x + 1)2 − 2(x + 1) + 3(y − 1)|
≤
|x + 1|2 + 2|x + 1| + 3|y − 1|
<
δ 2 + 2δ + 3δ, if |x + 1| < δ & |y − 1| < δ
<
δ + 2δ + 3δ, if 0 < δ < 1
= 6δ < if δ < .
6
n o
that means δ < 1 and δ < .
Thus, δ = min 1,
6
n 6 o
> 0 such that
Therefore, for given > 0, there exist δ = min 1,
6
0 < |x + 1| < δ & 0 < |y − 1| < δ ⇒ |x2 + 3y − 4| < .
lim
x2 + 3y = 4.
M
Hence,
(x,y)→(−1,1)
Example 3.9 Prove that
3
2x2 + y 2
=
5
(x,y)→(1,1) x + 4y
lim
2x2 + y 2
3
, (a, b) = (1, 1) and L = .
x + 4y
5
Let > 0 be given, we want to find δ > 0 such that
SP
solution: Here f (x, y) =
0 < |x − 1| < δ & 0 < |y − 1| < δ ⇒
Consider
|f (x, y) − L| =
=
=
2x2 + y 2
3
−
< .
x + 4y
5
3
2x2 + y 2
−
x + 4y
5
2
2
5(2x + y ) − 3(x + 4y)
5(x + 4y)
2
10x + 5y 2 − 3x − 12y
.
5x + 20y)
We have
0 < |x − 1| < δ & 0 < |y − 1| < δ
That means
which can be written as
1 − δ < x < 1 + δ;
1 − δ < y < 1 + δ,
(3)
1 − δ < x < 1 + δ;
4 − 4δ < 4y < 4 + 4δ.
Adding above two inequalities, we obtain
5 − 5δ < x + 4y < 5 + 5δ.
Therefore,
Using second inequality in (4), we have
1
1
1
<
<
.
5 + 5δ
x + 4y
5 − 5δ
1
1
<
.
x + 4y
5 − 5δ
(4)
Now, if
1
5 − 5δ
∴ 5 − 5δ
∴ −5δ
∴δ
<
1
> 1
> −4
4
<
.
5
Now,
10x2 + 5y 2 − 3x − 12y
5x + 20y
4
1
10x2 + 5y 2 − 3x − 12y if δ <
5
5
1
=
10(x − 1 + 1)2 + 5(y − 1 + 1)2 − 3(x − 1 + 1) − 12(y − 1 + 1)
5
1 =
10(x − 1)2 + 20(x − 1) + 10 + 5(y − 1)2 + 10(y − 1) + 5 − 3(x − 1) − 3 − 12(y − 1) − 12
5
1
=
10(x − 1)2 + 17(x − 1) + 5(y − 1)2 − 2(y − 1)
5
1
≤
10|x − 1|2 + 17|x − 1| + 5|y − 1|2 + 2|y − 1|
5
1 2
10δ + 17δ + 5δ 2 + 2δ , if 0 < |x − 1| < δ, 0 < |y − 1| < δ
<
5
1
<
{10δ + 17δ + 5δ + 2δ} , if δ < 1
5
34δ
5
=
< if δ <
5
34
4 5
Thus, taking δ = min
,
.
5 34
4 5
,
> 0 such that
Therefore, for given > 0, there exist δ = min
5 34
SP
M
<
0 < |x − 1| < δ & 0 < |y − 1| < δ ⇒ f (x, y) −
Hence,
3
< .
5
2x2 + y 2
3
= .
5
(x,y)→(1,1) x + 4y
lim
Example 3.10 Prove that
x2 + y 2
= 1.
(x,y)→(1,1) x + y
lim
x2 + y 2
, (a, b) = (1, 1) and L = 1.
x+y
Let > 0 be given, we want to find δ > 0 such that
Solution: Here f (x, y) =
0 < |x − 1| < δ & 0 < |y − 1| < δ ⇒
x2 + y 2
− 1 < .
x+y
Consider
|f (x, y) − L| =
≤
x2 + y 2
−1
x+y
x2 + y 2 − x − y
.
(x + y)
We have
0 < |x − 1| < δ & 0 < |y − 1| < δ
That means
1 − δ < x < 1 + δ;
1 − δ < y < 1 + δ,
Adding above two inequalities, we obtain
2 − 2δ < x + y < 2 + 2δ.
(5)
Therefore,
1
1
1
<
<
.
2 + 2δ
x+y
2 − 2δ
Using second inequality in (6), we have
1
1
<
.
x+y
2 − 2δ
Now, if
1
2 − 2δ
∴ 2 − 2δ
∴ −2δ
∴δ
<
1
> 1
> −1
1
<
.
2
Now,
x2 + y 2 − x − y if δ <
=
=
(x − 1 + 1)2 + (y − 1 + 1)2 − (x − 1 + 1) − (y − 1 + 1)
(x − 1)2 + 2(x − 1) + 1 + (y − 1)2 + 2(y − 1) + 1 − (x − 1) − 1 − (y − 1) − 1
=
(x − 1)2 + (x − 1) + (y − 1)2 + (y − 1)
<
δ 2 + δ + δ 2 + δ, if 0 < |x − 1| < δ, 0 < |y − 1| < δ
<
δ + δ + δ + δ, if δ < 1
4δ < if δ < 4
=
Thus, taking δ = min
1
2
<
M
x2 + y 2 − x − y
x+y
1 1 ,
. Therefore, for given > 0, there exist δ = min
,
> 0 such that
2 4
2 4
0 < |x − 1| < δ & 0 < |y − 1| < δ ⇒ |f (x, y) − 1| < .
x2 + y 2
= 1.
(x,y)→(1,1) x + y
lim
Example 3.11 Prove that
SP
Hence,
1
1
x sin + y sin
= 0.
y
x
(x,y)→(0,0)
lim
1
1
+ y sin , (a, b) = (0, 0) and L = 0.
y
x
Let > 0 be given, we want to find δ > 0 such that
1
1
0 < |x − 0| < δ & 0 < |y − 0| < δ ⇒ x sin + y sin
− 0 < .
y
x
solution: Here f (x, y) = x sin
Consider
1
1
+ y sin − 0
y
x
1
1
≤ x sin + y sin
y
x
1
1
≤ |x| sin + |y| sin
y
x
≤ |x| + |y|( as | sin θ| < 1 for all θ ∈ R)
|f (x, y) − L| =
x sin
δ + δ if |x| < δ, |y| < δ
< 2δ < if δ < .
2
Thus, taking δ =, . Therefore, for given > 0, there exist δ = > 0 such that
2
2
<
0 < |x| < δ & 0 < |y| < δ ⇒ |f (x, y) − 0| < .
Hence,
lim
(x,y)→(0,0)
x sin
1
1
+ y sin
y
x
= 0.
(6)
3.1
Path Dependent of the limit
To understand this topic, please watch the video https://youtu.be/zHY2lm3XreM.
Theorem 3.12 Let D be a subset of R2 , which contains neighborhood of point (a, b) ∈ R2 . Let f : D → R2 be such that
lim
f (x, y) = L exists, If g : I → D ⊂ R2 is any path such that
(x,y)→(a,b)
limx→a g(x) = b, and let h(x) := f (x, g(x)) , x ∈ I
then lim h(x) = L.
SP
M
x→a
Figure 6: Existence of limit along every path.
From the above theorem, if lim(x,y)→(a,b) f (x, y) = L exist, then f (x, g(x)) → L as x → a along every path y = g(x)
through (a, b). That is, a limit of f (x, y) exists and is same along every path to (a, b). This gives us a test for the
nonexistence of a limit.
Corollary 3.13 (Test for non-existence of limit) If for a function f (x, y) either the limit of f (x, y) at (a, b) does not
exist along at least one path through (a, b) or f (x, y) has different limits along at least two different paths through (a, b),
then
lim
f (x, y) does not exist.
(x,y)→(a,b)
Remarks 3.14
1. If
lim
(x,y)→(a,b)
f (x, y) = L exists then the limit (L) is independent of any path y = g(x) (or x = h(y)).
2. The converse of remark-1 is not true.
3. If
lim
(x,y)→(a,b)
f (x, y) depends upon given path y = g(x) or x = h(y), then the double limit does not exist.
Example 3.15 Show that
lim
(x,y)→(0,0)
x+y
does not exist.
x−y
Solution: Let us find this limit along some simple paths through (0, 0). For example, along path y = g(x) = mx, the
line through the origin with slope m. Note that y → 0 as x → 0. Since
f (x, mx) =
x + mx
1+m
=
,
x − mx
1−m
we have
1+m
1+m
=
,
1−m
1−m
x+y
which is different along different lines. Hence,
lim
does not exist.
(x,y)→(0,0) x − y
lim f (x, mx) = lim
x→0
x→0
Example 3.16 Show that
x4 y 4
does not exist.
+ y 4 )3
lim
(x,y)→(0,0) (x2
Solution: Let us find this limit along some simple paths through (0, 0). For example, along path x = h(y) = my 2 ,
the curve through the origin with parameter m. Note that x → 0 as y → 0. Since
f (my 2 , y) =
(my 2 )4 y 4
((my 2 )2 + y 4 )
3
=
m4 y 8 y 4
3
(m2 y 4 + y 4 )
we have
=
m4 y 12
3
(m2 + 1) y 12
=
m4
(m2 + 1)
3,
m4
m4
= 2
,
2
y→0 m + 1
m +1
lim f (my 2 , y) = lim
y→0
which is different with different values of m. Hence,
Example 3.17 Show that
lim
(x,y)→(0,0) x2 y 2
x4 y 4
does not exist.
(x,y)→(0,0) (x2 + y 4 )3
lim
x2 y 2
does not exist.
+ (x − y)4
Solution: Let us find this limit along some simple paths through (0, 0). For example, along path y = g(x) = mx, the
line through the origin with slope m. Note that y → 0 as x → 0. Since
m2 x4
m4
x2 (m2 x2 )
=
=
,
(x2 (m2 x2 ) + (x − mx)4 )
(m2 x4 + x4 (1 − m)4 )
(m2 + (1 − m)4 )
we have
m4
m4
=
,
x→0 (m2 + (1 − m)4 )
(m2 + (1 − m)4 )
lim f (x, mx) = lim
x→0
which is different with different values of m. Hence,
Example 3.18 Show that
M
f (x, mx) =
lim
(x,y)→(0,0) x2 y 2
x2 y 2
does not exist.
+ (x − y)4
x5 y
does not exist.
(x,y)→(0,0) x10 + y 2
lim
SP
Solution: Let us find this limit along some simple paths through (0, 0). For example, along path y = g(x) = mx5 ,
the curve through the origin with parameter m. Note that y → 0 as x → 0. Since
f (x, mx) =
we have
x5 (mx5 )
(mx10 )
m
= 10
=
,
2
10
+ (m x ))
(x (1 + m2 ))
(1 + m2 )
(x10
lim f (x, mx5 ) = lim
x→0
x→0
which is different with different values of m. Hence,
Example 3.19 Show that
lim
m
m
=
,
(1 + m2 )
(1 + m2 )
x5 y
does not exist.
(x,y)→(0,0) x10 + y 2
lim
xy 3
does not exist.
+ y6
(x,y)→(0,0) x2
Solution: Let us find this limit along some simple paths through (0, 0). For example, along path x = h(y) = my 3 ,
the curve through the origin with parameter m. Note that x → 0 as y → 0. Since
f (my 3 , y) =
we have
my 3 y 3
m2
=
,
2
6
6
(m y + y )
(1 + m2 )
y→0
which is different with different values of m. Hence,
Example 3.20 Show that
m
m
=
,
2
y→0 (1 + m )
(1 + m2 )
lim f (my 3 , y) = lim
xy 3
does not exist.
(x,y)→(0,0) x2 + y 6
lim
x3 + y 3
does not exist.
(x,y)→(0,0) x − y
lim
Solution: Let us find this limit along some simple paths through (0, 0). For example, along path x − y = mx3 , which
is same as y = x − mx3 the curve through the origin with parameter m. Note that y → 0 as x → 0. Since
3
3
x3 + x3 1 − mx2
1 + 1 − mx2
x3 + (x − mx3 )3
3
=
=
f (x, x − mx ) =
,
mx3
mx3
m
we have
3
1 + 1 − mx2
1 + (1 − m · 0)3
2
=
= ,
lim f (x, x − mx ) = lim
x→0
x→0
m
m
m
x3 + y 3
which is different with different values of m. Hence,
lim
does not exist.
(x,y)→(0,0) x − y
3
3.2
Repeated or Iterated limits
To understand this topic, please watch the video https://youtu.be/kBJFkb71KLI
Definition
3.21 Let
of R2 such that neighborhood of (a, b) contain in D. Let f : D ⊂ R2 → R be a function.
D be subset
n
o
IF lim lim f (x, y) and lim lim f (x, y) exist, then they are called iterated limits or repeated limits.
x→a
y→b
y→b
x→a

 x−y
Example 3.22 If f (x, y) =
x+y

2
x + y 6= 0,
Find the repeated limit of f (x, y) at the point (0, 0).
x + y = 0.
Solution: Here
x→0
lim f (x, y)
y→0
x−y
x→0 y→0 x + y
x−0
= lim
x→0 x + 0
x
= lim = 1.
x→0 x
=
lim
lim
n
y→0
lim
SP
Similarly,
Remarks 3.23
M
lim
o
lim f (x, y)
x→0
x−y
y→0 x→0 x + y
0−y
= lim
y→0 0 + y
−y
= −1.
= lim
x→0 y
=
lim
lim
1. If lim(x,y)→(0,0) f (x, y) = L exists then both the repeated limits exists and are equal. That means
n
o
lim lim f (x, y) = lim lim f (x, y) =
lim
f (x, y) = L.
x→0
y→0
y→0
x→0
(x,y)→(a,b)
2. The converse of above statement is not true. That means both repeated limits are exist and
n
o
lim lim f (x, y) 6= lim lim f (x, y) .
x→0
Then
lim
(x,y)→(0,0)
y→0
y→0
x→0
f (x, y) does not exist.
Example 3.24 Prove that lim(x,y)→(0,0)
3x + 4y
does not exist.
3x − 4y
Solution: Here
lim
x→0
lim f (x, y)
y→0
3x + 4y
x→0 y→0 3x − 4y
3x + 0
= lim
x→0 3x − 0
3x
= lim
= 1.
x→0 3x
=
lim
lim
Similarly,
lim
n
y→0
o
lim f (x, y)
=
x→0
=
=
3x + 4y
y→0 x→0 3x − 4y
0 + 4y
lim
y→0 0 − 4y
−4y
lim
= −1.
x→0 4y
lim
lim
3x + 4y
does not exist. As in case of
3x − 4y
function of single variable, the following theorem is useful for computing limits.
Thus both the repeated limits exists and are not equal. Therefore, lim(x,y)→(0,0)
Theorem 3.25 Let D be subset of R2 such that neighborhoods of (a, b) contain in D. Let f, g : D ⊂ R2 → R be a
functions. If
lim
f (x, y) = L and lim(x,y)→(a,b) g(x, y) = M and r ∈ R, then
(x,y)→(a,b)
(f ± g)(x, y) = L ± M,
lim
(x,y)→(a,b)
lim
(rf )(x, y) = rL,
(x,y)→(a,b)
lim
(f g)(x, y) = LM,
(x,y)→(a,b)
and if M 6= 0, then
L
f
(x, y) =
.
g
M
(x,y)→(a,b)
Example 3.26 Compute lim(x,y)→(1,2)
M
lim
x+y
.
x−y
Since for f (x, y) = x and g(x, y) = y,
lim
Using Theorem 3.25,
f (x, y) = 1 and
lim
g(x, y) = 2,
(x,y)→(1,2)
SP
(x,y)→(1,2)
1+2
x+y
=
lim
= −3.
(x,y)→(1,2) 1 − 2
(x,y)→(1,2) x − y
lim
Theorem 3.27 Let D be subset of R2 such that neighborhoods of (a, b) contain in D. Let f, g, h : D ⊂ R2 → R be a
functions. Let there exists some r > 0 such that g(x, y) ≤ f (x, y) ≤ h(x, y) for all (x, y) ∈ N ∗ ((a, b), δ) ∩ D. Then if
lim
g(x, y) = L =
(x,y)→(a,b)
lim
h(x, y),
(x,y)→(a,b)
then
lim
f (x, y) = L.
(x,y)→(a,b)
Example 3.28 Compute
lim
(x,y)→(0,0)
tan−1 (xy)
.
xy
From the Manchurian series expansion of tan−1 (φ), it follows that for φ in a neighborhood of φ = 0 except 0,
1−
φ2
tan−1 φ
<
< 1.
3
φ
Hence, in the neighborhood of (0, 0), except the axes, we have
x2 y 2
tan−1 (xy)
<
< 1.
3
xy
tan−1 (xy)
Thus, using sandwich theorem,
lim
= 1.
xy
(x,y)→(0,0)
1−
4
Continuity of function of Two variables
To understand this topic, please watch the video https://youtu.be/qJ_phrAeV1w.
Let D ⊂ R2 , (a, b) ∈ D and f : D → R. We say that f is continuous at (a, b) if for every > 0, there exists some δ > 0
such that (x, y) ∈ D,
|x − a| < δ and |y − b| < δ implies |f (x, y) − f (a, b)| < .
Remark 4.1 In order to check continuity of f (x, y) at a point (a, b) verify that
1.
lim
(x,y)→(a,b)
f (x, y) exists.
2. f (a, b) is well defined.
3.
lim
f (x, y) = f (a, b)
(x,y)→(a,b)
Theorem 4.2 If both f and g are continuous at (a, b) and r ∈ R, then the functions f + g, f − g rf and f g are also
continuous at (a, b). Further, in case g(a, b) 6= 0, the function f /g is defined in neighborhood of (a, b) and is continuous
there.
Theorem 4.3 Let f : D → R be continuous at (a, b) and E be a subset of R containing the range of f . If g : E → R is
continuous at f (a, b), then the composite function g ◦ f : D → R is also continuous at (a, b).
M
Example 4.4 Prove that the function f : R2 → R defined as f (x, y) = sin(xy) is continuous every where.
Solution: Define g : R → R as g(θ) = sin θ. and h : R2 → R as h(x, y) = xy.
Now, we know that g is continuous on R and h is continuous on R2 as product of two continuous function is
continuous.
Therefore, The composition map g ◦ h(x, y) = sin(xy) is continuous on R2 .
SP
Example 4.5 Discuss the continuity of the following function at (0,0).

 p xy
(x, y) 6= (0, 0),
x2 + y 2
f (x, y) =

0
(x, y) = (0, 0).
Solution I: Here f (0, 0) = 0. If f is continuous at (0, 0) then we must have
lim
f (x, y) = f (0, 0).
(x,y)→(0,0)
That means
xy
lim
(x,y)→(0,0)
p
x2 + y 2
= f (0, 0) = 0.
Let > 0 be given. We want to find δ > 0 such that
0 < |x − 0| < δ & 0 < |y − 0| < δ ⇒ |f (x, y) − 0| < .
We have
x2 ≥ 0
⇒ x2 + y 2 ≥ y 2
p
⇒
x2 + y 2 ≥ |y|
|y|
⇒ p
≤1
x2 + y 2
Consider
xy
p
x2 + y 2
|f (x, y) − 0| =
y
= |x| p
2
x + y2
≤
|x|
<
δ
(using inequality (7))
if |x| < δ.
(7)
Taking δ = > 0, we have |f (x, y) − 0| < .
Therefore, we say that for every > 0 there exist δ = > 0 such that
0 < |x − 0| < δ & 0 < |y − 0| < δ ⇒ |f (x, y) − 0| < .
Hence,
lim
f (x, y) = 0.
(x,y)→(0,0)
Therefore, f is continuous at (0, 0).
xy
Solution II: We want to show that lim(x,y)→(0,0) p
= f (0, 0) = 0. Let > 0 be given. We want to find δ > 0 such
x2 + y 2
that
0 < |x − 0| < δ & 0 < |y − 0| < δ ⇒ |f (x, y) − 0| < .
Let x = r cos θ and y = r sin θ
Consider
xy
|f (x, y) − 0| =
p
x2 + y 2
=
r2 sin θ cos θ
r
|r sin θ cos θ|
≤
|r|| sin θ|| cos θ|
≤
|r|,
=
SP
M
as | cos θ| ≤ 1 and | sin θ| ≤ 1.
Now, |x| < δ, |y| < δ is given. By squaring both side, we write x2 < δ 2 and y 2 < δ 2 .
Adding these inequalities, we obtain
x2 + y 2 < 2δ 2 ,
p
√
√
which gives x2 + y 2 < 2 δ. That means |r| < 2δ.
Therefore,
√
|f (x, y) − 0| ≤ |r| < 2δ < , if δ = > 0.
2
Hence, for every > 0 there exists δ = √ > 0 such that
2
0 < |x − 0| < δ & 0 < |y − 0| < δ ⇒ |f (x, y) − 0| < .
Therefore, f is continuous at (0, 0).
Example 4.6 Discuss the continuity of the following function at (0,0).

x4 y 4

(x, y) 6= (0, 0),
2
4 3
f (x, y) =
 (x + y )
0
(x, y) = (0, 0).
Solution If f is continuous at (0, 0), then we must have lim(x,y)→(0,0) f (x, y) exist and
lim
f (x, y) = f (0, 0).
(x,y)→(0,0)
But, we note that
lim
(x,y)→(0,0)
f (x, y) =
lim
(x,y)→(0,0)
x4 y 4
3
(x2 + y 4 )
lim
does not exist. To prove this taking x = my 2 for m ∈ R, we get
x4 y 4
(x,y)→(0,0) (x2
+
3
y4 )
=
=
=
=
Therefore, the limit is depends on path. Hence
lim
(x,y)→(0,0)
lim
y→0
lim
y→0
lim
(my 2 )4 y 4
3
((my 2 )2 + y 4 )
m4 y 12
(m4 + 1)
m4
3.
(m4 + 1)
y→0
3
(m4 y 4 + y 4 )
m4
3
f (x, y) does not exist.
Example 4.7 Discuss the continuity of the following function at (0,0).

sin(x2 + y 2 )

p
(x, y) 6= (0, 0),
2 + y 2 ))
f (x, y) =
1
−
cos(
(x

0
(x, y) = (0, 0).
Solution: Let x2 + y 2 = t. As (x, y) → (0, 0), we have t → 0. Now,
lim
f (x, y)
=
(x,y)→(0,0)
=
=
=
=
=
=
sin(t2 )
t→0 1 − cos(t)
1 + cos t
sin(t2 )
×
lim
t→0 1 − cos(t)
1 + cos t
2
sin(t )(1 + cos t)
lim
t→0
1 − cos2 (t)
sin(t2 )(1 + cos t)
lim
t→0
sin2 t
t2
sin(t2 )
lim (1 + cos t) 2 ×
t→0
t
sin2 t
sin θ
lim (1 + cos t) as lim
=1
t→0
θ→0 θ
2 6= f (0, 0).
lim
M
=
sin(x2 + y 2 )
p
(x,y)→(0,0) 1 − cos( (x2 + y 2 ))
lim
Hence, f is not continuous at (0,0).
Example 4.8 Discuss the continuity of the following function at (0,0).

 xy(x2 − y 2 )
(x, y) 6= (0, 0),
f (x, y) =
x2 + y 2

0
(x, y) = (0, 0).
SP
Solution: We shall show that lim(x,y)→(0,0) f (x, y) = f (0, 0) = 0. Let > 0 be given, we want to find δ > 0 such that
0 < |x − 0| < δ & 0 < |y − 0| < δ ⇒ |f (x, y) − 0| < .
Take x = r cos θ and y = r sin θ. Consider
|f (x, y) − 0| =
xy(x2 − y 2 )
x2 + y 2
=
r2 cos θ sin θ r2 cos2 θ − r2 sin2 θ
r2
=
r4 cos θ sin θ cos 2θ
r2
=
r2 cos θ sin θ cos 2θ
=
=
=
≤
r2
2 cos θ sin θ cos 2θ
2
r2
sin 2θ cos 2θ
2
r2
sin 4θ
4
r2
.
4
(8)
Now, |x| < δ and |y| < δ gives |x|2 < δ 2 and |y|2 < δ 2 .
Adding these inequalities, we achieve
r2 = x2 + y 2 < 2δ 2 .
Form inequity (8), we have
|f (x, y) − 0|
≤
δ2
r2
1
< 2δ 2 = .
4
4
2
(9)
Taking δ =
√
2 > 0, we have for every given > 0, there exist δ > 0 such that
0 < |x − 0| < δ & 0 < |y − 0| < δ ⇒ |f (x, y) − 0| < .
Therefore
lim
f (x, y) = f (0, 0) = 0.
(x,y)→(0,0)
Hence, f is continuous at (0, 0).
Example 4.9 Discuss the continuity of the following function at (0,0).

x2 y 2

(x, y) =
6 (0, 0),
f (x, y) =
x2 y 2 + (x − y)4

0
(x, y) = (0, 0).
Solution: If f is continuous at (0, 0) then we must have
lim
f (x, y) = f (0, 0).
(x,y)→(0,0)
That means
lim
(x,y)→(0,0) x2 y 2
x2 y 2
= f (0, 0) = 0.
+ (x − y)4
Consider the path y = mx, for m ∈ R, the limit
lim
x2 y 2
+ (x − y)4
x2 (mx)2
+ (x − mx)4
m2
m2
=
lim
= 2
,
2
4
m + (1 − m)4
(x,y)→(0,0) m + (1 − m)
=
lim
(x,y)→(0,0) x2 (mx)2
M
(x,y)→(0,0) x2 y 2
which depend on m. Therefore, the limit depends on path. Hence, the limit
lim
(x,y)→(0,0) x2 y 2
x2 y 2
does not exist.
+ (x − y)4
Example 4.10 Give an example to show that both the repeated limit of a function exist at a point (a, b) but
lim
f (x, y) does not exist.
SP
(x,y)→(a,b)
Solution: Let f : R2 → R be defined as
f (x, y) =


x2 y 2

x2 y 2
+ (x − y)4
0
(x, y) 6= (0, 0),
(x, y) = (0, 0).
We have calculated repeated limits as
lim
lim f (x, y)
x→0
y→0
= lim
x→0
0
0 + x4
=0
Also,
lim
n
y→0
Exercise 4.11
(a)
o
lim f (x, y) = lim
x→0
y→0
0
0 + (−y)4
= 0.
1. Evaluate the following limits by definition if exist:
lim
(x,y)→(3,3)
xy
(b)
lim
x + 2y
(x,y)→(1,2)
2. Evaluate the iterated limit at given point.

 x − y , x + y 6= 0
(a) f (x, y) =
at (0, 0).
x+y

3,
x+y =0

 sin(x − y)
, x − y 6= 0
(c) f (x, y) =
at (0, 0).
x−y

1,
x−y =0
(c)
x3 − y 3
(x,y)→(1,1) x2 + y 2
(d)
lim
(
(b) f (x, y) =
tan−1
4,
y
,
x
 3
 x + y3
,
(d) f (x, y) =
x+y

2,
(xy − 3x + 4)
lim
(x,y)→(2,1)
x 6= 0
x=0
at (0, 1).
x + y 6= 0
x+y =0
at (1, 1).
3. Evaluate the following limits if exists:
x3 − y 3
(x,y)→(0,0) x2 + y 2
xy
(f)
lim
(g)
(x,y)→(0,0) x3 + y 3
x2 + y 2
(x,y)→(0,0) x2 − y 2
xy
(e)
lim
(x,y)→(0,0) x2 + y 2
(a)
(b)
lim
lim
tan(x + y)
(x2 y 2 )
(d)
lim
x+y
(x,y)→(0,0)
(x,y)→(0,0) x2 + y 2
y
1
1
lim
x2 sin
(h)
lim
y sin + x sin
x
(x,y)→(0,0)
x
y
(x,y)→(0,0)
(c)
lim
4. Discuss the continuity for the following functions at given point:
 2
 x − y2
, (x, y) 6= (0, 0)
at (0, 0).
(b) f (x, y) =
x2 + y 2
(a) f (x, y) = x2 + y 2 at (0, 0).

4,
(x, y) = (0, 0)


 p xy
 xy 2
, (x, y) 6= (0, 0)
, x3 + y 3 6= 0
2
2
x +y
(c) f (x, y) =
at (0, 0).
(d) f (x, y) =
at (0, 0).
x3 + y 3


1,
(x, y) = (0, 0)
2,
x3 + y 3 = 0

(
y
 tan x − tan y
, sin x 6= sin y
x2 sin , x 6= 0
(e) f (x, y) =
at (0, 0).
(f) f (x, y) =
at (0, 0).
sin x − sin y
x

1,
x=0
2,
sin x = sin y
5. Marks the following statements True/False:
(c)
lim
(x,y)→(0,0)
lim
(x,y)→(0,0)
(d)
π
0,
4
ey
lim
(x,y)→(0,0)
(f) If
2
does not exist as the required limits are different along the lines y = 0 and y = x.
ln x2 + y 2 does not exist.
lim
(x,y)→
(e)
x2 − y 2
x2 + y 2
sec x tan y does not exist.
sin x
= 1.
x
lim
(x,y)→(x0 ,y0 )
M
(b)
f (x, y) = 1, then limx→a f (x, b) = 1.
lim
(x,y)→(a,b)
f (x, y) = L exists, then (x0 , y0 ) is in the domain of f .
SP
(a) If
f
(g) If
lim
f (x, y) = 2 and
lim (f g)(x, y) = 1 then
lim
(x, y) = 2.
g
(x,y)→(0,0)
(x,y)→(0,0)
(x,y)→(0,0)
(h) If
5
lim
(x,y,z)→(0,0,0)
f (x, y, z) and
lim
(x,y,z)→(0,0,0)
g(x, y, z) exist, then
lim
(x,y,z)→(0,0,0)
g 2 (x, y, z) is exists.
Partial Derivatives
To understand this topic, please watch the video https://youtu.be/7Iki2AT6HqY In this section we will develop the
mathematical tools for studying rates of change that involve two or more independent variables.
If z = f (x, y), then one can inquire how the value of z changes if y is held fixed and x is allowed to vary, or if x is held
fixed and y is allowed to vary. For example, the ideal gas law in physics states that under appropriate conditions
the pressure exerted by a gas is a function of the volume of the gas and its temperature. Thus, a physicist studying
gases might be interested in the rate of change of the pressure if the volume is held fixed and the temperature is
allowed to vary, or if the temperature is held fixed and the volume is allowed to vary. We now define a derivative that
describes such rates of change.
Suppose that (x0 , y0 ) is a point in the domain of a function f (x, y). If we fix y = y0 , then f (x, y0 ) is a function of the
variable x alone. The value of the derivative
d
[f (x, y0 )]
dx
at x0 then gives us a measure of the instantaneous rate of change of f with respect to x at the point (x0 , y0 ). Similarly,
the value of the derivative
d
[f (x0 , y)]
dy
at y0 gives us a measure of the instantaneous rate of change of f with respect to y at the point (x0 , y0 ). These
derivatives are so basic to the study of differential calculus of multi-variable functions that they have their own
name and notation.
Definition 5.1 If z = f (x, y) and (x0 , y0 ) is a point in the domain of f , then the partial derivative of f with respect to x
at (x0 , y0 ) [also called the partial derivative of z with respect to x at (x0 , y0 )] is the derivative at x0 of the function that
results when y = y0 is held fixed and x is allowed to vary. This partial derivative is denoted by fx (x0 , y0 ) and is given
by
f (x0 + ∆x, y0 ) − f (x0 , y0 )
d
= lim
fx (x0 , y0 ) =
f (x, y0 )
.
(10)
∆x→0
dx
∆x
x=x0
Similarly, the partial derivative of f with respect to y at (x0 , y0 ) [also called the partial derivative of z with respect to y
at (x0 , y0 )] is the derivative at y0 of the function that results when x = x0 is held fixed and y is allowed to vary. This
partial derivative is denoted by fy (x0 , y0 ) and is given by
fy (x0 , y0 ) =
d
f (x0 , y)
dy
= lim
∆y→0
y=y0
f (x0 , y0 + ∆y) − f (x0 , y0 )
.
∆y
(11)
Remark 5.2 The limits in (10) and (11) show the relationship between partial derivatives and derivatives of functions
of one variable. In practice, our usual method for computing partial derivatives is to hold one variable fixed and then
differentiate the resulting function using the derivative rules for functions of one variable.
Example 5.3 Find fx (1, 2) and fy (1, 2) for the function f (x, y) = x3 y 2 + x2 + y 2 + 2.s
Solution: Since
d 3
d
[f (x, 2)] =
4x + x2 + 4 + 2 = 12x2 + 2x
dx
dx
we have fx (1, 2) = 12 + 2 = 14. Also, Since
fx (x, 2) =
d 2
d
[f (1, y)] =
y + 1 + y 2 + 2 = 4y
dy
dy
we have fy (1, 2) = 4(2) = 8.
THE PARTIAL DERIVATIVE FUNCTIONS
M
fy (1, y) =
SP
Formula (10) and (11) define the partial derivatives of a function at a specific point (x0 , y0 ). However, often it will
be desirable to omit the subscripts and think of the partial derivatives as functions of the variables x and y. These
functions are
f (x + ∆x, y) − f (x, y)
f (x, y + ∆y) − f (x, y)
, a nd fy (x, y) = lim
.
∆x→0
∆y→0
∆x
∆y
fx (x, y) = lim
(12)
Example 5.4 Find fx (x, y) and fy (x, y) for the function f (x, y) = x3 y 2 + x2 + y 2 + 2 and use those partial derivatives to
compute fx (1, 2) and fy (1, 2).
Solution: Keeping y fixed and differentiating with respect to x yields
fx (x, y) = 3x2 y 2 + 2x
and keeping x fixed and differentiating with respect to y yields
fy (x, y) = 2x3 y + 2y.
Thus,
fx (1, 2) = 3(1)2 (2)2 + 2(1) = 14 and fy (1, 2) = 2(1)3 (2) + 2(2) = 8.
PARTIAL DERIVATIVE NOTATION
If z = f (x, y), then the partial derivatives fx and fy are also denoted by the symbols
∂f ∂z
∂f ∂z
,
and
,
∂x ∂x
∂y ∂y
Some typical notations for the partial derivatives of z = f (x, y) at a point (x0 , y0 ) are
∂f
∂x
,
x=x0 ,y=y0
∂z
∂x
,
(x0 ,y0 )
∂f
∂x
and
(x0 ,y0 )
∂f
∂y
,
x=x0 ,y=y0
∂z
∂y
,
(x0 ,y0 )
∂f
∂y
(x0 ,y0 )
(b) Cross-section is displayed in a twodimensional plot
(a) Tangent lines at a selected point are drawn. The slopes of the tangents at
that point are equal to the partial derivatives fx (x0 , y0 )
Figure 7: This plot shows a surface z = f (x, y) and intersecting vertical planes through the x axes
Solution:
and
∂z
∂z
and
if z = cos(xy).
∂x
∂y
∂
∂
∂z
=
[cos(xy)] = − sin(xy)
[xy] = −y sin(xy)
∂x
∂x
∂x
M
Example 5.5 Find
∂z
∂
∂
=
[cos(xy)] = − sin(xy)
[xy] = −x sin(xy).
∂y
∂y
∂y
PARTIAL DERIVATIVES VIEWED AS RATES OF CHANGE AND SLOPES
SP
Recall that if y = f (x), then the value of f 0 (x0 ) can be interpreted either as the rate of change of y with respect to x
at x0 or as the slope of the tangent line to the graph of f at x0 .
Partial derivatives have analogous interpretations. To see that this is so, suppose that C1 is the intersection of the
surface z = f (x, y) with the plane y = y0 and that C2 is its intersection with the plane x = x0 (Figure 7). Thus, fx (x, y0 )
can be interpreted as the rate of change of z with respect to x along the curve C1 , and fy (x0 , y) can be interpreted
as the rate of change of z with respect to y along the curve C2 . In particular, fx (x0 , y0 ) is the rate of change of z with
respect to x along the curve C1 at the point (x0 , y0 ), and fy (x0 , y0 ) is the rate of change of z with respect to y along
the curve C2 at the point (x0 , y0 ). We will call fx (x0 , y0 ) the slope of the surface in the x− direction at (x0 , y0 ) and
fy (x0 , y0 ) the slope of the surface of in the y-direction at (x0 , y0 )
Geometrically, fx (x0 , y0 ) can be viewed as the slope of the tangent line to the curve C1 at the point (x0 , y0 ), and
fy (x0 , y0 ) can be viewed as the slope of the tangent line to the curve C2 at the point (x0 , y0 ) (Figure 7 and 8). We
will call fx (x0 , y0 ) the slope of the surface in the x−direction at (x0 , y0 ) and fy (x0 , y0 ) the slope of the surface in the
y-direction at (x0 , y0 ).
2 1 2
2 1 2
2
2
2
Example 5.6 Find the slop of the sphere x + y + z = 1 in the y-direction at point
, ,
and
, ,− .
3 3 3
3 3 3
p
2 1 2
2 1 2
2
2
Solution: The point
, ,
lies on the upper hemisphere z = 1 − x − y , and the point
, ,−
lies on the
3p
3 3
3 3 3
lower hemisphere z = − 1 − x2 − y 2 . We could find the slopes by differentiating each expression for z separately
2
1
with respect to y and then evaluating the derivatives at x = and y = . However, it is more efficient to differentiate
3
3
the given equation
x2 + y 2 + z 2 = 1
implicitly with respect to y, since this will give us both slopes with one differentiation. To perform the implicit
differentiation, we view z as a function of x and y and differentiate both sides with respect to y, taking x to be fixed.
(b) Cross-section is displayed in a twodimensional plot
(a) Tangent lines at a selected point are drawn. The slopes of the tangents at
that point are equal to the partial derivatives fy (x0 , y0 )
The computations are as follows:
M
Figure 8: This plot shows a surface z = f (x, y) and intersecting vertical planes through the y axes
SP
∂ 2
∂
x + y2 + z2 =
[1]
∂y
∂y
∂z
0 + 2y + 2z
= 0
∂y
∂z
y
= −
∂y
z
2 1 2
2 1 2
Substituting the y− and z− coordinates of the points
, ,
and
, ,−
in this expression, we find that the
3 3 3
3 3 3
2 1 2
2 1 2
1
1
slope at the point
, ,
, ,−
is − and the slope at
is .
3 3 3
2
3 3 3
2
Example 5.7 Let f (x, y) = x2 y + 5y 3 .
(a) Find the slope of the surface z = f (x, y) in the x−direction at the point (1, −2).
1. Find the slope of the surface z = f (x, y) in the y−direction at the point (1, −2).
Solution (a): Differentiating f with respect to x with y held fixed yields
fx (x, y) = 2xy
Thus, the slope in the x-direction is fx (1, −2) = −4; that is, z is decreasing at the rate of 4 units per unit increase in
x.
Solution (b). Differentiating f with respect to y with x held fixed yields
fy (x, y) = x2 + 15y 2
Thus, the slope in the y-direction is fy (1, −2) = 61; that is, z is increasing at the rate of 61 units per unit increase
in y.
√
Exercise 5.8
1. Let f (x, y) = 3x + 2y.
(a) Find the slope of the surface z = f (x, y) in the x−direction at the point (4, 2).
(b) Find the slope of the surface z = f (x, y) in the y−direction at the point (4, 2).
2. Let f (x, y) = xe−y + 5y.
(a) Find the slope of the surface z = f (x, y) in the x−direction at the point (3, 0).
(b) Find the slope of the surface z = f (x, y) in the y−direction at the point (3, 0).
6
HIGHER-ORDER PARTIAL DERIVATIVES
To understand this topic, please watch the video https://youtu.be/zUxEX5BN64c [1em]
∂f
∂f
Suppose that f is a function of two variables x and y. Since the partial derivatives
and
are also functions
∂x
∂y
of x and y, these functions may themselves have partial derivatives. This gives rise to four possible second-order
partial derivatives of f , which are defined by
∂ ∂f
∂2f
=
= fxx ;
∂x2
∂x ∂x
∂2f
∂ ∂f
=
= fyy ;
∂y 2
∂y ∂y
∂2f
∂ ∂f
=
= fxy ;
∂y∂x
∂y ∂x
∂ ∂f
∂2f
=
= fyx .
∂x∂y
∂x ∂y
M
The last two cases are called the mixed second-order partial derivatives or the mixed second partials. Also, the
∂f
∂f
derivatives
and
are often called the first-order partial derivatives when it is necessary to distinguish them
∂x
∂y
from higher-order partial derivatives. Similar conventions apply to the second-order partial derivatives of a function
of three variables.
WARNING: Observe that the two notations for the mixed second partials have opposite conventions for the order of
differentiation. In the “∂” notation the derivatives are taken right to left, and in the “subscript” notation they are
taken left to right. The conventions are logical if you insert parentheses:
∂ ∂f
∂2f
=
, Right to left. Differentiate inside the parentheses first
∂y∂x
∂y ∂x
and
SP
fxy = (fx )y , Left to right. Differentiate inside the parentheses first.
Example 6.1 Find the second-order partial derivatives of f (x, y) = x2 y 3 + x4 y.
Solution: We have
so that
∂f
∂f
= 2xy 3 + 4x3 y and
= 3x2 y 2 + x4 ,
∂x
∂x
∂2f
∂x2
∂2f
∂y 2
∂2f
∂x∂y
∂2f
∂y∂x
=
=
=
=
∂ ∂f
=
∂x ∂x
∂ ∂f
=
∂y ∂y
∂ ∂f
=
∂x ∂y
∂ ∂f
=
∂y ∂x
∂
∂x
∂
∂y
∂
∂x
∂
∂y
2xy 3 + 4x3 y = 2y 3 + 12x2 y
3x2 y 2 + x4 = 6x2 y
3x2 y 2 + x4 = 6xy 2 + 4x3
2xy 3 + 4x3 y = 6xy 2 + 4x3 .
Third-order, fourth-order, and higher-order partial derivatives can be obtained by successive differentiation. Some
possibilities are
∂3f
∂ ∂2f
=
= fxxx
∂x3
∂x ∂x2
∂3f
∂ ∂2f
=
= fyyy
∂y 3
∂y ∂y 2
2 ∂3f
∂
∂ f
=
= fxyy
∂y 2 ∂x
∂y ∂y∂x
∂4f
∂ ∂3f
=
= fyyyy .
∂y 4
∂y ∂y 3
Example 6.2 Let f (x, y) = y 2 ex + y. Find fxyy .
Solution:
fxyy =
∂3f
∂2
=
2
∂y ∂x
∂y 2
∂f
∂x
=
∂2
∂
y 2 ex =
(2yex ) = 2ex .
2
∂y
∂y
PARTIAL DERIVATIVES OF FUNCTIONS WITH MORE THAN TWO VARIABLES
For a function f (x, y, z) of three variables, there are three partial derivatives:
fx (x, y, z), fy (x, y, z), fz (x, y, z).
The partial derivative fx is calculated by holding y and z constant and differentiating with respect to x. For fy the
variables x and z are held constant, and for fz the variables x and y are held constant. If a dependent variable
w = f (x, y, z)
is used, then the three partial derivatives of f can be denoted by
∂w ∂w ∂w
,
,
.
∂x ∂y ∂z
Example 6.3 If f (x, y, z) = x3 y 2 z 2 + 2xy + z. Find fx (1, 1, 1), fy (1, 1, 1) and fz (1, 1, 1).
Solution: We have
M
∂f
= 3x2 y 2 z 2 + 2y
∂x
∂f
= 2x2 yz 2 + 2x
∂y
∂f
= 2x3 y 2 z + 1
∂z
Therefore, fx (1, 1, 1) = 5, fy (1, 1, 1) = 4 and fz (1, 1, 1) = 3.
SP
In general, if f (v1 , v2 , . . . , vn ) is a function of n variables, there are n partial derivatives of f , each of which is obtained
by holding n − 1 of the variables fixed and differentiating the function f with respect to the remaining variable. If
w = f (v1 , v2 , . . . , vn ), then these partial derivatives are denoted by
∂w ∂w
∂w
,
,...,
,
∂v1 ∂v2
∂vn
∂w
is obtained by holding all variables except vi fixed and differentiating with respect to vi , i = 1, 2, . . . , n.
∂vi
Example 6.4 If
x
2
−1 y
2
−1
u = x tan
− y tan
, xy 6= 0,
x
y
prove that
x2 − y 2
∂2u
= 2
.
∂x∂y
x + y2
where
Solution: We have
∂u
∂y
=
x2
1
1+
=
=
y2
x2
·
1
− 2y tan−1
x
x3
− 2y tan−1
x2 + y 2
x
−1
x − 2y tan
y
x
− y2
y
1
1+
x2
y2
·
−x
y2
xy 2
x
+ 2
y
x + y2
Now,
∂2u
∂x∂y
=
=
x
y
2
1
1
2y
x2 − y 2
1 − 2y
= 2
2 = 1 − 2
2
y
x +y
x + y2
x
1+
y
∂
∂x
∂u
∂y
=
∂
∂x
x − 2y tan−1
Example 6.5 If
1
u= p
, x2 + y 2 + z 2 6= 0,
x2 + y 2 + z 2
show that
∂2u ∂2u ∂2u
+ 2 + 2 = 0.
∂x2
∂y
∂z
Solution: We have
∂u
∂x
∂2u
∂x2
Similarly,
3
3
−
−
1
= − (x2 + y 2 + z 2 ) 2 (2x) = −x(x2 + y 2 + z 2 ) 2
2
3
5
−
−
3x 2
= −(x2 + y 2 + z 2 ) 2 +
(x + y 2 + z 2 ) 2 (2x)
2
3
5
−
−
2
2
2
2
2
2
2
= −(x + y + z ) 2 + 3x (x + y + z ) 2
3
5
−
−
∂2u
2
2
2
2 2
2
2
2
2
= −(x + y + z )
+ 3y (x + y + z )
∂y 2
5
3
−
−
∂2u
2
2
2
2
2
2
2
= −(x + y + z ) 2 + 3z (x + y + z ) 2
∂z 2
M
Adding, we obtain
3
5
−
−
∂2u ∂2u ∂2u
2
2
2
2
2
2
2
2
2
+ 2 + 2 = −3 − (x + y + z ) 2 + 3(x + y + z )(x + y + z ) 2 = 0.
∂x2
∂y
∂z
x2 + y 2
, x + y 6= 0, Prove that
x+y
SP
Example 6.6 If z =
∂z
∂z
−
∂x ∂y
2
∂z
∂z
−
=4 1−
.
∂x ∂y
Solution: Taking partial derivative with respect to x and y of the given equation, we have
(2x + 0)(x + y) − (x2 + y 2 )(1)
x2 + 2xy − y 2
∂z
=
=
2
∂x
(x + y)
(x + y)2
Similarly,
∂z
y 2 + 2xy − x2
=
.
∂x
(x + y)2
Now,
∂z
∂z
−
∂x ∂y
2
=
x2 + 2xy − y 2
y 2 + 2xy − x2
−
(x + y)2
(x + y)2
=
2x2 − 2y 2
(x + y)2
2
=
2(x − y)
(x + y)
On the other side,
2
∂z
∂z
4xy
(x + y)2 − 4xy
2(x − y)
4 1−
−
=4 1−
.
=4
=
∂x ∂y
(x + y)2
(x + y)2
(x + y)
Therefore, we obtain the result as given.
r2
Example 6.7 Consider θ = t e 4t . Find the value of n such that
∂θ
1 ∂
2 ∂θ
r
=
.
2
r ∂r
∂r
∂t
n
−
2
.
Solution: We have


r2  −  −r
= tn e 4t 
2t
∂θ
∂r

r2
 − 4t 
.
e

= −r
tn−1
2
Therefore,

r2
∂θ
∂r
n−1
= −r3
t


2

r2
 − 4t 
e
.
Now,
∂
∂r
r2
∂θ
∂r
n−1
= −
t
2




r2
r2
−
 2  − 4t 
 −2r 
3
3r e
 + r e 4t 

4t

r2 r4
 4t 
2
−3r
+
e
.


2t

2
Therefore,
−
M
t
=
n−1

On the other side, we have
∂
∂r
r2
∂θ
∂r
n−1
=
t
2

r2 2
 4t  r
e
−
3
.


2t
−
SP
1
r2



r2 2 r2
 −  r
 − 
ntn−1 e 4t  + tn e 4t 
4t2

∂θ
∂t
=

=
Now, given that
n−1
⇒
t
2

r2 −
r2

n−2 
4t
t
e
 nt +
4
1 ∂
2 ∂θ
r
=
r2 ∂r
∂r


r2 2
−
 4t  r
−3
=
e

2t
⇒
t r2
−3
=
2 2t
r2 − 6t
⇒
=
4t
⇒ r2 − 6t =
∂θ
∂t


r2 −
r2

n−2 
t
e 4t  nt +
4
r2
4
4nt + r2
4t
4nt + r2
nt +
3
Therefore n = − .
2
Exercise 6.8
1. If z = x2 y + ex
2 2
y
. Find zx , zy , zxx , zxy , zyx , zxx .
y
x
2. If f (x, y) = x2 tan−1
− y 2 tan−1
. Prove that fxy = fyx .
x
y
3. Find
∂z
∂z
and
if z = x4 sin xy 3 .
∂x
∂x
4. If z = eax+by f (ax − by) prove that b
∂z
∂z
+a
= 2abz.
∂x
∂y
5. Given u = er cos θ cos(r sin θ), v = er cos θ sin(r sin θ) prove that
1 ∂v
∂v
1 ∂u
∂u
=
and
=−
.
∂r
r ∂θ
∂r
r ∂θ
6. Verify that fxy = fyx , when f is equal to
y
(b) log(x) tan−1 x2 + y 2
(a) sin−1
x
∂2u ∂2u
+ 2 = 0 if
∂x2
∂y
2xy
−1
(a) u = tan
x2 − y 2
7. Prove that
(b) log(x2 + y 2 ) + tan−1
8. If u = log(x3 + y 3 + z 3 − 3xyz), show that
9. If u = xy , show that
∂
∂
∂
+
+
∂x ∂y ∂z
x
2
u=
−9
.
(x = y + z)2
∂3u
∂3u
=
.
2
∂x ∂y
∂x∂y∂x
∂2z
= −(x log(ex))−1 .
∂x∂y
M
10. If xx y y z z = c , show that at x = y = z,
7
y
PARTIAL DERIVATIVES AND CONTINUITY
To understand this topic, please watch the video https://youtu.be/Qj_2DwnhO9E.
SP
In contrast to the case of functions of a single variable, the existence of partial derivatives for a multi-variable
function does not guarantee the continuity of the function. This fact is shown in the following examples.
Example 7.1 Find the partial derivative of f with respect to x and y at the point (0, 0), where

 xy(x2 − y 2 )
(x, y) 6= (0, 0)
f (x, y) =
x2 + y 2

0
(x, y) = (0, 0).
Solution: Here,
fx (0, 0) =
∂f
∂x
=
(0,0)
=
=
lim
∆x→0
f (0 + ∆x, 0) − f (0, 0)
∆x
f (∆x, 0)
∆x→0
∆x
∆x · 0((∆x)2 − 0)
lim
= 0;
∆x→0
(∆x)2 + 0
lim
and
fy (0, 0) =
∂f
∂y
=
(0,0)
=
=
Example 7.2 Consider
(
f (x, y) =
x2
f (0, 0 + ∆y) − f (0, 0)
∆y→0
∆y
lim
f (0, ∆y)
∆y
0 · ∆y(0 − (∆y)2 )
= 0.
lim
∆y→0
0 + (∆y)2
lim
∆y→0
xy
+ y2
0
(x, y) 6= (0, 0)
(x, y) = (0, 0).
Find the partial derivative of f with respect to x and y at the point (0, 0). Also, prove that f is not continuous at (0, 0).
Solution: Here,
fx (0, 0) =
∂f
∂x
=
(0,0)
=
=
lim
∆x→0
f (0 + ∆x, 0) − f (0, 0)
∆x
f (∆x, 0)
∆x→0
∆x
∆x · 0
lim
= 0;
∆x→0 (∆x)2 + 0
lim
and
fy (0, 0) =
∂f
∂y
=
(0,0)
=
=
lim
∆y→0
f (0, 0 + ∆y) − f (0, 0)
∆y
f (0, ∆y)
∆y→0
∆y
0 · ∆y
= 0.
lim
∆y→0 0 + (∆y)2
lim
Now, we shall show that f is not continuous at (0, 0). Along the path y = mx, we have
lim
f (x, y) =
(x,y)→(0,0)
lim
(x,y)→(0,0)
x(mx)
m
m
=
lim
=
,
x2 + (mx)2
1 + m2
(x,y)→(0,0) 1 + m2
(x,y)→(0,0)
Example 7.3 Let f (x, y) =
p
M
which is depend on m.
Therefore,
lim
f (x, y) does not exists. Hence the f is not continuous at (0, 0).
x2 + y 2 . Prove that fx (0, 0) and fy (0, 0) does not exists. and f is continuous at (0, 0).
Solution: Here,
=
(0,0)
f (0 + ∆x, 0) − f (0, 0)
∆x→0
∆x
lim
SP
fx (0, 0) =
∂f
∂x
f (∆x, 0)
∆x
p
(∆x)2 + 0
|∆x|
= lim
,
= lim
∆x→0 ∆x
∆x→0
∆x
=
which can be written as
lim
∆x→0+
lim
∆x→0
∆x
= 1 and
∆x
lim
∆x→0−
−∆x
= −1.
∆x
Therefore, fx (0, 0) does not exists.
Similarly,
fy (0, 0) =
∂f
∂y
=
(0,0)
=
=
which can be written as
f (0, 0 + ∆y) − f (0, 0)
∆y→0
∆y
lim
f (0, ∆y)
∆y
p
= 0 + (∆y)2
|∆y|
lim
= lim
,
∆y→0
∆y→0 ∆y
∆y
lim
∆y→0
∆y
= 1 and
∆y→0+ ∆y
lim
−∆y
= −1.
∆y→0− ∆y
lim
Therefore, fy (0, 0) does not exists.
Now, we shall show that f is continuous at (0, 0). It is sufficient to prove that for every > 0 there exists δ > 0 such
that
|f (x, y) − f (0, 0)| < whenever |x − 0| < δ, |y − 0| < δ.
As |x| < δ and |y| < δ, we have x2 + y 2 < 2δ 2 . Now,
p
√
|f (x, y) − f (0, 0)| = | x2 + y 2 | < 2δ, when δ < √ .
2
Therefore, for every > 0 there exists δ < √ > 0 such that
2
|f (x, y) − f (0, 0)| < whenever |x − 0| < δ, |y − 0| < δ.
Example 7.4 Consider
 3
 x + y3
f (x, y) =
x−y

0
(x, y) 6= (0, 0)
(x, y) = (0, 0)
Check whether f is continuous. Also, check that fx (0, 0) and fy (0, 0) exist or not?
Solution: Here,
fx (0, 0) =
∂f
∂x
=
(0,0)
=
=
=
Similarly, we have
fy (0, 0) =
∂f
∂y
=
(0,0)
f (0 + ∆x, 0) − f (0, 0)
∆x
f (∆x, 0)
∆x→0
∆x
p
(∆x)2 + 0
lim
∆x→0
∆x
(∆x)3 + 0
,
lim
∆x→0 ∆x · ∆
lim ∆x = 0.
lim
∆x→0
M
=
lim
∆x→0
f (0, 0 + ∆y) − f (0, 0)
∆y
f (0, ∆y)
∆y→0
∆y
0 + (∆y)3
,
lim
∆y→0 −∆y · ∆y
lim −∆y = 0.
lim
SP
=
lim
∆y→0
=
=
∆y→0
Now, we will show that f is not continuous at (0, 0).
Along the path y = x − mx3 , we have
f (x, x − mx3 ) =
So,
x3 + (x − mx3 )3
1 + (1 − mx2 )3
=
.
x − (x − mx3 )
m
1 + (1 − mx2 )3
2
= ,
x→0
m
m
lim
which depend on m, therefore
x3 + y 3
does not exist. Hence the f is not continuous at (0, 0).
(x,y)→(0,0) x − y
lim
This examples, show that the existence of both the partial derivative at a point need not imply continuity of the
function at that point. The reason being that the partial derivative only exhibit the rate of change of f (x, y) only
along two particular paths, namely the ones parallel to axes. A more general concept of rate of change of function,
which takes into account all directions is describe in the next section.
8
Differentiability
To understand this topic, please watch the video https://youtu.be/qFFDGL-HTho.
Recall that for a function y = f (x) of one variables, the concept of differentiability at a point x0 allowed us to approximate the function f by linear function in the neighborhood of x0 . Analytically, f differentiable at x0 is equivalent to
saying that
f (x0 + h) = f (x0 ) + hf 0 (x0 ) + h1 (h),
for all h sufficiently small, where 1 (h) → 0 as h → 0. The expression
L(x) = f (x0 ) + hf 0 (x0 )
is linear (or tangent line) approximation and 1 (h) is the error for the linear approximation.
For function of two variables, this motivates the following definition:
Definition 8.1 A function f : D ⊂ R2 → R is said to be differentiable at (x0 , y0 ) ∈ D if the following hold:
(a) Both fx (x0 , y0 ), fy (x0 , y0 ) exists,
(b) (x0 , y0 ) is an interior of the domain D, that is there exist δ > 0 such that
(x0 + h, y0 + k) ∈ D for all |h| < δ, |k| < δ.
(c) There exist function 1 (h, k) and 2 , |h| < δ, |k| < δ such that
f (x0 + h, y0 + k) = f (x0 , y0 ) + hfx (x0 , y0 ) + kfy (x0 , y0 ) + h1 (h, k) + k2 (h, k).
One can also define as
Definition 8.2 Let z = f (x, y) be real valued function of variables x and y defined on D ⊂ R2 and (x0 , y0 ) ∈ D. There
exits A and B such that
f (x0 + h, y0 + k) = f (x0 , y0 ) + hA + kB + h1 (h, k) + k2 (h, k),
M
where A and B are independent of h and k and 1 (h, k) and 2 (h, k) → 0 as (h, k) → (0, 0). Then the function is
differentiable at point (x0 , y0 ).
Remark 8.3 (Necessary and sufficient condition for differentiability:) Let f be differentiable at (x0 , y0 ). Then by
definition
f (x0 + h, y0 + k) − f (x0 , y0 ) − hfx (x0 , y0 ) − kfy (x0 , y0 )
√
≤ |1 (h, k)| + |2 (h, k)|.
h2 + k 2
Hence,
f (x0 + h, y0 + k) − f (x0 , y0 ) − hfx (x0 , y0 ) − kfy (x0 , y0 )
√
h2 + k 2
SP
lim
(h,k)→(0,0)
= 0.
In fact, the converse also holds, that means if (13) holds then f is differentiable. We assume this fact.
Example 8.4 Prove that the function f (x, y) = x2 + xy + y 2 for every x, y ∈ R is differentiable.
Solution: We note that
f (x + h, y + k)
(x + h)2 + (x + h)(y + k) + (y + k)2
=
= x2 + y 2 + xy + h2 + 2hx + yh + kx + kh + k 2 + 2ky
= (x2 + xy + y 2 ) + h(2x + y) + k(x + 2y) + (h2 + hk + k 2 )
= f (x, y) + hA + kB + h1 (h, k) + k2 (h, k),
where
f (x, y) = x2 + xy + y 2 ;
A = 2x + y;
B = 2y + x;
1 (h, k) = h + k → 0 as (h, k) → (0, 0);
2 (h, k) = k → 0 as (h, k) → (0, 0).
Therefore, f is differentiable at any point (x, y) ∈ R2 .
Example 8.5 Prove that the function
f (x, y) =
is differentiable at (0, 0).


(x + y ) sin

0
2
2
1
x2 + y 2
if (x, y) 6= (0, 0)
if (x, y) = (0, 0)
(13)
Solution: We have
2
fx (0, 0)
=
=
f (0 + ∆x, 0) − f (0, 0)
= lim
∆x→0
∆x→0
∆x 1
lim ∆x sin
= 0.
∆x→0
(∆x)2
(∆x) + 0
2
lim
1
sin
(∆x)2 + 0
∆x
−0
Similarly,
f (0, 0 + ∆y) − f (0, 0)
fy (0, 0) = lim
= lim
∆y→0
∆y→0
∆y
1
= 0.
= lim ∆y sin
∆y→0
(∆y)2
1
0 + (∆y)2
∆y
02 + (∆y)2 sin
−0
Now, we shall show that f is differentiable at (0, 0). It is sufficient to prove that
f (x0 + h, y0 + k) − f (x0 , y0 ) − hfx (x0 , y0 ) − kfy (x0 , y0 )
√
lim
= 0.
(h,k)→(0,0)
h2 + k 2
Consider

lim
(h,k)→(0,0)
f (0 + h, 0 + k) − f (0, 0) − hfx (0, 0) − kfy (0, 0)
√
h2 + k 2
2
2
 (h + k ) sin

=
lim
(h,k)→(0,0) 
M
=
p
lim

1
−
0
−
h(0)
−
k(0)

h2 + k 2

√

h2 + k 2
1
h2 + k 2
h2 + k 2 sin
(h,k)→(0,0)
we note that
p
and
h2 + k 2 sin
1
h2 + k 2
≤
p
,
h2 + k 2
SP
p
h2 + k 2 → 0 as (h, k) → (0, 0).
Hence, f is differentiable at (0, 0).
Theorem 8.6 Let f : D ⊂ R2 → R be such that f is differentiable at (x0 , y0 ) ∈ D. Then prove that fx and fy exists.
Proof: Suppose that f is is differentiable at (x0 , y0 ) ∈ D. Therefore, there exist A and B such that
f (x0 + h, y0 + k) = f (x0 , y0 ) + hA + kB + h1 (h, k) + k2 (h, k),
where A and B are independent of h and k and 1 (h, k) and 2 (h, k) → 0 as (h, k) → (0, 0).
Putting k = 0 in equation (14), we have
f (x0 + h, y0 ) = f (x0 , y0 ) + hA + h1 (h, 0),
which can be written as
f (x0 + h, y0 ) − f (x0 , y0 )
= A + 1 (h, 0).
h
Taking limit both side with h → 0, we obtain
lim
h→0
f (x0 + h, y0 ) − f (x0 , y0 )
= lim (A + 1 (h, 0)) .
h→0
h
Hence, fx (x0 , y0 ) = A as 1 (h, 0) → 0 as h → 0.
Therefore, fx (x0 , y0 ) is exists. Putting h = 0 in equation (14), we have
f (x0 , y0 + k) = f (x0 , y0 ) + kB + k2 (0, k),
which can be written as
f (x0 , y0 + k) − f (x0 , y0 )
= B + 2 (0, k).
k
Taking limit both side with k → 0, we obtain
lim
k→0
f (x0 , y0 + k) − f (x0 , y0 )
= lim (B + 2 (0, k)) .
k→0
k
Hence, fy (x0 , y0 ) = B as 1 (0, k) → 0 as k → 0.
Therefore, fy (x0 , y0 ) is exists.
(14)
Remark 8.7 Converse of the above result is not true, which can be seen by following example:
Example 8.8 Define f : R2 → R as

 p xy
, (x, y) 6= (0, 0)
x2 + y 2
f (x, y) =

0,
(x, y) = (0, 0).
Prove that fx and fy exists at (0, 0) but f is not differentiable at (0, 0).
Solution: Here,
fx (0, 0) =
∂f
∂x
=
(0,0)
=
=
lim
∆x→0
f (0 + ∆x, 0) − f (0, 0)
∆x
f (∆x, 0)
∆x→0
∆x
∆x · 0
lim
= 0;
∆x→0 (∆x)2 + 0
lim
and
fy (0, 0) =
∂f
∂y
=
(0,0)
=
f (0, ∆y)
∆y
0 · ∆y
lim
= 0.
∆y→0 0 + (∆y)2
lim
∆y→0
M
=
f (0, 0 + ∆y) − f (0, 0)
∆y→0
∆y
lim
Now, we shall show that f is not differentiable at (0, 0).
Suppose that f is differentiable at (0, 0).
There exist constant A and B such that
f (x0 + h, y0 + k) = f (x0 , y0 ) + hA + kB + h1 (h, k) + k2 (h, k),
SP
where A and B are independent of h and k and 1 (h, k) and 2 (h, k) → 0 as (h, k) → (0, 0).
Here (x0 , y0 ) = (0, 0) and fx (0, 0) and fy (0, 0) exists. So,
A = fx (0, 0) = 0 and B = fy (0, 0) = 0.
Therefore,
f (0 + h, 0 + k) = f (0, 0) + hfx (0, 0) + kfy (0, 0) + h1 (h, k) + k2 (h, k)
hk
√
= h1 (h, k) + k2 (h, k).
h2 + k 2
Along the path h = mk, we have
(mk)k
p
(mk)2 + k 2
= mk1 (mk, k) + k2 (mk, k).
Therefore,
√
m
m2 + 1
=
m1 (mk, k) + 2 (mk, k).
Taking limit k → 0 both side, we get
√
m
m2 + 1
=
0
as 1 (h, k) and 2 (h, k) → 0 as (h, k) → (0, 0).
Therefore, f is not differentiable at (0, 0).
Method-II
Using (13), we have
lim
(h,k)→(0,0)
f (h, k) − f (0, 0) − hfx (0, 0) − kfy (0, 0)
√
h2 + k 2
f (h, k)
√
(h,k)→(0,0)
h2 + k 2
hk
=
lim
2
(h,k)→(0,0) h + k 2
=
lim
Along the path h = mk, we have
(mk)k
m2
m2
=
lim
=
,
k→0 1 + m2
(mk,k)→(0,0) (mk)2 + k 2
1 + m2
lim
which is depend on m. Hence the function f is not differentiable at (0, 0).
Theorem 8.9 Let f : D ⊂ R2 → R be such that f is differentiable at (x0 , y0 ) ∈ D Then prove that f
is continuous at (x0 , y0 ).
Proof: Given that f is differentiable at point (x0 , y0 ). Therefore, there exist A and B such that
f (x0 + h, y0 + k) = f (x0 , y0 ) + hA + kB + h1 (h, k) + k2 (h, k),
(15)
where A and B are independent of h and k and 1 (h, k) and 2 (h, k) → 0 as (h, k) → (0, 0).
Consider
lim
f (x, y).
(x,y)→(x0 ,y0 )
Putting x = x0 + h and y = y0 + k and using (15), we have
lim
(h,k)→(0,0)
f (x0 + h, y0 + k) =
lim
(h,k)→(0,0)
f (x0 , y0 ) + hA + kB + h1 (h, k) + k2 (h, k)
M
= f (x0 , y0 ) + (0)A + (0)B + 0 + 0
= f (x0 , y0 ).
Therefore lim(x,y)→(x0 ,y0 ) f (x, y) = f (x0 , y0 ). Hence, f is continuous at (x0 , y0 ).
Remark 8.10 The converse of the above theorem is not true.
Example 8.11 Define f : R2 → R as
SP

 p xy
, (x, y) 6= (0, 0)
x2 + y 2
f (x, y) =

0,
(x, y) = (0, 0).
Prove that fx and fy exists at (0, 0), f is continuous at (0, 0) and f is not differentiable at (0, 0).
Solution: We already prove that fx and fy exists at (0, 0), f is not differentiable at (0, 0) in example
8.8.
We shall show that f is continuous at (0,0). It is sufficient to prove that for every > 0 there
exists δ > 0 such that
|f (x, y) − f (0, 0)| < whenever |x − 0| < δ, |y − 0| < δ.
Putting x = r cos θ and y = r sin θ, we have
|f (x, y) − f (0, 0)| = p
xy
x2 + y 2
p
r2 sin θ cos θ
√
=
≤ |r| = x2 + y 2 < r2
when δ < √ . Hence, for every > 0 there exists δ < √ > 0 such that
2
2
|f (x, y) − f (0, 0)| < whenever |x − 0| < δ, |y − 0| < δ.
Therefore, f is continuous at (0, 0).
Exercise 8.12
1. Define f : R2 → R as


x2 y 2
(x, y) 6= (0, 0)
f (x, y) =
(x2 + y 2 )2

0,
(x, y) = (0, 0)
Prove that f is not differentiable at (0, 0).
2. Discuss the differentiability of the following function :

xy 4

(x, y) 6= (0, 0)
(a) f (x, y) =
at (0,0)
(x2 + y 2 )2

0,
(x, y) = (0, 0)

3 3
xy

(x, y) 6= (0, 0)
(b) f (x, y) =
at (0,0)
(x2 + y 2 )3

0,
(x, y) = (0, 0)
9
Sufficient condition for differentiability
To understand this topic, please watch the video https://youtu.be/9s_hZYlLY8A
Theorem 9.1 Let f : D ⊂ R2 → R and (x0 , y0 ) ∈ R2 such that the following hold:
(a) Both fy exists at all points in some neighborhood of (x0 , y0 ).
(b) both fx are continuous at the point (x0 , y0 ).
Then prove that f is differentiable at (x0 , y0 ).
M
Proof: Given that fy exists at (x0 , y0 ). Therefore,
f (x0 , y0 + k) − f (x0 , y0 )
= fy (x0 , y0 ),
k→0
k
lim
which can be written as
f (x0 , y0 + k) − f (x0 , y0 ) = k [fy (x0 , y0 ) + ξ(k)] ,
(16)
SP
where ξ(k) → 0 as k → 0. Given that fx is continuous at (x0 , y0 ) and fx is exists in the neighborhood
of (x0 , y0 ).
Therefore, f is differentiable with respect to x in the interval (x0 , x0 + h).
Using Lagrange’s mean value theorem, there exist point θ ∈ (0, 1) such that
f (x0 + h, y0 + k) − f (x0 , y0 + k)
= fx (x0 + θh, y0 + k)
h
Taking limit h → 0 both side, we have
f (x0 + h, y0 + k) − f (x0 , y0 + k)
= lim fx (x0 + θh, y0 + k)
h→0
h→0
h
lim
(17)
Also, using hypothesis that fx is continuous at (x0 , y0 ), we have
lim
(h,k)→(0,0)
fx (x0 + θh, y0 + k) = fx (x0 , y0 ),
which can be written as
fx (x0 + θh, y0 + k) = fx (x0 , y0 ) + φ(h, k),
(18)
where φ(h, k) → 0 as (h, k) → (0, 0).
Using value of (18) in (17), we obtain
f (x0 + h, y0 + k) − f (x0 , y0 + k)
= fx (x0 , y0 ) + φ(h, k),
h
which is same as
f (x0 + h, y0 + k) − f (x0 , y0 + k) = hfx (x0 , y0 ) + hφ(h, k).
(19)
Now, using equations (16) and (19), we have
f (x0 + h, y0 + k) − f (x0 , y0 + k) = f (x0 + h, y0 + k) − f (x0 , y0 + k) + f (x0 , y0 + k) − f (x0 , y0 + k)
= hfx (x0 , y0 ) + hφ(h, k) + kfy (x0 , y0 ) + kξ(k),
where φ(h, k) and ξ(k) → 0 as h, k → 0.
Hence, f is differentiable at (x0 , y0 ).
Theorem 9.2 Let f : D ⊂ R2 → R and (x0 , y0 ) ∈ R2 such that the following hold:
(a) Both fy exists at all points in some neighborhood of (x0 , y0 ).
(b) both fx are continuous at the point (x0 , y0 ).
Then prove that f is differentiable at (x0 , y0 ).
The proof of these theorem left to readers.
Remark 9.3 The converse of the above results is not true, which can be seen by following example.
if x 6= 0, y 6= 0
M
Example 9.4 Define f : R2 → R as

1
1

2
2
 x sin
+ y sin
,



x
y



1

x2 sin
f (x, y) =
x


1


y 2 sin



y


0
if y 6= 0
if (x, y) = (0, 0).
SP
Prove that
if x 6= 0
(a) f is differentiable at (0, 0)
(b) fx and fy are not continuous at (0, 0).
Solution: Here,
f (∆x, 0) − f (0, 0)
f (0 + ∆x, 0) − f (0, 0)
= lim
∆x→0
∆x→0
∆x
∆x 1
∆x2 sin
1
∆x
= lim
= lim ∆x sin
= 0,
∆x→0
∆x→0
∆x
∆x
fx (0, 0) =
lim
and
f (0, 0 + ∆y) − f (0, 0)
f (0, ∆y) − f (0, 0)
= lim
∆y→0
∆y→0
∆y
∆x
1
∆y 2 sin
1
∆y
= lim
= lim ∆y sin
= 0,
∆y→0
∆y→0
∆y
∆y
fy (0, 0) =
lim
Therefore, for any (x, y) ∈ R2 , we have

1
1


− cos
,
 2x sin
x
x
fx (x, y) =
0



0
if x 6= 0, y 6= 0
if x = 0, y 6= 0
if (x, y) = (0, 0).
and

1
1


− cos
, if x 6= 0, y 6= 0
 2y sin
y
y
fy (x, y) =
0
if x 6= 0, y = 0



0
if (x, y) = (0, 0).
So, fx and fy exist at the point (0, 0)
Observe that
1
1
lim fx (x, y) = lim
2x sin
− cos
(x,y)→(0,0)
(x,y)→(0,0)
x
x
does not exist. Hence, fx is not continuous at (0, 0). Also,
1
1
lim fy (x, y) = lim
2y sin
− cos
(x,y)→(0,0)
(x,y)→(0,0)
y
y
does not exist. Hence, fy is not continuous at (0, 0).
We shall show that f is differentiable at (0, 0). It is enough to prove that
f (x0 + h, y0 + k) − f (x0 , y0 ) − hfx (x0 , y0 ) − kfy (x0 , y0 )
√
= 0.
(h,k)→(0,0)
h2 + k 2
lim
Consider,
f (x0 + h, y0 + k) − f (x0 , y0 ) − hfx (x0 , y0 ) − kfy (x0 , y0 )
√
(h,k)→(0,0)
h2 + k 2
f (h, k) − f (0, 0) − hfx (0, 0) − kfy (0, 0)
√
lim
(h,k)→(0,0)
h2 + k 2
1
1
2
2
h sin
+ k sin
− 0 − h(0) − k(0)
h
k
√
lim
(h,k)→(0,0)
h2 + k 2
1
1
2
2
h sin
+ k sin
− 0 − h(0) − k(0)
h
k
√
lim
(h,k)→(0,0)
h2 + k 2
1
1
2
2
h sin
+ k sin
h
k
√
lim
(h,k)→(0,0)
h2 + k 2
=
=
=
SP
=
M
lim
Now, we shall show that
1
1
2
h sin
+ k sin
h
k
√
lim
= 0.
(h,k)→(0,0)
h2 + k 2
We need to prove that for every > 0 there exists δ > 0 such that
1
1
2
2
+ k sin
h sin
h
k
√
< whenever |h| < δ, |k| < δ.
2
2
h +k
2
Consider
1
1
2
+ k sin
h sin
h
k
√
2
2
h +k
2
1
≤ √
h2 sin
2
2
h +k
≤ √
1
1
2
+ k sin
h
k
√
1
h2 + k 2 = h2 + k 2
h2 + k 2
√
√
Now, |h| < δ and |k| < δ gives h2 + k 2 < 2δ.
Hence, for every > 0 there exists δ < √ > 0 such that
2
1
1
2
2
+ k sin
h sin
h
k
√
< whenever |h| < δ, |k| < δ.
2
2
h +k
Therefore, f is differentiable at (0, 0).
Exercise 9.5
1. Define f : R2 → R as


x2 y 2
(x, y) 6= (0, 0)
f (x, y) =
(x3 + y 3 )

0,
(x, y) = (0, 0)
3. Define f : R2 → R as
M
Then prove that fx (0, 0) and fy (0, 0) exist and function f is continuous but not differentiable at
point (0, 0).
p
2. Define f : R2 → R as f (x, y) = |xy|. Then prove that fx (0, 0) and fy (0, 0) exist and function f is
continuous but not differentiable at point (0, 0).
 2
 x + y2
(x, y) 6= (0, 0)
f (x, y) =
(x − y)

0,
(x, y) = (0, 0)
SP
Then prove that fx (0, 0) and fy (0, 0) exist and function f is not continuous and f is not differentiable at point (0, 0).
4. Define f : R2 → R as

2
 p xy
(x, y) 6= (0, 0)
f (x, y) =
x2 + y 2

0,
(x, y) = (0, 0)
Then prove that fx (0, 0) and fy (0, 0) exist and function f is continuous and f is differentiable
at point (0, 0).
10
EQUALITY OF MIXED PARTIALS
To understand this topic, please watch the video https://youtu.be/HOmA-P5vwCM
It has been seen that two repeated second order partial derivative are generally equal. They are
not, however always equal as is shown below by considering two examples. It easy to see a priori
also why fyx (a, b) may different from fxy (a, b).
We have
fy (a + h, b) − fy (a, b)
fyx (a, b) = lim
.
h→0
h
Also,
f (a + h, b + k) − f (a + h, b)
fy (a + h, b) = lim
,
k→0
k
and
f (a, b + k) − f (a, b)
fy (a, b) = lim
.
k→0
k
Therefore
f (a + h, b + k) − f (a + h, b) − f (a, b + k) + f (a, b)
hk
φ(h, k)
,
= lim lim
h→0 k→0
hk
fyx (a, b) = lim lim
h→0 k→0
where φ(h, k) = f (a + h, b + k) − f (a + h, b) − f (a, b + k) + f (a, b). It may similarly be shown that
φ(h, k)
,
k→0 h→0
hk
fxyy (a, b) = lim lim
Thus, we see that fxy (a, b) and fyx (a, b) are repeated limits of the same expression taken in different
orders. Also, the two repeated limit may not equal, as for example
h−k
h
= lim = 1
h→0 k→0 h + k
h→0 h
lim lim
and
h−k
−k
= lim
= −1.
k→0 h→0 h + k
k→0 k
lim lim
Solution: We have
fyx (0, 0) = lim
Also,
fy (0 + h, 0) − fy (0, 0)
h
SP
h→0
M
Example 10.1 Prove that fyx (0, 0) 6= fxy (0, 0) for the function f given by

 xy(x2 − y 2 )
, (x, y) 6= (0, 0)
f (x, y) =
x2 + y 2

0
(x, y) = (0, 0)
f (0, 0 + k) − f (0, 0)
k→0
k
f (0, k) − f (0, 0)
0
= lim
= lim = 0
k→0
k→0 k
k
(20)
fy (0, 0) = lim
and
f (h, 0 + k) − f (h, 0)
k→0
k
f (h, k) − f (0, 0)
hk(h2 − k 2 )
= lim
= lim
= h.
k→0
k→0 k(h2 + k 2 )
k
(21)
fy (h, 0) = lim
(22)
Thus from (20), (21) and (22), we have
h−0
= 1.
h→0
h
(23)
fx (0, 0 + k) − fx (0, 0)
k→0
k
(24)
fyx (0, 0) = lim
Again
fxy (0, 0) = lim
Also,
f (0 + h, 0) − f (0, 0)
h→0
h
f (h, 0) − f (0, 0)
0
= lim
= lim = 0
h→0
h→0 h
h
fx (0, 0) = lim
(25)
and
f (0 + h, k) − f (0, k)
h
f (h, k) − f (0, 0)
hk(h2 − k 2 )
= lim
= lim
= −k.
h→0
h→0 h(h2 + k 2 )
h
fx (h, 0) = lim
h→0
(26)
hus from (24), (25) and (26), we have
−k − 0
= −1.
k→0
k
fxy (0, 0) = lim
(27)
Thus fxy (0, 0) 6= fyx (0, 0).
Example 10.2 Show that fxy (0, 0) 6= fyx (0, 0), where f (x, y) = 0 if xy = 0 and
y
x
f (x, y) = x2 tan−1 − y 2 tan−1 , if xy 6= 0.
x
y
We have
fy (0 + h, 0) − fy (0, 0)
h→0
h
fyx (0, 0) = lim
(28)
Also,
=
SP
=
=
=
and
M
f (h, 0 + k) − f (h, 0)
k→0
k
f (h, k) − f (0, 0)
lim
k→0
k
h
k
h2 tan−1 − k 2 tan−1
h
k
lim
k→0
k
k
tan−1
h − k tan−1 h
lim h
k
k→0
k
h
tan−1 t
h(1) − 0 = h as lim
= 1.
t→0
t
fy (h, 0) = lim
f (0, 0 + k) − f (0, 0)
k→0
k
f (0, k) − f (0, 0)
0
= lim
= lim = 0.
k→0
k→0 k
k
(29)
fy (0, 0) = lim
(30)
Thus from (28), (29) and (30), we have
h−0
= 1.
h→0
h
(31)
fx (0, 0 + k) − fx (0, 0)
k→0
k
(32)
fyx (0, 0) = lim
Again
fxy (0, 0) = lim
Also,
f (0 + h, 0) − f (0, 0)
h→0
h
f (h, 0) − f (0, 0)
0
= lim
= lim = 0
h→0
h→0 h
h
fx (0, 0) = lim
(33)
and
f (0 + h, k) − f (0, k)
h→0
h
f (h, k) − f (0, 0)
lim
h→0
h
h
k
h2 tan−1 − k 2 tan−1
h
k
lim
h→0
h
h
tan−1
k
k
lim h tan−1 − k
h
k→0
h
k
tan−1 t
0 − k(1) = −k as lim
= 1.
t→0
t
fx (h, 0) = lim
=
=
=
=
(34)
(35)
Thus from (32), (33) and (34), we have
−k − 0
= −1.
k→0
k
fxy (0, 0) = lim
(36)
M
Thus fxy (0, 0) 6= fyx (0, 0).
To understand this topic, please watch the video https://youtu.be/I-AFsTmqSCk.
(a) fy exists at (x0 , y0 ).
SP
Theorem 10.3 (Schwartz’s Theorem) Let f : D ⊂ R2 → R be function and (x0 , y0 ) ∈ D. If
(b) fxy is continuous at (x0 , y0 ).
Then prove that fyx also exists and fyx (x0 , y0 ) = fxy (x0 , y0 ).
Proof: Let N be neighborhood of (x0 , y0 ) such that (x0 + h, y0 + k) ∈ N ⊂ D.
Define
φ(h, k) = f (x0 + h, y0 + k) − f (x0 + h, y0 ) − f (x0 , y0 + k) + f (x0 , y0 ),
Now, define
(37)
g(x) = f (x, y0 + k) − f (x, y0 ).
(38)
g(x0 ) = f (x0 , y0 + k) − f (x0 , y0 ).
(39)
By putting x = x0 in (38), we have
Now, by putting x = x0 + h in (38), we have
g(x0 + h) = f (x0 + h, y0 + k) − f (x0 + h, y0 ).
(40)
Using (39) and (40) in (37), we get
φ(h, k) = g(x0 + h) − g(x0 ).
(41)
Given that fxy exist at point (x0 , y0 ).
Therefore, fx exists at (x0 , y0 ). Hence f is differentiable with respect to x in neighborhood N . That
gives, g is differentiable with respect to x in neighborhood of (x0 , x0 + h).
Applying Lagrange’s Mean value theorem, there exist θ ∈ (0, 1) such that
g(x0 + h) − g(x0 )
= g 0 (x0 + θh)
h
(42)
Using (41) in (42), we have
φ(h, k) = hg 0 (x0 + θh)
(43)
Now,
g 0 (x) = fx (x, y0 + k) − fx (x, y0 )
Replacing x by x0 + θh, we obtain
g 0 (x) = fx (x0 + θh, y0 + k) − fx (x0 + θh, y0 )
Putting this value in (42), we obtain
φ(h, k) = h [fx (x0 + θh, y0 + k) − fx (x0 + θh, y0 )] .
(44)
Also, given that fxy is continuous at (x0 , y0 ).
Therefore, fxy exist in the neighborhood of (x0 , y0 ). Hence, fx is differentiable with respect to y in
some neighborhood [y0 , y0 + k].
By Lagrange’s Theorem, there exist θ1 ∈ (0, 1).
fx (x0 + θh, y0 + k) − fx (x0 + h, y0 )
= fxy (x0 + θh, y0 + θ1 k)
k
M
Using (44), we get
(45)
φ(h, k)
= fxy (x0 + θh, y0 + θ1 h)
hk
φ(h, k) = hkfxy (x0 + θh, y0 + θ1 h)
by the hypothesis, fxy continuous at (x0 , y0 ).
fxy (x0 + θh, y0 + θ1 h) = fxy (x0 , y0 )
SP
lim
(h,k)→(0,0)
(46)
Using equation (46), we have
φ(h, k)
(h,k)→(0,0)
hk
f (x0 + h, y0 + k) − f (x0 + h, y0 ) − f (x0 , y0 + k) + f (x0 , y0 )
lim
(h,k)→(0,0)
hk
1 f (x0 + h, y0 + k) − f (x0 + h, y0 ) f (x0 , y0 + k) − f (x0 , y0 )
lim
−
(h,k)→(0,0) h
k
k
fy (x0 + h, y0 ) − fy (x0 , y0 )
lim
h→0
h
lim
= fxy (x0 , y0 )
= fxy (x0 , y0 )
= fxy (x0 , y0 )
= fxy (x0 , y0 ).
Therefore fyx (a, b) = fxy (a, b).
Remark 10.4 The converse of the Schwartz’s Theorem is not true, which can be seen by following
examples.
Example 10.5 Let f : R2 → R defined as

 x2 y 2
, (x, y) 6= (0, 0)
f (x, y) =
x2 + y 2

0,
(x, y) = (0, 0).
Prove that
(a) fxy (0, 0) = fyx (0, 0).
(b) fxy is not continuous at (0, 0).
Solution: We have
f (0 + h, 0) − f (0, 0)
0−0
= lim
= 0.
h→0
h→0
h
h
fx (0, 0) = lim
Now,
fx (0, 0 + k) − fx (0, 0)
fx (0, k) − f (0, 0)
= lim
.
k→0
k→0
k
k
fxy (0, 0) = lim
To calculate fx (0, k), we have
h2 k 2
−0
2 + k2
f (0 + h, k) − f (0, 0)
h2 k 2
hk 2
h
fx (0, k) = lim
= lim
= lim
=
lim
= 0.
h→0
h→0
h→0 h(h2 + k 2 )
h→0 (h2 + k 2 )
h
h
Therefore,
fx (0, k) − f (0, 0)
0 − 0)
= lim
= 0.
k→0
k→0
k
k
fxy (0, 0) = lim
M
Similarly,
f (0, 0 + k) − f (0, 0)
0−0
= lim
= 0.
k→0
k→0
k
k
fy (0, 0) = lim
Now,
fy (0 + h, 0) − fy (0, 0)
fy (h, 0) − f (0, 0)
= lim
.
h→0
h→0
k
h
SP
fyx (0, 0) = lim
To calculate fy (h, 0), we have
h2 k 2
−0
2 + k2
f (h, 0 + k) − f (0, 0)
h2 k 2
h2 k
h
fy (h, 0) = lim
= lim
= lim
= lim 2
= 0.
k→0
k→0
k→0 k(h2 + k 2 )
k→0 (h + k 2 )
k
k
Therefore,
0−0
fy (h, 0) − f (0, 0)
= lim
= 0.
h→0
h→0
h
h
fyx (0, 0) = lim
Hence, fxy (0, 0) = fyx (0, 0). Now, we shall show that fxy is not continuous at (0, 0).
For (x, y) 6= (0, 0), we have
fx (x, y) =
(x2 + y 2 )(2xy 2 ) − x2 y 2 (2x)
2x3 y 2 + 2xy 4 − 2x3 y 2
2xy 4
=
=
(x2 + y 2 )2
(x2 + y 2 )2
(x2 + y 2 )2
Hence,


2xy 4
(x, y) 6= (0, 0)
fx (x, y) =
(x2 + y 2 )2

0
(x, y) = (0, 0).
Now, if (x, y) 6= (0, 0), we have
fxy (x, y) =
=
=
=
=
(x2 + y 2 )2 (8xy 3 ) − 2xy 4 (2(x2 + y 2 )(2y))
(x2 + y 2 )4
(x2 + y 2 )2 (8xy 3 ) − 8xy 5 (x2 + y 2 )
(x2 + y 2 )4
8xy 3 [x4 + y 4 + 2x2 y 2 − x2 y 2 − y 4 ]
(x2 + y 2 )4
8xy 3 [x4 + x2 y 2 ]
(x2 + y 2 )4
8x3 y 3
(x2 + y 2 )3
Hence,


8x3 y 3
(x, y) 6= (0, 0)
fxy (x, y) =
(x2 + y 2 )3

0
(x, y) = (0, 0).
lim
(x,y)→(0,0)
fxy (x, y) =
lim
(x,mx)→(0,0)
M
We shall show that lim(x,y)→(0,0) fxy (x, y) does not exist.
Taking path along y = x, we have
8m2 x6
8m2
8x3 (mx)3
=
lim
=
,
x→0 x6 (1 + m2 )3
x→0 (x2 + (mx)2 )3
(1 + m2 )3
fxy (x, mx) = lim
(x,y)→(0,0)
SP
which depend on m.
Therefore, lim fxy (x, y) does not exist and fxy is not continuous at (0, 0).
To understand this topic, please watch the video https://youtu.be/_Wh9Ds5lbKQ.
Theorem 10.6 (Young’s Theorem) Let f : D ⊂ R2 → R be a function and (x0 , y0 ) ∈ D. Consider N
is neighborhood of (x0 , y0 ). if
1. fx and fy exists in N .
2. fx and fy both are differentiable at (x0 , y0 ).
Then prove that fxy (x0 , y0 ) and fyx (x0 , y0 ) exit and fxy (x0 , y0 ) = fyx (x0 , y0 ).
Proof: Given that N be neighborhood of (x0 , y0 ) such that (x0 + h, y0 + k) ∈ N ⊂ D.
Define
φ(h, k) = f (x0 + h, y0 + k) − f (x0 + h, y0 ) − f (x0 , y0 + k) + f (x0 , y0 ),
(47)
Now, define
g(x) = f (x, y0 + k) − f (x, y0 ).
(48)
φ(h, k) = g(x0 + h) − g(x0 ).
(49)
Now,
Since, fx and fy are differentiable at (x0 , y0 ), fxx , fyy , fxy and fyx are exists at (x0 , y0 ).
Since, fx is differentiable at (x0 , y0 ), g is differentiable in the interval (x0 , x0 + h).
By Lagrange’s Mean value theorem, There exists θ ∈ (0, 1) such that
g(x0 + h) − g(x0 )
= g 0 (x0 + θh).
h
(50)
As g(x) = f (x, y0 + k) − f (x, y0 ), we have
g 0 (x) = fx (x, y0 + k) − fx (x, y0 ).
(51)
Using (47) and (51) in (50), we get
φ(h, k)
= fx (x0 + θh, y0 + k) − fx (x0 + θh, y0 ).
h
Therefore,
φ(h, k) = h [fx (x0 + θh, y0 + k) − fx (x0 + θh, y0 )] .
(52)
Now, given that fx is differentiable at (x0 , y0 ), there exist A and B such that
fx (x0 + h, y0 + k) = fx (x0 , y0 ) + hA + kB + h1 (h, k) + k2 (h, k)
Replacing h by θh and A = fxx (x0 , y0 ) and B = fxy (x0 , y0 ), we have
fx (x0 + θh, y0 + k) = fx (x0 , y0 ) + θhfxx (x0 , y0 ) + kfxy (x0 , y0 ) + θh1 (θh, k) + k2 (θh, k).
(53)
Putting k = 0, we get
fx (x0 + θh, y0 ) = fx (x0 , y0 ) + θhfxx (x0 , y0 ) + θh3 (θh, 0).
(54)
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Now, using (53) and (54) in (52), we have
φ(h, k) = h [fx (x0 , y0 ) + θhfxx (x0 , y0 ) + kfxy (x0 , y0 ) + θh1 (θh, k) + k2 (θh, h)
−fx (x0 , y0 ) − θhfxx (x0 , y0 ) − θh3 (θh, 0)]
= hkfxy (x0 , y0 ) + θh2 1 (θh, k) + hk2 (θh, k) − θh2 3 (θh, 0)
Therefore,
SP
φ(h, h) = h2 fxy (x0 , y0 ) + θh2 1 (θh, h) + h2 2 (θh, h) − θh2 3 (θh, 0),
which can be express as
φ(h, h)
= fxy (x0 , y0 ) + θ1 (θh, h) + 2 (θh, h) − θ3 (θh, 0).
h2
Taking limit as h → 0, we have
lim
h→0
φ(h, h)
= lim [fxy (x0 , y0 ) + θ1 (θh, h) + 2 (θh, h) − θ3 (θh, h)] .
h→0
h2
Hence,
φ(h, h)
= fxy (x0 , y0 )
h→0
h2
lim
as 1 (h, k), 2 (h, k) and 3 (h, k) tends to 0 as h → 0.
Now, define
G(y) = f (x0 + h, y) − f (x0 , y).
From equation (47), we have
φ(h, k) = G(y0 + k) − G(y0 )
and
G0 (y) = fy (x0 + h, y) − fy (x0 , y)
Following producer as above, we have
φ(k, k) = k 2 fyx (x0 , y0 ) + θ1 k 2 4 (k, θ1 k) + k 2 5 (k, θ1 k) − θ1 k 2 6 (0, θ1 k),
(55)
where 4 (h, k), 5 (h, k) and 6 (h, k) tends to 0 as h → 0.
Therefore,
φ(k, k)
= fyx (x0 , y0 )
k→0
k2
lim
(56)
Using (55) and (56), we get
fyx (x0 , y0 ) = fxy (x0 , y0 ).
Remark 10.7 The converse of Young’s theorem is not true, which can be seen by following example.
Example 10.8 Let f : R2 → R defined as

 x2 y 2
, (x, y) 6= (0, 0)
f (x, y) =
x2 + y 2

0,
(x, y) = (0, 0).
Prove that
(a) fxy (0, 0) = fyx (0, 0).
(b) fx and fy are not differentiable at (0, 0).
M
Solution: In example 10.5, we have prove that fxy (0, 0) = fyx (0, 0) = 0.
Also, we obtain

2xy 4

(x, y) 6= (0, 0)
fx (x, y) =
(x2 + y 2 )2

0
(x, y) = (0, 0).
fx (h, 0) − fx (0, 0)
0−0
= lim
= 0.
h→0
h
h
SP
Also, we have
fxx (0, 0) = lim
h→0
Now, we shall show that fx is not differentiable at (0, 0). It is enough to prove that
fx (0 + h, 0 + k) − hfxx (0, 0) − kfxy (0, 0) − fx (0, 0)
√
(h,k)→(0,0)
h2 + k 2
lim
does not exist.
Consider
2hk 4
− h(0) − k(0) − (0)
fx (0 + h, 0 + k) − hfxx (0, 0) − kfxy (0, 0) − fx (0, 0)
(h2 + k 2 )2
√
√
lim
=
lim
(h,k)→(0,0)
(h,k)→(0,0)
h2 + k 2
h2 + k 2
2hk 4
=
lim
.
(h,k)→(0,0) (h2 + k 2 )5/2
Taking path along k = mh, we obtain
2hk 4
2h(mh)4
2m4 h5
2m4
=
lim
=
lim
=
,
h→0 (h2 + (mh)2 )5/2
h→0 h5 (1 + m2 )5/2
(h,k)→(0,0) (h2 + k 2 )5/2
(1 + m2 )5/2
lim
which depends on m. Hence, the limit does not exist.
Therefore, fx (0, 0) is not differentiable at (0, 0).
Observe that

2yx4

(x, y) 6= (0, 0)
fy (x, y) =
(x2 + y 2 )2

0
(x, y) = (0, 0).
ow, we shall show that fy is not differentiable at (0, 0). It is enough to prove that
fy (0 + h, 0 + k) − hfyy (0, 0) − kfyx (0, 0) − fy (0, 0)
√
(h,k)→(0,0)
h2 + k 2
lim
does not exist.
We note that,
0−0
fy (0, k) − fy (0, 0)
= lim
= 0.
k→0
k→0
k
k
fyy (0, 0) = lim
Consider
2kh4
− h(0) − k(0) − (0)
fy (0 + h, 0 + k) − hfyy (0, 0) − kfyx (0, 0) − fy (0, 0)
(h2 + k 2 )2
√
√
lim
=
lim
(h,k)→(0,0)
(h,k)→(0,0)
h2 + k 2
h2 + k 2
2kh4
=
lim
.
(h,k)→(0,0) (h2 + k 2 )5/2
Taking path along h = mk, we obtain
2k(mk)4
2m4 k 5
2m4
2kh4
=
lim
=
lim
=
,
k→0 ((mk)2 + k 2 )5/2
k→0 k 5 (1 + m2 )5/2
(h,k)→(0,0) (h2 + k 2 )5/2
(1 + m2 )5/2
M
lim
which depends on m. Hence, the limit does not exist.
Therefore, fy (0, 0) is not differentiable at (0, 0).
Directional Derivative
SP
11
To understand this topic, please watch the video https://youtu.be/QrcevYqyJ5E.
In this section we extend the concept of a partial derivative to the more general notion of a
directional derivative. We have seen that the partial derivatives of a function give the instantaneous rates of change of that function in directions parallel to the coordinate axes. Directional
derivatives allow us to compute the rates of change of a function with respect to distance in any
direction.
Suppose that we wish to compute the instantaneous rate of change of a function f (x, y) with respect to distance from a point (x0 , y0 ) in some direction. Since there are infinitely many different
directions from (x0 , y0 ) in which we could move, we need a convenient method for describing a
specific direction starting at (x0 , y0 ). One way to do this is to use a unit vector
u = u1 î + u2 ĵ,
that has its initial point at (x0 , y0 ) and points in the desired direction (Figure 9). This vector
determines a line l in the xy-plane that can be expressed parametrically as
x = x0 + su1 ,
y = y0 + su2
Since u is a unit vector, s is the arc length parameter that has its reference point at (x0 , y0 ) and
has positive values in the direction of u. For s = 0, the point (x, y) is at the reference point (x0 , y0 ),
and as s increases, the point (x, y) moves along l in the direction of u. On the line l the variable
dz
z = f (x0 + su1 , y0 + su2 ) is a function of the parameter s. The value of the derivative
at s = 0
ds
then gives an instantaneous rate of change of f (x, y) with respect to distance from (x0 , y0 ) in the
direction of u.
Figure 9
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Definition 11.1 If f (x, y) is a function of x and y, and if u = u1 î + u2 ĵ is a unit vector, then the
directional derivative of f in the direction of u at (x0 , y0 ) is denoted by Du f (x0 , y0 ) and is defined by
d
[f (x0 + su1 , y0 + su2 )]s=0
ds
f (x0 + su1 , y0 + su2 ) − f (x0 , y0 )
= lim
s→0
s
Du f (x0 , y0 ) =
SP
provided this derivative exists.
(57)
(58)
Geometrically, Du f (x0 , y0 ) can be interpreted as the slope of the surface z = f (x, y) in the direction
of u at the point (x0 , y0 , f (x0 , y0 )). Usually the value of Du f (x0 , y0 ) will depend on both the point
(x0 , y0 ) and the direction u. Thus, at a fixed point the slope of the surface may vary with the
direction (Figure 10). Analytically, the directional derivative represents the instantaneous rate
of change of f (x, y) with respect to distance in the direction of u at the point (x0 , y0 ).
Example 11.2 Let f (x, y) = xy. Find and interpret Du f (1, 2) for the unit vector
√
3
1
u=
î + ĵ.
2
2
Solution: It follows from Equation (57) that
"
!#
√
d
s 3
s
Du f (1, 2) =
f 1+
,2 +
ds
2
2
s=0
Since
f
we have
√
√ !
s 3
s 3
1+
,1 +
=
2
2
√ !
√
s 3 s
3 2
1 √
1+
1+
=
s +
+ 3 s + 2,
2
2
4
2
"√
#
d
3 2
1 √
Du f (1, 2) =
s +
+ 3 s+2
ds 4
2
s=0
"√
#
√
3
1
1 √
=
s+
+ 3
= + 3.
2
2
2
s=0
M
SP
Figure 10: The slope of the surface varies with the direction of u.
1 √
+ 3 = 2.23 we conclude that if we move a small distance from the point (1, 2) in the
2
direction of u, the function f (x, y) = xy will increase by about 2.23 times the distance moved.
Since, =
Example 11.3 Define f : R2 → R as


xy 2
(x, y) 6= (0, 0)
f (x, y) =
x2 + y 4

0
(x, y) = (0, 0).
Find the directional derivative of f at point (0, 0) along the direction of the vector
1 1
√ ,√ .
2 2
Solution: It follows from the definition of directional derivative that
s
s
− f (0, 0)
f 0 + √ ,0 + √
f (x0 + tu1 , y0 + tu2 ) − f (x0 , y0 )
2
2
Du f (0, 0) = lim
= lim
s→0
s→0
s
s
s
s
− f (0, 0)
f √ ,√
2 2
= lim
s→0
s
2
s
s
√
√
2
2
= lim
4 !
2
s→0
s
s
√
s
+ √
2
2
1
2
1
= √ =√
= lim
2
s→0 √
1 s
2 2
2
2 2
+
2
4
Theorem 11.4 If f (x, y) is differentiable at (x0 , y0 ), and if u = u1 î + u2 ĵ is a unit vector, then the
directional derivative Du f (x0 , y0 ) exists and is given by
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Du f (x0 , y0 ) = fx (x0 , y0 )u1 + fy (x0 , y0 )u2 .
Example 11.5 Find the directional derivative of f (x, y) = exy at (−2, 0) in the direction of the unit
π
vector that makes an angle of with the positive x-axis.
3
Solution: The partial derivatives of f are
fy (x, y) = xexy
SP
fx (x, y) = yexy ,
fx (−2, 0) = 0, fy (−2, 0) = −2
π
The unit vector u that makes an angle of with the positive x-axis is
3
√
π π 1
3
î + sin
ĵ = î +
ĵ.
u = cos
3
3
2
2
Thus, from
Du f (x0 , y0 ) = fx (x0 , y0 )u1 + fy (x0 , y0 )u2 ,
we have
Du f (−2, 0) = fx (−2, 0) cos
π 3
+ fy (−2, 0) sin
π 3
1
= (0)
+ (−2)
2
√ !
√
3
= − 3.
2
Exercise11.6 Find the directional derivative of f at point (0, 0) along the direction of the vector
1 1
√ , √ , where
2 2

 xy 2
(x, y) 6= (0, 0)
f (x, y) =
x2 + y 2

0
(x, y) = (0, 0).
Exercise
11.7 Find the directional derivative of f at point (0, 0) along the direction of the vector
1 2
√ , √ , where
3 6

 xy 3
(x, y) 6= (0, 0)
f (x, y) =
x2 + y 6

0
(x, y) = (0, 0).
Exercise 11.8 Let f (x, y) =
y
. Find a unit vector u for which Du f (2, 3) = 0.
x+y
Exercise 11.9 Find the directional derivative of f (x, y) = 4x3 y 2 at P (2, 1) in the direction of 4î − 3ĵ.
12
Differential
To understand this topic, please watch the video https://youtu.be/ghEHdb8KOpg.
Differential for one variable
Let y = f (x) be a real valued function of real variable x. Suppose that f is differentiable at x then
f (x + ∆x) − f (x)
.
∆x→0
∆x
f 0 (x) = lim
M
The value of x changed from x + ∆x to x and the value of y = f (x) changed from f (x + ∆x) = y + ∆y
to y.
y + ∆y − y
∆y
f 0 (x) = lim
= lim
.
∆x→0
∆x→0 ∆x
∆x
where ∆x is called differential of x and ∆y is called differential of y and it is denoted by dx and
dy respectively.
Differential for two variables
SP
Let z = f (x, y) be a real valued function of real variables x and y. Suppose f is differentiable at
(x, y). Then
f (x + ∆x, y + ∆y) = f (x, y) + ∆xfx (x, y) + ∆yfy (x, y) + ∆x 1 (∆x, ∆y) + ∆y 2 (∆x, ∆y),
where 1 (∆x, ∆y) and 2 (∆x, ∆y) → 0 as (∆x, ∆y) → 0. Here, we note that x changes from x + ∆x to
x, x changes from y + ∆y to y and z changes from x + ∆z to z.
Therefore,
z + ∆z = z + ∆xfx (x, y) + ∆yfy (x, y) + ∆x 1 (∆x, ∆y) + ∆y 2 (∆x, ∆y),
Hence,
∆z = ∆xfx (x, y) + ∆yfy (x, y) + ∆x 1 (∆x, ∆y) + ∆y 2 (∆x, ∆y),
(59)
Here in the equation (59), the principal part fx (x, y)dx + fy (x, y)dy is called the total differential of
z and denoted by dz. Therefore,
dz = fx (x, y)dx + fy (x, y)dy.
Similarly, for a function w = f (x, y, z) of three variables we have the total differential of w at
(x, y, z),
dw = fx (x, y, z)dx + fy (x, y, z)dy + fz (x, y, z)dz
In the two-variable case, the approximation
∆f ≈ fx (x, y)∆x + fy (x, y)∆y
can be written in the form
∆f ≈ df
(60)
for dx = ∆x and dy = ∆y. Equivalently, we can write approximation (60) as
∆z ≈ dz
(61)
In other words, we can estimate the change ∆z in z by the value of the differential dz where, dx
is the change in x and dy is the change in y. Furthermore, it follows from (13) that if ∆x and ∆y
are close to 0,pthen the magnitude of the error in approximation (61) will be much smaller than
the distance (∆x)2 + (∆y)2 between (x, y) and (x + ∆x, y + ∆y).
Example 12.1 Approximate the change in z = xy 2 from its value at (0.5, 1.0) to its value at
(0.503, 1.004). Compare the magnitude of the error in this approximation with the distance between
the points (0.5, 1.0) and (0.503, 1.004).
Solution: For z = xy 2 we have dz = y 2 dx + 2xydy. Evaluating this differential at (x, y) = (0.5, 1.0),
dx = ∆x = 0.503 − 0.5 = 0.003, and dy = ∆y = 1.004 − 1.0 = 0.004 yields
dz = 1.02 (0.003) + 2(0.5)(1.0)(0.004) = 0.007.
Since z = 0.5 at (x, y) = (0.5, 1.0) and z = 0.507032048 at (x, y) = (0.503, 1.004), we have
∆z = 0.507032048 − 0.5 = 0.007032048
and the error in approximating ∆z by dz has magnitude
|dz − ∆z| = |0.007 − 0.007032048| = 0.000032048.
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Since the distance between (0.5, 1.0) and (0.503, 1.004) = (0.5 + ∆x, 1.0 + ∆y) is
p
p
√
(∆x)2 + (∆y)2 = (0.003)2 + (0.004)2 = 0.000025 = 0.005,
we have
SP
|dz − ∆z|
0.000032048
1
p
=
= 0.0064096 <
.
0.005
150
(∆x)2 + (∆y)2
1
Thus, the magnitude of the error in our approximation is less than
of the distance between
150
the two points.
Example 12.2 The length, width, and height of a rectangular box are measured with an error of
at most 5%. Use a total differential to estimate the maximum percentage error that results if these
quantities are used to calculate the diagonal of the box.
Solution: The diagonal D of a box with length x, width y, and height z is given by
p
D = x2 + y 2 + z 2
p
Let x0 , y0 , z0 , and D0 = x20 + y02 + z02 denote the actual values of the length, width, height, and
diagonal of the box. The total differential dD of D at (x0 , y0 , z0 ) is given by
y0
z0
x0
p
p
dx
+
dy
+
dz
dD = p 2
x0 + y02 + z02
x20 + y02 + z02
x20 + y02 + z02
p
If x, y, z and D =
x2 + y 2 + z 2 are the measured and computed values of the length, width,
height, and diagonal, respectively, then
∆x = x − x0 ,
∆y = y − y0 , ∆z = z − z0
and
∆y
∆z
∆x
≤ 0.05,
≤ 0.05,
≤ 0.05
x0
y0
z0
∆D
We are seeking an estimate for the maximum size of
. With the aid of Equation
D0
dw = fx (x, y, z)dx + fy (x, y, z)dy + fz (x, y, z)dz,
we have
∆D
dD
1
[x0 ∆x + y0 ∆y + z0 ∆z]
≈
= 2
D0
D0
x0 + y02 + z02
1
2 ∆x
2 ∆y
2 ∆z
= 2
x
+ y0
+ z0
x0 + y02 + z02 0 x0
y0
z0
Since
dD
D0
∆y
∆z
∆x
1
+ y02
+ z02
x20
2
2
+ y0 + z0
x
y0
z0
0
1
2 ∆x
2 ∆y
2 ∆z
x0
≤ 2
+ y0
+ z0
x0 + y02 + z02
x0
y0
z0
1
2
2
2
x
(0.05)
+
y
(0.05)
+
z
(0.05)
= 0.05,
≤ 2
0
0
x0 + y02 + z02 0
=
x20
we estimate the maximum percentage error in D to be 5%.
LOCAL LINEAR APPROXIMATIONS
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We now show that if a function f is differentiable at a point, then it can be very closely approximated by a linear function near that point. For example, suppose that f (x, y) is differentiable at
the point (x0 , y0 ). Then approximation
∆f ≈ fx (x0 , y0 )∆x + fy (x0 , y0 )∆y
can be written in the form
SP
f (x0 + ∆x, y0 + ∆y) ≈ f (x0 , y0 ) + fx (x0 , y0 )∆x + fy (x0 , y0 )∆y
If we let x = x0 + ∆x and y = y0 + ∆y, this approximation becomes
f (x, y) ≈ f (x0 , y0 ) + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 )
(62)
which yields a linear approximation of f (x, y). Since the error in this approximation is equal to
the error in approximation
∆f ≈ fx (x0 , y0 )∆x + fy (x0 , y0 )∆y,
we conclude that for (x, y) close to (x0 , y0 ), the error in (62) will be much smaller than the distance
between these two points. When f (x, y) is differentiable at (x0 , y0 ), we get
L(x, y) = f (x0 , y0 ) + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 )
and refer to L(x, y) as the local linear approximation to f at (x0 , y0 ).
Example 12.3 Let L(x, y) denote the local linear approximation to f (x, y) =
(3, 4). Compare the error in approximating
p
f (3.04, 3.98) = (3.04)2 + (3.98)2
p
x2 + y 2 at the point
by L(3.04, 3.98) with the distance between the points (3, 4) and (3.04, 3.98).
Solution: We have
x
y
fx (x, y) = p
and fy (x, y) = p
x2 + y 2
x2 + y 2
3
4
with fx (3, 4) = and fy (3, 4) = . Therefore, the local linear approximation to f at (3, 4) is given by
5
5
3
4
L(x, y) = 5 + (x − 3) + (y − 4)
5
5
Consequently,
3
4
f (3.04, 3.98) ≈ L(3.04, 3.98) = 5 + (0.04) + (−0.02) = 5.08
5
5
Since
f (3.04, 3.98) =
p
(3.04)2 + (3.98)2 ≈ 5.00819
1
the error in the approximation is about 5.00819 − 5.008 = 0.00019. This is less than
of the
200
distance
p
(3.04 − 3)2 + (3.98 − 4)2 ≈ 0.045
between the points (3, 4) and (3.04, 3.98).
Exercise 12.4 Suppose that a function f (x, y, z) is differentiable at the point (2, 1, −2), fx (2, 1, −2) =
−1, fy (2, 1, −2) = 1, and fz (2, 1, −2) = −2. If f (2, 1, −2) = 0, estimate the value of f (1.98, 0.99, −1.97).
Exercise 12.5 Compute the differential dz of the function z = exy .
13
THE CHAIN RULE
M
In this section we will derive versions of the chain rule for functions of two or three variables.
These new versions will allow us to generate useful relationships among the derivatives and
partial derivatives of various functions.
CHAIN RULES FOR DERIVATIVES
To understand this topic, please watch the video https://youtu.be/hghiy--fn1s.
SP
If y is a differentiable function of x and x is a differentiable function of t, then the chain rule for
functions of one variable states that, under composition, y becomes a differentiable function of
t with
dy
dy dx
=
.
dt
dx dt
We will now derive a version of the chain rule for functions of two variables.
Assume that z = f (x, y) is a function of x and y, and suppose that x and y are in turn functions
of a single variable t, say
x = x(t), y = y(t)
The composition z = f (x(t), y(t)) then expresses z as a function of the single variable t. Thus, we
dz
∂z ∂z
can ask for the derivative
and we can inquire about its relationship to the derivatives
,
,
dt
∂x ∂y
dx
dy
, and . Letting ∆x, ∆y, and ∆z denote the changes in x, y, and z, respectively, that correspond
dt
dt
to a change of ∆t in t, we have
dz
∆z dx
∆x
dy
∆y
= lim
,
= lim
and
= lim
∆t→0 ∆t
dt ∆t→0 ∆t dt
dt ∆t→0 ∆t
It follows that
∆z ≈
∂z
∂z
∆x +
∆y,
∂x
∂y
(63)
where the partial derivatives are evaluated at (x(t), y(t)). Dividing both sides of (63) by ∆t yields
∆z
∂z ∆x ∂z ∆y
≈
+
,
∆t
∂x ∆t
∂y ∆t
(64)
Similarly, we can produce the analog of (64) for functions of three variables as follows: assume
that w = f (x, y, z) is a function of x, y, and z, and suppose that x, y, and z are functions of a
single variable t . As above we define ∆w, ∆x, ∆y, and ∆z to be the changes in w, x, y and z that
correspond to a change of ∆t in t. Then
∆w
∂w ∆x ∂w ∆y ∂w ∆z
≈
+
+
,
∆t
∂x ∆t
∂y ∆t
∂z ∆t
(65)
Taking the limit as ∆t → 0 of both sides of (64) and (65) suggests the following results.
Theorem 13.1 (Chain Rules for Derivatives) If x = x(t) and y = y(t) are differentiable at t, and
if z = f (x, y) is differentiable at the point (x, y) = (x(t), y(t)), then z = f (x(t), y(t)) is differentiable at
t and
dz
∂z dx ∂z dy
=
+
,
dt
∂x dt
∂y dt
(66)
where the ordinary derivatives are evaluated at t and the partial derivatives are evaluated at (x, y).
Proof: Let ∆x, ∆y, and ∆z denote the changes in x, y, and z, respectively, that correspond to a
change of ∆t in t. Then
dx
∆x
= lim
,
∆t→0 ∆t
dt
dy
∆y
= lim
dt ∆t→0 ∆t
M
dz
∆z
= lim
,
dt ∆t→0 ∆t
Since f (x, y) is differentiable at (x(t), y(t)), we have
p
f (x + ∆x, y + ∆y) = f (x, y) + fx (x, y)∆x + fy (x, y)∆y + (∆x, ∆y) (∆x)2 + (∆y)2 ,
which can be written as
p
∂z
∂z
∆x +
∆y + (∆x, ∆y) (∆x)2 + (∆y)2
∂x
∂y
SP
∆z =
(67)
where the partial derivatives are evaluated at (x(t), y(t)) and where (∆x, ∆y) satisfies (∆x, ∆y) →
0 as (∆x, ∆y) → (0, 0) and (0, 0) = 0. Dividing both sides of (67) by ∆t yields
p
∆z
∂z ∆x ∂z ∆y (∆x, ∆y) (∆x)2 + (∆y)2
=
+
+
(68)
∆t
∂x ∆t
∂y ∆t
∆t
Since
p
(∆x)2 + (∆y)2
lim
=
∆t→0
∆t
s
2
2
(∆y)
(∆x)
+
=
2
∆t→0
(∆t)
(∆t)2
s 2
2
dx
dy
=
+
,
dt
dt
lim
s
(∆x)2
(∆y)2
+
lim
∆t→0 (∆t)2
∆t→0 (∆t)2
lim
we have
lim
∆t→0
(∆x, ∆y)
p
(∆x)2 + (∆y)2
∆t
p
|(∆x, ∆y)| (∆x)2 + (∆y)2
= lim
∆t→0
|∆t|
p
(∆x)2 + (∆y)2
= lim |(∆x, ∆y)| lim
∆t→0
∆t→0
|∆t|
s 2
2
dx
dy
= 0·
+
= 0.
dt
dt
M
Figure 11
Therefore
p
(∆x, ∆y) (∆x)2 + (∆y)2
lim
= 0.
∆t→0
∆t
Taking the limit as ∆t → 0 of both sides of (68) then yields the equation
SP
dz
∂z dx ∂z dy
=
+
.
dt
∂x dt
∂y dt
Similarly, one can have
Theorem 13.2 If each of the functions x = x(t), y = y(t), and z = z(t) is differentiable at t, and if w =
f (x, y, z) is differentiable at the point (x, y, z) = (x(t), y(t), z(t)), then the function w = f (x(t), y(t), z(t))
is differentiable at t and
∂w dx ∂w dy ∂w dz
dw
=
+
+
,
dt
∂x dt
∂y dt
∂z dt
where the ordinary derivatives are evaluated at t and the partial derivatives are evaluated at
(x, y, z).
Example 13.3 Suppose that
z = x2 y, x = t2 , y = t3
dz
Use the chain rule to find
, and check the result by expressing z as a function of t and differendt
tiating directly.
Solution: By Chain Rule,
dz
∂z dx ∂z dy
=
+
= (2xy)(2t) + (x2 )(3t2 )
dt
∂x dt
∂y dt
5
= (2t )(2t) + (t4 )(3t2 ) = 7t6 .
Alternatively, we can express z directly as a function of t,
z = x2 y = (t2 )2 (t3 ) = t7
and then differentiate to obtain
dz
= 7t6 . However, this procedure may not always be convenient.
dt
p
Example 13.4 Suppose that w = x2 + y 2 + z 2 , x = cos θ, y = sin θ and z = tan θ. Use the chain rule
π
du
when θ = .
to find
dθ
4
Solution: Using the chain rule, we have
Chain Rule for Partial Derivatives
M
dw
∂z dx ∂z dy ∂w dz
=
+
+
dt
∂x dθ ∂y dθ
∂z dθ
1 2
1
1
=
(x + y 2 + z 2 )−1/2 (2x)(− sin θ) + (x2 + y 2 + z 2 )−1/2 (2y)(cos θ) + (x2 + y 2 + z 2 )−1/2 (2z)(sec2 θ).
2
2
2
π
When θ = , we have
4
π
1
π
1
π
x = cos = √ , y = sin = √ , z = tan = 1
4
4
4
2
2
1
dw
1
π
Substituting x = √ , y = √ , z = 1, θ = in the formula for
yields
4
dθ
2
2
√
√
√
dw
1
1
1
1
1
1
1
1
√ ( 2) − √
√ ( 2) √
√ (2) (2) = 2.
=
+
+
dt θ= π
2
2
2
2
2
2
2
2
4
To understand this topic, please watch the video https://youtu.be/Z1zrcRLky9Q.
SP
In above section, the variables x and y are each functions of a single variable t. We now consider
the case where x and y are each functions of two variables. Let z = f (x, y) and suppose that x
and y are functions of u and v, say
x = x(u, v), y = y(u, v)
The composition z = f (x(u, v), y(u, v)) expresses z as a function of the two variables u and v. Thus,
∂z
∂z
we can ask for the partial derivatives
and
; and we can inquire about the relationship
∂u
∂u
∂z ∂z ∂x ∂x ∂y
∂y
between these derivatives and the derivatives
,
,
,
,
, and
.
∂x ∂y ∂u ∂v ∂u
∂v
Similarly, if w = f (x, y, z) and x, y, and z are each functions of u and v, then the composition
w = f (x(u, v), y(u, v), z(u, v)) expresses w as a function of u and v. Thus we can also ask for the
∂w
∂w
and
; and we can investigate the relationship between these derivatives, the
derivatives
∂u
∂v
∂w ∂w
∂w
partial derivatives
,
, and
, and the partial derivatives of x, y, and z with respect to u
∂x ∂y
∂z
and v.
Theorem 13.5 Let z = f (x, y) be a differentiable at (x, y). Also, x = x(u, v) and y = y(u, v) are also
differentiable function. Then prove that z is differentiable at (u, v) and the differential of z as a
function of (u, v) is same as function of x and y.
Proof: Here x = x(u, v) and y = y(u, v) both are differentiable at any point on (u, v). Thus, the
differential of (x, y) is given by
dx =
∂x
∂x
∂y
∂y
du +
dv and dy =
du +
dv
∂u
∂v
∂u
∂v
Since, x = x(u, v) is differentiable at (u, v), we have
∆x = xu ∆u + xv ∆v + ∆u1 (∆u, ∆v) + ∆v2 (∆u, ∆v),
(69)
where 1 (∆u, ∆v) → 0 and 2 (∆u, ∆v) → 0 as (∆u, ∆v) → (0, 0).
Similarly, as y = y(u, v) is differentiable at (u, v), we have
∆y = yu ∆u + yv ∆v + ∆u3 (∆u, ∆v) + ∆v4 (∆u, ∆v),
(70)
where 3 (∆u, ∆v) → 0 and 4 (∆u, ∆v) → 0 as (∆u, ∆v) → (0, 0).
Also, given that z = f (x, y) is differentiable at (x, y), we have
∆z = zx ∆x + zy ∆y + ∆x5 (∆x, ∆y) + ∆y6 (∆x, ∆y),
(71)
where 5 (∆x, ∆y) → 0 and 6 (∆x, ∆y) → 0 as (∆x, ∆y) → (0, 0).
Using equations (69) and (70) in equation (71), we obtain
M
∆z = zx (xu ∆u + xv ∆v + ∆u1 (∆u, ∆v) + ∆v2 (∆u, ∆v))
+zy (yu ∆u + yv ∆v + ∆u3 (∆u, ∆v) + ∆v4 (∆u, ∆v))
+5 (∆x, ∆y) (xu ∆u + xv ∆v + ∆u1 (∆u, ∆v) + ∆v2 (∆u, ∆v))
+6 (∆x, ∆y) (yu ∆u + yv ∆v + ∆u3 (∆u, ∆v) + ∆v4 (∆u, ∆v)) ,
= zx (xu ∆u + xv ∆v) + zy (yu ∆u + yv ∆v)
+zx ∆u1 (∆u, ∆v) + zx ∆v2 (∆u, ∆v)
+zy ∆u3 (∆u, ∆v) + zy ∆v4 (∆u, ∆v)
+5 (∆x, ∆y) (xu ∆u + xv ∆v + ∆u1 (∆u, ∆v) + ∆v2 (∆u, ∆v))
+6 (∆x, ∆y) (yu ∆u + yv ∆v + ∆u3 (∆u, ∆v) + ∆v4 (∆u, ∆v)) ,
= zx (xu ∆u + xv ∆v) + zy (yu ∆u + yv ∆v) + ∆uφ1 (∆u, ∆v) + ∆vφ2 (∆u, ∆v),
where
and
SP
φ1 (∆u, ∆v) = zx 1 (∆u, ∆v) + zx ∆v2 (∆u, ∆v)
+5 (∆x, ∆y) (xu ∆u + xv ∆v + ∆u1 (∆u, ∆v) + ∆v2 (∆u, ∆v))
φ2 (∆u, ∆v) = zy ∆u3 (∆u, ∆v) + zy ∆v4 (∆u, ∆v)
+6 (∆x, ∆y) (yu ∆u + yv ∆v + ∆u3 (∆u, ∆v) + ∆v4 (∆u, ∆v)) .
We note that, φ1 (∆u, ∆v) → (0, 0) and φ2 (∆u, ∆v) → (0, 0) as (∆u, ∆v) → (0, 0). Hence,
dz = zx dx + zy dy.
This complete the proof of the theorem.
Theorem 13.6 (Chain Rules for Partial Derivatives) If x = x(u, v) and y = y(u, v) have firstorder partial derivatives at the point (u, v), and if z = f (x, y) is differentiable at the point (x, y) =
(x(u, v), y(u, v)), then z = f (x(u, v), y(u, v)) has first order partial derivatives at the point (u, v) given
by
∂z
∂z ∂x ∂z ∂y
∂z
∂z ∂x ∂z ∂y
=
+
and
=
+
∂u
∂x ∂u ∂y ∂u
∂v
∂x ∂v ∂y ∂v
Proof: Given that z is differentiable function with respect to x and y. Also, given that x = x(u, v)
and y = y(u, v) are differentiable functions of u and v. Therefore, z is differentiable function with
respect to u and v.
Hence, we can write differential of z with respect to u and v as
∂z
∂z
du +
dv.
(72)
∂u
∂v
Now, x = x(u, v) is differentiable function with respect to u and v, we can write differential in
terms of u and v as
∂x
∂x
dx =
du +
dv.
(73)
∂u
∂v
dz =
M
Figure 12
SP
Similarly, Now, y = y(u, v) is differentiable function with respect to u and v, we can write differential in terms of u and v as
∂y
∂y
du +
dv.
(74)
dy =
∂u
∂v
Also, z = f (x, y) is differentiable function with respect to x and y. We can write differential of z
in terms of x and y as
∂z
∂z
dz =
dx +
dy.
(75)
∂x
∂y
Now, using (73) and (74) in (75), we obtain
∂z ∂x
∂x
∂z ∂y
∂y
dz =
du +
dv +
du +
dv
∂x ∂u
∂v
∂y ∂u
∂v
∂z ∂x ∂z ∂y
∂z ∂x ∂z ∂y
=
+
du +
+
dv.
(76)
∂x ∂u ∂y ∂u
∂x ∂v ∂y ∂v
Comparing equations (72) and (76), we have
∂z
∂z ∂x ∂z ∂y
∂z
∂z ∂x ∂z ∂y
=
+
and
=
+
.
∂u
∂x ∂u ∂y ∂u
∂v
∂x ∂v ∂y ∂v
This complete the proof of the theorem.
Remark 13.7 Figure 12 show tree diagrams for the formulas in Theorem 13.6 As illustrated in
∂z
figure 12, the formula for
can be obtained by tracing all paths through the tree that start with z
∂u
∂z
and end with u, and the formula for
can be obtained by tracing all paths through the tree that
∂v
start with z and end with v.
Example 13.8 Given that z = exy , where x = 2u + v and y =
u
∂z
∂z
. Find
and
using chain rule.
v
∂u
∂v
Solution: Here,
∂z ∂x ∂z ∂y
1
∂z
xy
xy
=
+
= (ye )(2) + (xe )
∂u
∂x ∂u ∂y ∂u
v
!
(2u+v) u
h
i
x xy
2u 2u + v
v
= 2y +
e =
+
e
v
v
v
!
(2u+v) u
4u
v
+1 e
=
v
and
∂z ∂x ∂z ∂y
∂z
=
+
∂v
∂x ∂v ∂y ∂v
=
=
SP
=
xy
M
=
u
(ye )(1) + (xe ) − 2
v
h
u i xy
y−x 2 e
v
!
(2u+v) u
u (2u + v)u
v
−
e
v
v2
u!
2u2 (2u+v) v
.
− 2 e
v
xy
Figure 13
Example 13.9 Suppose that
w = exyz ,
x = 3u + v,
y = 3u − v,
z = u2 v.
∂w
∂w
and
.
∂u
∂v
Solution: From the tree diagram and corresponding formulas in Figure 13, we obtain
Use appropriate forms of the chain rule to find
∂w
∂w ∂x ∂w ∂y ∂w ∂z
=
+
+
∂u
∂x ∂u
∂y ∂u
∂z ∂u
= yzexyz (3) + xzexyz (3) + xyexyz (2uv) = exyz (yz + xz + 2xyuv)
and
∂w
∂w ∂x ∂w ∂y ∂w ∂z
=
+
+
∂v
∂x ∂v
∂y ∂v
∂z ∂v
xyz
xyz
= yze (1) + xze (−1) + xyexyz (u2 ) = exyz yz − xz + xyu2 .
Example 13.10 Suppose that w = x2 + y 2 − z 2 and
x = ρ sin φcosθ, y = ρ sin φ sin θ, z = ρ cos φ
∂w
∂w
and
.
∂ρ
∂θ
SP
M
Use appropriate forms of the chain rule to find
Figure 14
Solution: From the tree diagram and corresponding formulas in Figure 14, we obtain
∂w ∂x ∂w ∂y ∂w ∂z
+
+
∂x ∂ρ
∂y ∂ρ
∂z ∂ρ
(2x)(sin φ cos θ) + (2y)(sin φ sin θ) + (−2z)(cos φ)
2ρ sin2 φ cos2 θ + 2ρ sin2 φ sin2 θ − 2ρ cos2 φ
2ρ sin2 φ cos2 θ + sin2 θ − 2ρ cos2 φ
= 2ρ sin2 φ − cos2 φ
= −2ρ cos 2φ
∂w
=
∂ρ
=
=
=
and
∂w
∂w ∂x ∂w ∂y ∂w ∂z
=
+
+
∂θ
∂x ∂θ
∂y ∂θ
∂z ∂θ
= (2x)(−ρ sin φ sin θ) + (2y)(ρ sin φ sin θ)
= −2ρ2 sin2 φ sin θ cos θ + 2ρ2 sin2 φ sin θ cos θ = 0.
Example 13.11 Suppose that
w = xy + yz, y = sin x, z = ex
dw
Use an appropriate form of the chain rule to find
.
dx
Figure 15
Solution: From the tree diagram and corresponding formulas in Figure 15, we obtain
dw
∂w dx ∂w dy ∂w dz
=
+
+
dx
∂x dx
∂y dx
∂z dx
x
= sin x + (x + e ) cos x + ex sin x.
This result can also be obtained by first expressing w explicitly in terms of x as
M
w = x sin x + ex sin x
and then differentiating with respect to x; however, such direct substitution is not always possible. To understand the following examples, please watch the video https://youtu.be/QprggLEH_aY.
Solution: Here
SP
Example 13.12 If z = f (x, y), where x = r cos θ and y = r sin θ, prove that
2 2 2
2
∂z
∂z
∂z
1 ∂z
+ 2
=
+
.
∂r
r
∂θ
∂x
∂y
∂z ∂x ∂z ∂y
∂z
=
+
∂r
∂x ∂r ∂y ∂r
∂z
∂z
=
(cos θ) +
(sin θ)
∂x
∂y
(77)
and
∂z
∂z ∂x ∂z ∂y
=
+
∂θ
∂x ∂θ ∂y ∂θ
∂z
∂z
=
(−r sin θ) +
(r cos θ).
∂x
∂y
Using (77) and (78), we obtain
2
2
2
2
1 ∂z
∂z
1 ∂z
∂z
∂z
∂z
+ 2
=
(cos θ) +
(sin θ) + 2
(−r sin θ) +
(r cos θ)
∂r
r
∂θ
∂x
∂y
r
∂x
∂y
2 2
∂z
∂z
∂z
∂z
=
cos θ +
sin θ +
cos θ −
sin θ
∂x
∂y
∂y
∂x
2 2
∂z
∂z
=
+
∂x
∂y
(78)
Example 13.13 If z = eax+by f (ax − by). Prove that bzx + azy = 2abz.
Solution: Assume that u = ax + by and v = ax − by. Then
z = eu f (v).
We note that,
∂z
∂z ∂u ∂z ∂v
∂z
∂z
=
+
=a
+a
∂x
∂u ∂x ∂v ∂x
∂u
∂v
(79)
∂z
∂z ∂u ∂z ∂v
∂z
∂z
=
+
=b
−b .
∂y
∂u ∂y ∂v ∂y
∂u
∂v
(80)
and
Now, using (79) and (80), we get
bzx + azy
∂z ∂z
∂z
∂z
= ab
+
+ ab
−
∂u ∂v
∂u ∂v
∂z
= ab 2
= 2abzu = 2abz.
∂u
M
Example 13.14 If u = f (r), where r2 = x2 + y 2 + z 2 . Then prove that
2 2 2 2
∂u
∂u
du
∂u
+
+
=
.
(i)
∂x
∂y
∂z
dr
∂ 2u ∂ 2u ∂ 2u
2
+ 2 + 2 = f 00 (r) + f 0 (r).
2
∂x
∂y
∂z
r
SP
(ii)
Solution: Here, r2 = x2 + y 2 + z 2 , we have
2r
Therefore,
We note that
∂r
∂r
∂r
= 2x and 2r
= 2y and 2r
= 2z.
∂x
∂y
∂z
∂r
x
∂r
y
∂r
z
=
and
=
and
= .
∂x
r
∂y
r
∂z
r
∂u
du dr
x du
=
=
∂x
∂r ∂x
r dr
(81)
and
∂u
du ∂r
y du
∂u
du ∂r
z du
=
=
and
=
=
.
∂y
dr ∂y
r dr
∂z
dr ∂y
r dr
Now, using (81) and (82), we have
2 2 2
2 2 2
∂u
∂u
∂u
x du
y du
z du
+
+
=
+
+
∂x
∂y
∂z
r dr
r dr
r dr
2 2
2
2
2
du
du
x +y +z
=
=
.
2
dr
r
dr
(82)
To prove (ii), taking partial derivative of (81) and (82), we obtain
∂ x 0 ∂ 2u
∂ x du
=
f (r)
=
∂x2
∂x r dr
∂x r
∂
∂r
r
(f 0 (r)x) − f 0 (r)x
∂x
∂x
=
2
r
x
∂
0
0
r x (f (r)) + f (r) − f 0 (r)x
∂x
r
=
2
r
i
h
x2
x
r xf 00 (r) + f 0 (r) − f 0 (r)
r
r
=
r2
rx2 f 00 (r) + r2 f 0 (r) − x2 f 0 (r)
=
r3
(83)
Similarly , we have
and
M
ry 2 f 00 (r) + r2 f 0 (r) − y 2 f 0 (r)
∂ 2u
=
∂y 2
r3
∂ 2u
rz 2 f 00 (r) + r2 f 0 (r) − z 2 f 0 (r)
=
∂z 2
r3
(84)
(85)
Adding equations (83), (84) and (85), we have
Hence proved.
SP
∂ 2u ∂ 2u ∂ 2u
r(x2 + y 2 + z 2 )f 00 (r) + 3r2 f 0 (r) − (x2 + y 2 + z 2 )f 0 (r)
+
+
=
∂x2 ∂y 2 ∂z 2
r3
2
r3 f 00 (r) + 3r2 f 0 (r) − r2 f 0 (r)
= f 00 (r) + f 0 (r).
=
3
r
r
Example 13.15 If u = (x2 + y 2 + z 2 )m/2 , then prove that
m−2
∂ 2u ∂ 2u ∂ 2u
+ 2 + 2 = m(m + 1)u m .
2
∂x
∂y
∂z
Solution: Suppose that x2 + y 2 + z 2 = r2 . Therefore,
u = (r2 )m/2 = rm .
Taking partial derivative with respect to x, we have
∂u
du ∂r
2x
xmrm−1
=
= mrm−1 p
=p
= xmrm−2 .
2
2
2
2
2
2
∂x
dr ∂x
2 x +y +z
x +y +z
Again taking partial derivative with respect to x, we have
∂ 2u
m−3 ∂r
m−2
= m x(m − 2)r
+r
∂x2
∂x
h
i
x
= m x(m − 2)rm−3 + rm−2
r
h
i
2
m−4 x
m−2
= m x (m − 2)r
+r
= mrm−4 x2 (m − 2) + r2
r
(86)
(87)
Similarly, we obtain
2
∂ 2u
2
m−4
y
(m
−
2)
+
r
=
mr
∂y 2
(88)
2
∂ 2u
m−4
2
=
mr
z
(m
−
2)
+
r
∂z 2
(89)
and
Adding equations (87), (88) and (89), we have
2
∂ 2u ∂ 2u ∂ 2u
2
2
2
m−4
(x
+
y
+
z
)(m
−
2)
+
3r
+
+
=
mr
∂x2 ∂y 2 ∂z 2
m−2
= mrm−4 mr2 − 2r2 + 3r2 = m(m + 1)rm−2 = m(m + 1)u m .
Exercise 13.16
y2
z2
x2
+
+
=. Then prove that
a2 + u b2 + u c2 + u
2 2 2
∂u
∂u
∂u
∂u
∂u
∂u
+
+
=2 x
+y
+z
∂x
∂y
∂z
∂x
∂y
∂z
3. If u = log(x2 + y 2 + z 2 ) then prove that
∂ 2u
∂ 2u
∂ 2u
=y
=z
,
∂z∂y
∂z∂x
∂x∂y
∂ 2u ∂ 2u ∂ 2u
2
(b)
+ 2+ 2 = 2
.
2
∂x
∂y
∂z
x + y2 + z2
SP
(a) x
M
2. If
1. If u = log(x3 + y 3 − x2 y − x2 y) then prove that uxx + 2uxy + uyy = −
4. If z = f (x, y) and x + y = 2eu cos v and x − y = 2ieu cos v. Then prove that
∂ 2v
∂ 2z ∂ 2z
+
=
uxy
.
∂x2 ∂y 2
∂x∂y
5. If z is function of x and y. Prove that if x = eu + e−v , y = eu + e−v , then prove that
∂z
∂z
∂z
∂z
−
=x
−y .
∂u ∂v
∂x
∂y
6. If H = f (y − z, z − x, x − y), prove that
∂H ∂H ∂H
+
+
= 0.
∂x
∂y
∂z
7. If z = f (x, y) and u = lx + my, v = ly − mx, then prove that
2
∂ 2z ∂ 2z
∂ z ∂ 2z
2
2
,
+
= (l + m )
+
∂x2 ∂y 2
∂u2 ∂v 2
where l, m are constants.
8. If u = x2 − y 2 , v = 2xy, f (x, y) = g(u, v), then prove that
2
∂ 2f
∂ g ∂ 2g
∂ 2f
2
2
+ 2 = 4(x + y )
+
.
∂x2
∂y
∂u2 ∂v 2
y−x z−x
∂u
∂u
∂u
If u = u
,
, show that x2
+ y2
+ z2
= 0.
xy
xz
∂x
∂y
∂z
4
.
(x + y)2
14
IMPLICIT DIFFERENTIATION
To understand the following examples, please watch the video https://youtu.be/b6QUtEYlm3E.
Consider the special case where z = f (x, y) is a function of x and y and y is a differentiable
function of x. Equation (66) then becomes
dz
∂f dx ∂f dy
∂f
∂f dy
=
+
=
+
.
dx
∂x dx ∂y dx
∂x ∂y dx
(90)
This result can be used to find derivatives of functions that are defined implicitly. For example,
suppose that the equation
f (x, y) = c
(91)
dy
defines y implicitly as a differentiable function of x and we are interested in finding
. Differdx
entiating both sides of (91) with respect to x and applying (90) yields
∂f
∂f dy
+
= 0.
∂x ∂y dx
Thus, if
∂f
6= 0, we obtain
∂y
SP
M
∂f
fx
dy
= − ∂x = −
∂f
dx
fy
∂y
∂f
∂f
and
, we get
Differentiating again with respect to x, regarding
∂x
∂y
2
2
∂ f
∂ 2 f dy ∂f
∂ f
∂ 2 f dy ∂f
+
−
+
d2 y
∂x2 ∂y∂x dx ∂y
∂x∂y ∂y 2 dx ∂y
= −
2
dx2
∂f
∂x
fx
fx
fxx + fxy −
fy − fxy + fyy −
fx
fy
fy
= −
[fy ]2
fxx (fy )2 − 2fx fy fxy + fyy (fx )2
=
.
[fy ]3
(92)
(93)
(94)
In summary, we have the following result.
Theorem 14.1 If the equation f (x, y) = c defines y implicitly as a differentiable function of x, and
∂f
6= 0, then
∂y
∂f
dy
fx
= − ∂x = − .
(95)
∂f
dx
fy
∂y
and
d2 y
fxx (fy )2 − 2fx fy fxy + fyy (fx )2
=
(96)
dx2
[fy ]3
Example 14.2 Given that
x3 + y 2 x − 3 = 0.
Find
dy
and check the result using implicit differentiation.
dx
Solution: Using (95), with f (x, y) = x3 + y 2 x − 3,
∂f
dy
3x2 + y 2
= − ∂x = −
.
∂f
dx
2yx
∂y
Alternatively, differentiating implicitly yields
3x2 + y 2 + x(2y)
dy
dy
3x2 + y 2
− 0 = 0 or
=−
,
dx
dx
2yx
which agrees with the result obtained by (95).
SP
M
The chain rule also applies to implicit partial differentiation. Consider the case where w =
f (x, y, z) is a function of x, y, and z and z is a differentiable function of x and y. It follows from
Theorem 13.6 that
∂f ∂z
∂f
+
= 0.
(97)
∂x ∂z ∂x
∂f
If
6= 0, then
∂z
∂f
∂z
= − ∂x .
(98)
∂f
∂x
∂z
∂z
A similar result holds for
.
∂y
Theorem 14.3 If the equation f (x, y, z) = c defines z implicitly as a differentiable function of x and
∂f
6= 0, then
y, and if
∂z
∂f
∂f
∂z
∂z
∂y
= − ∂x
and
=−
.
∂f
∂f
∂x
∂y
∂z
∂z
∂z
∂z
2 1 2
2
2
2
Example 14.4 Consider the sphere x + y + z = 1. Find
and
at the point
, ,
.
∂x
∂y
3 3 3
Solution: By Theorem 14.3 with f (x, y, z) = x2 + y 2 + z 2 , we have
∂f
∂f
∂z
2x
x
∂z
2y
y
∂y
= − ∂x = −
=−
and
=−
=− =− .
∂f
∂f
∂x
2z
z
∂y
2z
z
∂z
∂z
2 1 2
∂z
∂z
1
At the point
, ,
, evaluating these derivatives gives
= −1 and
=− .
3 3 3
∂x
∂y
2
Example 14.5 If ax2 + 2hxy + by 2 + 2gx + 2f y + c = 0. Find
d2 y
.
dx2
Solution: Here f (x, y, z) = ax2 + 2hxy + by 2 + 2gx + 2f y + c. Then
fx = 2(ax + hy + g), fy = 2(hx + by + f )
and
fxx = 2a, fyy = 2b, fxy = 2h.
d2 y
fxx (fy )2 − 2fx fy fxy + fyy (fx )2
=
−
dx2
[fy ]3
=
=
=
=
2a · 4(ax + hy + g)2 − 2 · 2h · 2(ax + hy + g) · 2(hx + by + f ) + 2b · 4(ax + hy + g)2
8(hx + by + f )3
a(ax + hy + g)2 − h(ax + hy + g)(hx + by + f ) + b(hx + by + f )2
(hx + by + f )3
(h2 − ab)(ax2 + 2hxy + by 2 + 2gx + 2f y) + 2f gh − bg 2 − hf 2
(hx + by + f )3
(h2 − ab)(−c) + 2f gh − bg 2 − hf 2
(hx + by + f )3
abc − ch2 + 2f gh − bg 2 − hf 2
.
(hx + by + f )3
Exercise 14.6 Find
M
=
d2 y
in the following cases:dx2
2. x4 + y 4 = 4a2 xy
3. x5 + y 5 = 5a3 xy
5. xy = y x
6. (tan x)y + y cot x = a
d2 y 2a2 x2
+
= 0.
dx2
y5
SP
1. x3 + y 3 = 3axy
4. x5 + y 5 = 5a3 x2
Exercise 14.7 Prove that if y 3 − 3ax2 + x3 = 0, then
p
√
d2 y
a
2
2
.
Exercise 14.8 If x 1 − y + y 1 − x = a, show that 2 =
dx
(1 − x2 )3/2
Exercise 14.9 If f (x, y) = 0 and g(y, z) = 0, show that
∂f ∂φ
∂f ∂φ dz
=
∂y ∂z dx
∂x ∂y
Exercise 14.10 If A, B, C are the angles of the triangle such that sin2 A + sin2 B + sin2 C = constant.
dA
tan C − tan B
Prove that
=
.
dB
tan A − tan C
15
Which variable is to be treated as constant
To understand the this topic, please watch the video https://youtu.be/F6NPj703ZaA.
Consider the relation
x = r cos θ
and
y = r sin θ
(99)
∂r
To find
, we need a relation between r and x. Such a relation will contain one more variable
∂x
θ or y, for we can eliminate only one variable out of four from the relation (99). Thus the two
possible relations are
r = x sec θ
(100)
and
r 2 = x2 + y 2 .
(101)
∂r
either from (100) by treating θ as constant or from (101) by regarding y as
∂x
∂r
constant. And there is no reason to suppose that the two values of
so found, are equal. To
∂x
avoid confusion
as to which variable is regarded constant, we introduce the following:
∂r
Notation:
means the partial derivative of r with respect to x keeping θ constant in a
∂x θ
relation expressing
r as
a function of x and θ.
∂r
Thus form (100),
= secθ.
∂x θ
When no indication
is given regarding
the
to be kept
then according
variable
constant, to
con∂
∂
∂
∂
∂
∂
vention
always means
and
means
. Similarly,
mean
and
∂x
∂x y
∂y
∂y x
∂r
∂r θ
∂
∂
means
.
∂θ
∂θ r
∂u
∂x
1
∂v
∂y
2
2
Example 15.1 If x = au + bv, y = au − bv, prove that
= =
.
∂x y ∂u v 2
∂y x ∂v u
Now, we can find
M
Solution: Partial differentiating x2 = au + bv with respect to u keeping v constant, we get
∂x
a
=
∂u v 2x
SP
and partial differentiating y 2 = au − bv with respect to v keeping u constant, we get
b
∂y
=− .
∂v u
2y
∂u
∂v
We note that, To compute
and
, we need u and v as function of x and y.
∂x y
∂y x
By adding x2 = au + bv and y 2 = au − bv, we have
2au = x2 + y 2
and by subtracting x2 = au + bv and y 2 = au − bv, we have
2bv = x2 − y 2 .
Partial differentiating u with respect to x keeping y constant, we get
2x
x
∂u
=
=
∂x y
2a
a
and partial differentiating v with respect to y keeping x constant, we get
∂v
−2y
−y
=
=
∂y x
2b
b
Hence,
∂u
∂x
y
∂x
∂u
v
1
= =
2
∂v
∂y
x
∂y
∂v
.
u
Exercise 15.2 If u = lx + my, v = mx − ly, show that
∂u
∂x
l2
∂y
∂v
l2 + m2
= 2
,
=
.
∂x y ∂u v l + m2 ∂v x ∂y u
l2
16
Jacobian
To understand the this topic, please watch the video https://youtu.be/_vcDnZYSf4s.
If u and v are functions of two independent variables x and y, then the determinant
u, v
∂(u, v)
or J
.
called the Jacobian of u, v with respect to x, y and written as
∂(x, y)
x, y
ux uy uz
∂(u, v, w)
Similarly the Jacobian of u, v, w with respect to x, y, z is
= vx vy vz .
∂(x, y, z)
wx wy wz
Example 16.1 In polar coordinate, x = r cos θ, y = r sin θ, show that
ux uy
vx vy
is
∂(x, y)
=r
∂(r, θ)
Solution: We have xr = cos θ, xθ = −r sin θ and yr = sin θ, yθ = r cos θ.
∂(x, y)
=
∂(r, θ)
xr x θ
yr yθ
=
cos θ −r sin θ
sin θ r cos θ
= r(cos2 θ + sin2 θ) = r.
Solution: We have
M
Example 16.2 In cylindrical coordinate, x = ρcosφ, y = ρ sin φ, z = z, show that
∂(x, y, z)
= ρ.
∂(ρ, φ, z)
xρ = cos φ, xφ = −ρ sin φ, xz = 0;
yρ = sin φ, yφ = ρ cos φ, yz = 0;
Now,
xρ xφ x z
yρ yφ yz
zρ zφ zz
=
zρ = 0, zφ = 0, zz = 1.
cos φ −ρ sin φ 0
sin φ ρ cos θ 0
0
0
1
SP
∂(x, y, z)
=
∂(ρ, φ, z)
and
= ρ(cos2 φ + sin2 φ) = ρ.
Exercise 16.3 In spherical polar coordinates x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, show that
∂(x, y, z)
= r2 sin θ.
∂(r, θ, φ)
x2 x3
x3 x1
x1 x2
Exercise 16.4 If y1 =
, y2 =
and y3 =
, show that the Jacobian of y1 , y2 , y3 with respect
x1
x2
x3
to x1 , x2 , x3 is 4.
17
Euler’s Theorem for Homogeneous functions
To understand the this topic, please watch the video https://youtu.be/mpYgQMcn5pA.
Definition 17.1 An expression of the from
a0 xn + a1 xn−1 y + a2 xn−2 y 2 + . . . + an−1 xy n−1 + an y n
in which every term is of the nth degree, is called a homogeneous function of degree n. This can be
written as
y
y 2
y n n
x a0 + a1
+ a2
+ . . . + an
.
x
x
x
y
Thus any function f (x, y) which can be expression in the form xn φ
, is called a homogeneous
x
function of degree n in x and y.
Example 17.2 The function x3 cos
y
x
is a homogeneous function of degree 3 in x and y.
Definition 17.3 A function f (x, y) is called homogeneous of degree n in x and y if
f (tx, ty) = tn f (x, y)
for all x, y, t.
Example 17.4 The function f (x, y) =
x3 + y 3
is homogeneous function of degree 2 in variable x
xy
and y as
f (tx, ty) =
(tx)3 + (ty)3
x3 + y 3
= t2
= t2 f (x, y),
(tx)(ty)
xy
for all
x, y, t.
Definition 17.5 A function f (x1 , x2 , . . . , xn ) is called homogeneous of degree n in variables
x1 , x2 , . . . , xn if
f (tx1 , tx2 , . . . txn ) = tn f (x1 , x2 , . . . , xn )
for all x1 , x2 , . . . , xn , t.
xn
x2 x3
, ,...,
x1 x1
x1
for some function g.
M
It also can be written as f (x1 , x2 , . . . , xn ) =
xn1 g
Example 17.6 The function f (x, y, z) = x2 +y 2 +z 2 is homogeneous function of degree 2 in variables
x, y and z.
To understand the this topic, please watch the video https://youtu.be/7cNAfDZ2Pw0.
SP
Theorem 17.7 (Euler’s Theorem) Let z = f (x, y) be a real valued function of two variables x and
y defined on D. prove that f is homogeneous function of degree m if and only if
xfx + yfy = mf.
Solution: Suppose that z = f (x, y) is homogeneous function of degree m. Then for every (x, y) ∈ D,
there exists t ∈ D ⊂ R such that
f (tx, ty) = tm f (x, y).
Putting X = tx and Y = ty, we have
f (X, Y ) = tm f (x, y).
Differentiating equation (102) with respect to t, we get
∂f dX
∂f dY
+
∂X dt
∂Y dt
= mtm−1 f (x, y),
which can be written as
x
∂f
∂f
+y
∂X
∂Y
= mtm−1 f (x, y).
Multiplying both side by t, we obtain
tx
∂f
∂f
+ ty
∂X
∂Y
= mtm f (x, y).
The above statement is ture for all t ∈ R. In particular, taking t = 1 we have
x
∂f
∂f
+y
∂x
∂y
= mf (x, y).
(102)
Another way to prove this:
Suppose that z = f (x, y) is homogeneous function of degree m. Therefore
y f (x, y) = xm φ
) , where φ is some function.
x
Taking partial derivative with respect to x and y, we get
y y
y
m 0 y
m−1
m−2
0 y
m−1
+x φ
− 2 = mx
φ
−x
yφ
fx = mx
φ
x
x
x
x
x
and
fy = xm φ0
y 1
y
= xm−1 φ0
.
x x
x
Now
xfx + yfy = mxm φ
y
x
− xm−1 yφ0
y
x
+ yxm−1 φ0
y
x
= mxm φ
y
x
.
Hence,
xfx + yfy = mf.
Conversely, suppose that xfx (x, y) + yfy (x, y) = mf (x, y) for fixed (x, y) ∈ D. We shall show that f
is homogeneous function of degree m. It is sufficient to prove that
M
f (tx, ty) = tm f (x, y)
for t.
Consider u = f (tx, ty), v = f (x, y), X = tx and Y = ty. Therefore we say that u is function of X and
Y , that means u = f (X, Y ).
Applying chain rule for derivative, we have
SP
du
∂f dX
∂f dY
=
+
dt
∂X dt
∂Y dt
∂f
∂f
= x
+y
∂X
∂Y
Multiplying t both side, we achieve
t
∂f
∂f
du
= tx
+ ty
dt
∂X
∂Y
∂f
∂f
= X
+Y
= mf (X, Y ).
∂X
∂Y
Hence, we have
t
du
= mu.
dt
Taking integration both side, we have
Z
du
=m
u
Z
dt
.
t
Therefore, we have
ln u = m ln t + ln c.
Finally for every t, we have
u = ctm .
Taking t = 1, we obtain u = v = c. Therefore, u = tm v. Hence,
f (tx, ty) = tm f (x, y).
Hence the result.
Example 17.8 If f (x, y) =
x2019 + y 2019
. Find xfx + yfy .
x2020 + y 2020
Solution: We note that, for any t,
2019
+ y 2019
(tx)2019 + (ty)2019
−1 x
f (tx, ty) =
=t
.
(tx)2020 + (ty)2020
x2020 + y 2020
Therefore, f is homogeneous function of degree −1 in variables x and y. Hence, by Euler’s theorem, we have
x2019 + y 2019
.
xfx + yfy = mf = − 2020
x
+ y 2020
To understand the this topic, please watch the video https://youtu.be/OaJOmFxVhEA.
Theorem 17.9 (Euler’s Theorem for n variables ) Let f : D ⊂ Rn → R be a real valued function
of variables x1 , x2 , . . . , xn defined on D. prove that f is homogeneous function of degree m if and
only if
x1 fx1 + x2 fx2 + · · · + xn fxn = mf (x1 , x2 , . . . , xn ).
Proof: Suppose that z = f (x1 , x2 , . . . , xn ) is a homogeneous function of degree m having n variables
x1 , x2 , . . . xn . Therefore
M
f (tx1 , tx2 , . . . , txn ) = tm f (x1 , x2 , . . . , xn ), for every t ∈ R
By taking X1 = tx1 , X2 = tx2 , . . . , Xn = txn , we have
f (X1 , X2 , . . . , Xn ) = tm f (x1 , x2 , . . . , xm ).
Taking derivative both side with respect to t, we get
which gives
x1
SP
∂f dX1
∂f dX2
∂f dXn
+
+ ··· +
= mtm−1 f (x1 , x2 , . . . , xn ),
∂X1 dt
∂X2 dt
∂Xn dt
∂f
∂f
∂f
+ x2
+ · · · + xn
= mtm−1 f (x1 , x2 , . . . , xn ),
∂X1
∂X2
∂Xn
Multiply both side by t, we obtain
tx1
∂f
∂f
∂f
+ tx2
+ · · · + txn
= mtm f (x1 , x2 , . . . , xn ),
∂X1
∂X2
∂Xn
X1
∂f
∂f
∂f
+ X2
+ · · · + Xn
= mtm f (x1 , x2 , . . . , xn ),
∂X1
∂X2
∂Xn
Hence, we have
In particular t = 1, we get
x1
∂f
∂f
∂f
+ x2
+ · · · + xn
= mf (x1 , x2 , . . . , xn ).
∂x1
∂x2
∂xn
Conversely, suppose that
x1
∂f
∂f
∂f
+ x2
+ · · · + xn
= mf (x1 , x2 , . . . , xn ).
∂x1
∂x2
∂xn
It sufficient to prove that for every t ∈ R,
f (tx1 , tx2 , . . . , txn ) = tm f (x1 , x2 , . . . , xn )
Consider
u(x1 , x2 , . . . , xn ) = f (tx1 , tx2 , . . . , txn )
and
v(x1 , x2 , . . . , xn ) = f (x1 , x2 , . . . , xn ).
Also, assume that X1 = tx1 , X2 = tx2 , . . ., Xn = txn . Therefore,
u(x1 , x2 , . . . xn ) = f (X1 , X2 , . . . , Xn ).
Now,
du
∂f dX1
∂f dX2
∂f dXn
=
+
+ ··· +
dt
∂X1 dt
∂X2 dt
∂Xn dt
∂f
∂f
∂f
= x1
+ x2
+ · · · + xn
.
∂X1
∂X2
∂Xn
Multiply t both side, we get
∂f
∂f
∂f
du
= tx1
+ tx2
+ · · · + txn
dt
∂X1
∂X2
∂Xn
∂f
∂f
∂f
= X1
+ X2
+ · · · + Xn
∂X1
∂X2
∂Xn
= mf (X1 , X2 , . . . , Xn ).
Now, we have
t
Taking integration both side, we have
Therefore, we have
du
= mu
dt
du
=m
u
Z
dt
.
t
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Z
M
t
ln u = m ln t + ln c.
Finally for every t, we have
u = ctm .
Taking t = 1, we obtain u = v = c. Therefore, u = tm v. Hence,
f (tx1 , tx2 , . . . , txn ) = tm f (x1 , x2 , . . . , xn )
Hence the result.
Example 17.10 If f (x, y, x) = log
xy + yz + xz
. Find xfx + yfy + zfz .
x2 + y 2 + z 2
Solution: We note that, for any t,
(tx)(ty) + (ty)(tz) + (tx)(tz)
xy + yz + xz
f (tx, ty, tz) = log
log
.
(tx)2 + (ty)2 + (tz)2
x2 + y 2 + z 2
Therefore, f is homogeneous function of degree 0 in variables x,y and z. Hence, by Euler’s theorem, we have
xfx + yfy + zfz = mf = 0.
To understand the this topic, please watch the video https://youtu.be/OWhWPAZJ2hQ.
Corollary 17.11 Let z = f (x, y) be a homogeneous function of degree m having real valued in
variables x and y. Also, fx and fy are differentiable. Then prove that
x2
∂f
∂f
∂f
+ 2xy 2 + y 2 2 = m(m − 1)f.
2
∂x
∂x
∂y
Solution: Given that fx and fy are differentiable function. Therefore by Young’s theorem, we
have fxy = fyx .
Also, given that f is homogeneous function of degree m. By Euler’s Theorem, we have
xfx + yfy = mf
(103)
Partially differentiate equation (103) with respect to x, we have
xfxx + fx + fyx = mfx .
(104)
Partially differentiate equation (103) with respect to y, we have
fy + xfxy + yfyy = mfy .
(105)
Multiplying equation (104) by x, (105) by y and using fxy = fyx , we have
Hence,
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x2 fxx + xfx + xyfyx + yfy + xyfxy + y 2 fyy = mxfx + myfy .
x2 fxx + 2xyfyx + y 2 fyy = m2 f − mf = m(m − 1)f.
Hence the result.
x2 + y 2
then find the values of
x3 y
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Example 17.12 Let f (x, y) =
(a) xfx + yfy ;
(b) x2 fxx + 2xyfyx + y 2 fyy .
Solution: Here
f (tx, ty) =
2
2
(tx)2 + (ty)2
−2 x + y
=
t
= t−2 f (x, y).
(tx)3 (ty)
x3 y
Therefore, f is homogeneous function of degree −2 in variables x and y. By Euler’s Theorem
xfx + yfy = −2f = −2
x2 + y 2
x3 y
and
x2 fxx + 2xyfyx + y 2 fyy = m(m − 1)f = −2(−2 − 1)f = 6
x2 + y 2
.
x3 y
To understand the this topic, please watch the video https://youtu.be/xmYalhNLaCw.
Corollary 17.13 Let z = f (x, y) be homogeneous function of degree m. Also, if z = F (u) is one-one
onto function Then prove that
(a) xux + yuy = m
F (u)
F 0 (u)
(b) x2 uxx + 2xyuxy + y 2 uyy = G(u) (G0 (u) − 1) , where G(u) = m
F (u)
.
F 0 (u)
Solution: Given that f is homogeneous function of degree m. Therefore, by Euler’s theorem, we
have
xfx + yfy = mf.
Given that z = F (u) is one-one and onto, we can write as u = F −1 (z). Taking partial derivative of
u with respect to x and y, we have
∂u
du ∂z
=
∂x
dz ∂x
OR
∂u
du ∂z
=
.
∂y
dz ∂y
Now,
∂u
∂u
du ∂z
∂z
x
+y
=
x
+y
.
∂x
∂y
dz
∂x
∂y
(106)
Since, z is homogeneous function of degree m, we have
x
∂z
∂z
+y
= mz.
∂x
∂y
Using (106), we can write
∂u
∂u
du
+y
=
mz.
∂x
∂y
dz
(107)
M
x
Taking partial derivative of z = F (u) with respect to z, we have
1 = F 0 (u)
Hence,
du
.
dz
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1
du
= 0 ,
dz
F (u)
provided F 0 (u) 6= 0. Now, using (107), we have
x
Take G(u) = m
∂u
∂u
F (u)
+y
= m 0 .
∂x
∂y
F (u)
F (u)
. So, we have
F 0 (u)
xux + yuy = G(u).
(108)
Partially differentiate with respect to x, we obtain
xuxx + ux + yuyx = G0 (u)
∂u
.
∂x
Multiplying both side by x, we get
x2 uxx + xux + xyuyx = xG0 (u)
∂u
.
∂x
(109)
Partially differentiate with respect to y, we obtain
xuxy + yuyy + uy = G0 (u)
∂u
.
∂y
Multiplying both side by x, we get
xyuyx + y 2 uyy + yuy = yG0 (u)
∂u
.
∂y
(110)
Adding equations (109) and (110), we obtain
2
2
x uxx + xux + xyuyx + xyuxy + y uyy + yuy
∂u
∂u
= G (u) x
+y
.
∂x
∂y
0
Using Yonge’s theorem and equation (108), we obtain
∂u
∂u
x uxx + 2xyuxy + y uyy + G(u) = G (u) x
+y
∂x
∂y
0
= mG (u) [xux + yuy ] = mG0 (u)G(u).
2
0
2
Now, we have
x2 uxx + 2xyuxy + y 2 uyy = G(u) (G0 (u) − 1) .
Hence the theorem.
To understand the this topic, please watch the video https://youtu.be/eC4B5m_UVC0.
Example 17.14 If z = sin−1 (x2 + y 2 ) Then find the value of the following
1. xzz + yzy
2. x2 zxx + 2xyzxy + y 2 zyy .
xzx + yzy = 2
Finally, we get
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Solution: Suppose that u(x, y) = x2 + y 2 . The function f is homogeneous of degree 2. Then we
have
u = sin(z),
which is one and onto function of one variables. Now, using above result
sin(z)
= 2 tan z = G(z)
cos z
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x2 zxx + 2xyzxy + y 2 zyy = mG(z)(G0 (z) − 1) = 2 tan z(2sec2 z − 1).
Example 17.15 If z = tan
1. xzz + yzy
x2020 + y 2020
. Find the value of
x+y
2. x2 zxx + 2xyzxy + y 2 zyy .
x2020 + y 2020
. We note that f is homogeneous function of degree 2019
x+y
into variables x and y. Take z = tan(f ). Hence
Solution: Here f (x, y) =
f (z) = tan−1 (z).
Now, we get
xzx + yzy = m
Say
f (u)
tan−1 z
=
2019
= 2019(1 + z 2 ) tan−1 z.
1
f 0 (u)
1 + z2
G(u) = 2019(1 + z 2 ) tan−1 z.
Therefore,
0
−1
G (u) = 2019 2z tan
1
z + (1 + z )
= 2019 1 + 2z tan−1 z .
2
1+z
2
Hence, we have
x2 zxx + 2xyzxy + y 2 zyy = (2019)2 (1 + z 2 ) tan−1 z 1 + 2z tan−1 z .
Example 17.16 Verify the Euler’s Theorem for f (x, y) =
Solution: Here f (x, y) =
x2 + y 2
.
x+y
x2 + y 2
. We note that
x+y
f (tx, ty) =
x2 + y 2
t2 x2 + t2 y 2
=t
= tf (x, y)
tx + ty
x+y
Hence, the function f is homogeneous function of degree 1.
We shall show that
xfx + yfy = f.
Partial differentiating f with respect to x and y, we obtain
fx =
2x(x + y) − (x2 + y 2 )
2x2 + 2xy − x2 − y 2
x2 + 2xy − y 2
=
=
(x + y)2
(x + y)2
(x + y)2
and
fy =
xfx + yfy =
x3 + 2yx2 − xy 2 + y 3 + 2y 2 x − x2 y
(x + y)(x2 + y 2 )
x2 + y 2
=
=
= f (x, y).
(x + y)2
(x + y)2
x+y
Hence, the Euler’s theorem is verified.
Exercise 17.17
1. Verify the Euler’s theorem f (x, y) = xn ln
y
x
.
x2 + y 2
∂f
∂f
, then prove that x
+y .
xy
∂x
∂y
SP
2. If f (x, y) =
M
Now,
y 2 + 2xy − x2
2y(x + y) − (x2 + y 2 )
=
.
(x + y)2
(x + y)2
3. Find the value of xux + yuy and x2 uxx + 2xyuxy + y 2 uyy for following function
2
3
3
3
x + y2
x + y3
x
+
y
−1
(b) u = log
.
(c) u = cosec
.
(a) u = tan−1
x+y
x10 + y 10
x+y
s
x+y
x3 + y 3
x20 + y 20
(f) u = sin−1 √
√ .
(e) log u =
.
.
(d) u =
x+ y
3x + 4y
x+y
4. If u = sin−1
5. If u =
18
x + 2y + 3z
, find the value of xux + yuy + zuz .
x8 + y 8 + z 8
x3 + y 3 + z 3
xy + yz + zx
+ log 2
find the value of xux + yuy + zuz .
3
3
3
xy z
x + y2 + z2
MAXIMA AND MINIMA OF FUNCTIONS OF TWO VARIABLES
To understand the this topic, please watch the video https://youtu.be/8NT5RZJjmDs.
In your previous study, readers have learned how to find maximum and minimum values of a
function of one variable. In this section we will develop similar techniques for functions of two
variables.
Extrema
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If we imagine the graph of a function f of two variables to be a mountain range (Figure 16),
then the mountaintops, which are the high points in their immediate vicinity, are called relative
maxima of f , and the valley bottoms, which are the low points in their immediate vicinity, are
called relative minima of f .
Just as a geologist might be interested in finding the highest mountain and deepest valley in
an entire mountain range, so a mathematician might be interested in finding the largest and
smallest values of f (x, y) over the entire domain of f . These are called the absolute maximum
and absolute minimum values of f . The following definitions make these informal ideas precise.
Figure 16
Definition 18.1 A function f of two variables is said to have a relative maximum/local maximum at a point (x0 , y0 ) if there is a disk centered at (x0 , y0 ) such that f (x0 , y0 ) ≥ f (x, y) for all points
(x, y) that lie inside the disk, and f is said to have an absolute maximum/global maximum at
(x0 , y0 ) if f (x0 , y0 ) ≥ f (x, y) for all points (x, y) in the domain of f .
Definition 18.2 A function f of two variables is said to have a relative minimum/local minimum at a point (x0 , y0 ) if there is a disk centered at (x0 , y0 ) such that f (x0 , y0 ) ≤ f (x, y) for all points
(x, y) that lie inside the disk, and f is said to have an absolute minimum/global minimum at
(x0 , y0 ) if f (x0 , y0 ) ≤ f (x, y) for all points (x, y) in the domain of f .
If f has a relative maximum or a relative minimum at (x0 , y0 ), then we say that f has a relative
extremum at (x0 , y0 ), and if f has an absolute maximum or absolute minimum at (x0 , y0 ), then
we say that f has an absolute extremum at (x0 , y0 ).
Definition 18.3 A set of points in 2-space is called bounded if the entire set can be contained
within some rectangle, and is called unbounded if there is no rectangle that contains all the points
of the set. Similarly, a set of points in 3-space is bounded if the entire set can be contained within
some box, and is unbounded otherwise.
The extreme-value theorem
The following theorem, which we state without proof, is the result for functions of two variables.
Theorem 18.4 (Extreme-Value Theorem) If f (x, y) is continuous on a closed and bounded set
R, then f has both an absolute maximum and an absolute minimum on R.
Example 18.5 The square region R whose points satisfy the inequalities −4 ≤ x ≤ 4 and −5 ≤ y ≤ 5
is a closed and bounded set in the xy-plane. The function f whose graph is shown in Figure ?? is
continuous on R; thus, it is guaranteed to have an absolute maximum and minimum on R by the
last theorem.
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Remark 18.6 If any of the conditions in the Extreme-Value Theorem fail to hold, then there is
no guarantee that an absolute maximum or absolute minimum exists on the region R. Thus, a
discontinuous function on a closed and bounded set need not have any absolute extrema, and
a continuous function on a set that is not closed and bounded also need not have any absolute
extrema.
Finding Relative Extrema
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Recall that if a function g of one variable has a relative extremum at a point x0 where g is
differentiable, then g(x0 ) = 0. To obtain the analog of this result for functions of two variables,
suppose that f (x, y) has a relative maximum at a point (x0 , y0 ) and that the partial derivatives of
f exist at (x0 , y0 ). It seems plausible geometrically that the traces of the surface z = f (x, y) on the
planes x = x0 and y = y0 have horizontal tangent lines at (x0 , y0 ) (Figure ??), so
fx (x0 , y0 ) = 0 and fy (x0 , y0 ) = 0.
The same conclusion holds if f has a relative minimum at (x0 , y0 ), all of which suggests the
following result.
Theorem 18.7 If f has a relative extremum at a point (x0 , y0 ), and if the first order partial derivatives of f exist at this point, then
fx (x0 , y0 ) = 0 and fy (x0 , y0 ) = 0
Proof: Given that f have relative extreme at point (x0 , y0 ). Without loss of generality assume that
f has relative maximum at point (x0 , y0 ).
Therefore, there exists an disk N centered at (x0 , y0 ) such that f (x0 , y0 ) ≥ f (x, y) for all points
(x, y) ∈ N .
Also, given that fx is exists. Therefore,
f (x0 + h, y0 ) − f (x0 , y0 )
.
h→0
h
fx (x0 , y0 ) = lim
For some h > 0, (x0 + h, y0 ) ∈ N . Therefore,
f (x0 + h, y0 ) ≤ f (x0 , y0 ).
f (x0 + h, y0 ) − f (x0 , y0 )
≤ 0.
h
Taking limit h → 0 both side, we have
f (x0 + h, y0 ) − f (x0 , y0 )
≤ 0.
h→0
h
lim
Therefore
fx (x0 , y0 ) ≤ 0.
(111)
For some h < 0, (x0 + h, y0 ) ∈ N . Therefore,
f (x0 + h, y0 ) ≤ f (x0 , y0 ).
f (x0 + h, y0 ) − f (x0 , y0 )
≥ 0.
h
Taking limit h → 0 both side, we have
f (x0 + h, y0 ) − f (x0 , y0 )
≥ 0.
h→0
h
lim
Therefore
fx (x0 , y0 ) ≥ 0.
(112)
Using inequalities (111) and (112), we have
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fx (x0 , y0 ) = 0.
Similarly, one can have fy (x0 , y0 ) = 0. This completes the proof of the theorem.
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Recall that the critical points of a function f of one variable are those values of x in the domain
of f at which f (x) = 0 or f is not differentiable. The following definition is the analog for functions
of two variables.
Definition 18.8 A point (x0 , y0 ) in the domain of a function f (x, y) is called a critical point of the
function if fx (x0 , y0 ) = 0 and fy (x0 , y0 ) = 0 or if one or both partial derivatives do not exist at (x0 , y0 ).
It follows from this definition and Theorem 18.7 that relative extrema occur at critical points,
just as for a function of one variable. However, recall that for a function of one variable a relative extremum need not occur at every critical point. For example, the function might have an
inflection point with a horizontal tangent line at the critical point. Similarly, a function of two
variables need not have a relative extremum at every critical point. For example,
Example 18.9 consider the function
f (x, y) = y 2 − x2 .
This function, whose graph is the hyperbolic paraboloid shown in Figure 17, has a critical point at
(0, 0), since
fx (x, y) = −2x and fy (x, y) = 2y.
from which it follows that
fx (0, 0) = 0 and fy (0, 0) = 0.
However, the function f has neither a relative maximum nor a relative minimum at (0, 0). For obvious
reasons, the point (0, 0) is called a saddle point of f . In general, we will say that a surface z =
f (x, y) has a saddle point at (x0 , y0 ) if there are two distinct vertical planes through this point such
that the trace of the surface in one of the planes has a relative maximum at (x0 , y0 ) and the trace
in the other has a relative minimum at (x0 , y0 ).
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Figure 17: , The function f (x, y) = y 2 − x2 has neither a relative maximum nor a relative minimum at the critical
point (0, 0).
SP
p
(a) Here, f (x, y) = x2 + y 2 fx (0, 0) = (b) Here, f (x, y) = 1 − x2 − y 2 , fx (0, 0) = (c) Here, f (x, y) = x2 − y 2 , fx (0, 0) =
fy (0, 0) = 0 relative and absolute min fy (0, 0) = 0 relative and absolute max fy (0, 0) = 0 do not exist relative and abat (0, 0)
at (0, 0)
solute min at (0, 0)
Figure 18
Example 18.10 The three functions graphed in Figure 18 all have critical points at (0, 0). For the
paraboloids, the partial derivatives at the origin are zero. You can check this algebraically by evaluating the partial derivatives at (0, 0), but you can see it geometrically by observing that the traces
in the xz-plane and yz-plane have horizontal tangent lines at (0, 0). For the cone neither partial
derivative exists at the origin because the traces in the xz-plane and the yz-plane have corners
there. The paraboloid in part 18a and the cone in part 18c have a relative minimum and absolute
minimum at the origin, and the paraboloid in part 18b has a relative maximum and an absolute
maximum at the origin.
Sufficient Condition for extreme values
For functions of one variable the second derivative test was used to determine the behavior of a
function at a critical point. The following theorem, which is usually proved in advanced calculus,
is the analog of that theorem for functions of two variables.
Theorem 18.11 (The Second Partials Test) Let f be a function of two variables with continuous
second-order partial derivatives in some disk centered at a critical point (x0 , y0 ), and let
2
D = fxx (x0 , y0 )fyy (x0 , y0 ) − fxy
(x0 , y0 )
(a) If D > 0 and fxx (x0 , y0 ) > 0, then f has a relative minimum at (x0 , y0 ).
(b) If D > 0 and fxx (x0 , y0 ) < 0, then f has a relative maximum at (x0 , y0 ).
(c) If D < 0, then f has a saddle point at (x0 , y0 ).
(d) If D = 0, then no conclusion can be drawn.
To understand the following example, please watch the video https://youtu.be/cmyA_p3Rjp4.
Example 18.12 Locate all relative extrema and saddle points of
f (x, y) = 3x2 − 2xy + y 2 − 8y.
Solution: Since fx = 6x − 2y and fy = −2x + 2y − 8, the critical points of f satisfy the equations
6x − 2y = 0 and − 2x + 2y − 8 = 0.
Solving these equations for x and y, we have
x = 2, y = 6.
So, (2, 6) is the only critical point. We need the second order partial derivatives
fxx (x, y) = 6, fyy (x, y) = 2, fxy (x, y) = −2.
At the point (2, 6), we have
2
D = fxx (2, 6)fyy (2, 6) − fxy
(2, 6) = (6)(2) − (−2)2 = 8 > 0
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and
fxx (2, 6) > 0.
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so f has a relative minimum at (2, 6) by the second partials test. Figure 19 shows a graph of f in
the vicinity of the relative minimum.
Figure 19: The plot of the function 3x2 − 2xy + y 2 − 8y with minima at (2, 6) with red color.
To understand the following example, please watch the video https://youtu.be/5ZxLXAQbohY.
Example 18.13 Locate all relative extrema and saddle points of f (x, y) = 4xy − x4 − y 4 .
solution: Since
fx (x, y) = 4y − 4x3 and fy (x, y) = 4x − 4y 3
the critical points of f have coordinates satisfying the equations
4y − 4x3 = 0 ⇒ y = x3
and
(113)
4x − 4y 3 = 0 ⇒ x = y 3 .
(114)
3 3
9
8
Substituting the top equation in the bottom yields x = (x ) or, equivalently, x −x = 0 or x(x −1) =
0, which has solutions x = 0, x = 1, x = −1. Substituting these values in the equation (113), we
obtain the corresponding y-values y = 0, y = 1, y = −1. Thus, the critical points of f are (0, 0),
(1, 1), and (−1, −1). Also, we have
fxx (x, y) = −12x2 ,
fyy (x, y) = −12y 2 ,
fxy (x, y) = 4
which yield the following table:
At the points (1, 1) and (−1, −1), we have D > 0 and fxx < 0, so relative maxima occur at these
Critical Point (x0 , y0 )
(0,0)
(1,1)
(−1,−1)
fxx (x0 , y0 )
0
−12
−12
fyy (x0 , y0 )
0
−12
−12
fxy (x0 , y0 )
4
4
4
D = fxx fyy − fxy
-16
128
128
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critical points. At (0, 0) there is a saddle point since D < 0. The surface and a contour plot are
shown in Figure 20.
Figure 20: The graph of the function 4xy − x4 − y 4 with maxima at (1,1) and (−1, −1).
To understand the following example, please watch the video https://youtu.be/K_S_2NXjHqU.
Example 18.14 Find the extreme values of the function f (x, y) = x3 + y 3 − 3x − 12y + 5.
Solution: Here,
fx (x, y) = 3x2 − 3
and
fy (x, y) = 3y 2 − 12
fx (x, y) = 0
and
fy (x, y) = 0,
and
3y 2 − 12 = 0.
Consider
we have
3x2 − 3 = 0
Solving these two equations, we have
x = ±1
y = ±2.
Hence, there are four stationary points namely (1, 2), (1, −2), (−1, 2), (−1, −2). Also, we have
fxx (x, y) = 6x fyy (x, y) = 6y,
fxy = 0.
which yield the following table:
At points (1, 2), the function has a relative minimum as D > 0 and fxx (1, 2) = 6 > 0 and at the
Critical Point (x0 , y0 )
(1, 2)
(1, −2)
(−1, 2)
(−1, −2)
fxx (x0 , y0 )
6
6
−6
−6
fyy (x0 , y0 )
12
−12
12
−12
fxy (x0 , y0 )
0
0
0
0
D = fxx fyy − fxy
72
−72
−72
72
point (−1, −2), the function has a relative maximum as D > 0 and fxx = −6 < 0. One should
note that, the points (1, −2) and (−1, 2) are saddle points for the function f , that means at these
points either maximum nor minimum. The extreme values of the function are f (1, 2) = −13 and
f (−1, −2) = 23.
To understand the following example, please watch the video https://youtu.be/DdjEIEsj1Vo.
Example 18.15 Find the extreme value of the function f (x, y) = xy(a − x − y).
Solution: Here,
Consider
fx (x, y) = y(a − x − y) − xy = 0
Solving these two equations, we get
fy (x, y) = x(a − x − y) + xy(−1).
M
fx (x, y) = y(a − x − y) + xy(−1)
fy (x, y) = x(a − x − y) + xy(−1) = 0
y = 0 or 2x + y = 0 and x = 0 or x + 2y = a.
SP
We have following combinations,
1. If x = 0 and y = 0 then the point (0, 0) is the stationary point.
2. If x = 0 and x + 2y = 0 then the point (0, a) is the stationary point.
3. If a − 2y − x = 0 and y = 0, then the point (a, 0) is the stationary point.
a a
4. If a − 2y − x = 0 and a − 2x − y = 0, the point
, .
3 3
Also, we have
fx x(x, y) = −2y fyy (x, y) = −2x, fxy (x, y) = a − 2y − x
which yield the following table:
Critical Point (x0 , y0 )
(0, 0)
(a, 0)
(0, a)
a a
,
3 3
fxx (x0 , y0 )
0
0
−2a
2a
−
3
fyy (x0 , y0 )
0
−2a
0
2a
−
3
fxy (x0 , y0 )
a
−a
−a
a
−
3
D = fxx fyy − fxy
−a2
−a2
−a2
a2
3
a a
−a
At points
, , the function has a relative minimum as D > 0 and fxx (0, a) = −
> 0.
3 3
3
One should note that, the points (0, 0), (0, a) and (a, 0) are saddle points for the function f , that
means at these points either maximum nor minimum. The extreme values of the function are
a a a3
f
,
= .
3 3
27
To understand the following example, please watch the video https://youtu.be/WxbtCoWzb14.
Example 18.16 Find the three positive integer, whose product is 64 and sum is minimum.
Solution: Suppose that three positive integers are x, y and z.
Give that product of x, y and z is 64, therefore,
xyz = 64,
which can be written as
z=
64
xy.
Now, f (x, y) = x + y + z, we have
64
.
xy
Taking partial derivative with respect to x and y, we have
y
64
fx (x, y) = 1 + 64 − 2 2 = 1 − 2
xy
xy
f (x, y) = x + y +
x
64
fy (x, y) = 1 + 64 − 2 2 = 1 − 2
xy
xy
Consider, fx (x, y) = 0 and fy (x, y) = 0, we have
M
and
x2 y = 64 and xy 2 = 64
Solving these equation, we have
x=0
or
y=0
or
x = y.
SP
We note that x = 0 and y = 0 are not possible values, as they are positive integer. Therefore, by
x = y and x2 y = 64, we note that, (4, 4) is the only possible stationary point.
Now,
128
128
64
fxx = 3
fyy = 3 fxy = 2 2
xy
xy
xy
Now, at the point (4, 4), we obtain
1
fxx (4, 4) = ,
2
Now,
1
fyy (4, 4) = ,
2
fxy (4, 4) =
1
4
1 1
3
1
− =
> 0; fxx (4, 4) = > 0.
4 6
16
2
Therefore, f has minimum at (4, 4). Therefore, the required positive integer are (4, 4, 4).
D=
Exercise 18.17
1. Find the positive integer, whose sum is 24 and product is maximum.
π
2. Find the extreme value of f (x, y) = sin2 x + sin2 y, where x + y = a 6= (2k + 1) , k ∈ Z.
2
3. If f (x, y, z) = x2 + y 2 + z 2 with condition x + y + z = 3, find the extreme values.
4. Determine the dimensions of a rectangular box, open at the top, having a volume of 32 f t3 , and
requiring the least amount of material for its construction.
5. Examine the extreme values of the following function: f (x, y) = x4 + y 4 − 2x2 + 4xy − 2y 2 .
6. In a plane triangle, find the maximum value od cos A cos B cos C.
19
Lagrange’s method of undermined multiplies
To understand the this topic, please watch the video https://youtu.be/z--LL1Wuyb0
Sometimes it is required to find the stationary values of a function of several variables which are
not all independent but are connected by some given relations. Ordinarily, we try to convert the
given function to the one, having least number of independent variables with the help of given
relations. Then solve it by the above method. When such a procedure becomes impracticable,
Lagrange’s method* proves very convenient. Now we explain this method. Let
u = f (x, y, z)
(115)
be a function of three variables x, y, z which are connected by the relation.
φ(x, y, z) = 0.
(116)
For u to have stationary values, it is necessary that
∂u
= 0,
∂x
∂u
∂u
= 0,
= 0.
∂y
∂z
∂u
∂u
∂u
dx +
dy +
dz = du = 0.
∂x
∂y
∂z
M
(117)
Differentiating φ(x, y, z) = 0, we get
∂φ
∂φ
∂φ
dx +
dy +
dz = dφ = 0.
∂x
∂y
∂z
(118)
SP
Multiply (118) by a parameter λ and add to (117). Then
∂u
∂φ
∂φ
∂φ
∂u
∂u
+λ
+λ
+λ
dx +
dy +
dz = 0
∂x
∂x
∂y
∂y
∂z
∂z
This equation will be satisfied if
∂φ
∂u
+λ
=0
∂x
∂x
∂φ
∂u
+λ
∂y
∂y
=0
∂φ
∂u
+λ
∂z
∂z
=0
These three equations together with φ(x, y, z) = 0 will determine the values of x, y, z and λ for
which u is stationary.
Working Rules
1. Write F = f (x, y, z) + λφ(x, y, z)
2. Obtain the equations
∂F
∂F
∂F
= 0,
= 0,
= 0.
∂x
∂y
∂z
3. Solve the above equations together with φ(x, y, z) = 0.
The values of x, y, z so obtained will give the stationary value of f (x, y, z).
Remark 19.1 Although the Lagrange’s method is often very useful in application yet the drawback
is that we cannot determine the nature of the stationary point. This can sometimes, be determined
from physical considerations of the problem.
To understand the following example, please watch the video https://youtu.be/_ldiaXg3toA.
Example 19.2 A rectangular box open at the top is to have volume of 32 cubic ft Find the dimensions of the box requiring least material for its construction.
Solution: Let x, y and z ft. be the edges of the box and S be its surface. Then
S = xy + 2yz + 2zx
(119)
xyz = 32
(120)
∂F
= y + 2z + λyz = 0;
∂x
(121)
∂F
= x + 2z + λzx = 0;
∂y
(122)
and
Write F = xy + 2yz + 2zx + λ(xyz − 32) Then
∂F
= 2y + 2x + λxy = 0.
∂z
Multiplying (121) by x and (122) by y and subtracting, we get
or x = y.
M
2zx − 2zy = 0
(123)
SP
[The value z = 0 is neglected, as it will not satisfy xyz = 32)]
Again multiplying (122) by y and (123) by z and subtracting, we get y = 2z.
Hence the dimensions of the box are x = y = 2z = 4
Now, let us see what happens as z increases from a small value to a large one. When z is small, the
box is flat with a large base showing that S is large. As z increases, the base of the box decreases
rapidly and S also decreases. After a certain stage, S again starts increasing as z increases. Thus
S must be a minimum at some intermediate stage which is given by x = y = 2z = 4. Hence S is
minimum when x = y = 4 ft and z = 2 ft.
To understand the following example, please watch the video https://youtu.be/L5l_BbdRPOg.
Exercise 19.3 Given x + y + z = a, find the maximum valve of xm y n z p .
Solution: Let f (x, y, z) = xm y b z p and φ(x, y, z) = x + y + z − a.
Then
F (x, y, z) = f (x, y, z) + λφ(x, y, z)
= xm y n z p + λ(x + y + z − a).
For stationary values of F ,
∂F
= 0,
∂x
∂F
= 0,
∂y
∂F
= 0.
∂z
Therefore, we have
mxm−1 y n z p + λ = 0,
or
nxm y n−1 z p + λ = 0,
pxm y n z p−1 + λ = 0
−λ = mxn−1 y n z p = nxm y n−1 z p = pxm y n z p−1 .
Using x + y + z = a, we obtain
m
n
p
m+n+p
m+n+p
= = =
=
.
x
y
z
x+y+z
a
Therefore, the maximum value of f occurs, when
am
an
ap
x=
,y =
,z =
m+n+p
m+n+p
m+n+p
Hence, the maximum value of f (x, y, z) =
am+n+p mm nn pp
.
(m + n + p)m+n+p
To understand the following example, please watch the video https://youtu.be/6Fgpspatibs.
Example 19.4 Find the maximum and minimum distances of the point (3, 4, 12) from the sphere
x2 + y 2 + z 2 = 1.
Solution: Let P (x, y, z) be any point on the sphere and A(3, 4, 12) the given point so that
AP 2 = (x − 3)2 + (y − 4)2 + (2 − 12)2 = f (x, y, z), (say)
We have to find the maximum and minimum values of f (x, y, z) subject to the condition
φ(x, y, z) = x2 + y 2 + z 2 − 1 = 0
Let
Then
Therefore,
give
∂F
∂F
∂F
= 2(x − 3) + 2λx,
= 2(y − 4) + 2λy,
= 2(z − 12) + 2λz.
∂x
∂y
∂z
∂F
∂F
∂F
= 2(x − 3) + 2λx,
= 2(y − 4) + 2λy,
= 2(z − 12) + 2λz
∂x
∂y
∂z
SP
Then
M
F (x, y, z) = f (x, y, z) + λφ(x, y, z)
= (x − 3)2 + (y − 4)2 + (2 − 12)2 + λ(x2 + y 2 + z 2 − 4).
∂F
∂F
∂F
= 0,
= 0, and
=0
∂x
∂y
∂z
x − 3 + λx = 0, y − 4 + λy = 0, z − 12 + λz = 0.
which gives
x−3
y−4
z − 12
λ=−
=−
=
=±
x
y
z
p
√
(x − 3)2 + (y − 4)2 + (z − 12)2
f
p
=±
.
1
x2 + y 2 + z 2
Substituting for λ in (124), we get
x=
3
3
4
12
√ ,y =
√ ,z =
√ .
=
1+λ
1± f
1± f
1± f
Therefore,
x2 + y 2 + z 2 =
9 + 16 + 144
169
√ 2 =
√
(1 + f )
(1 ± f )2
Using x2 + y 2 + z 2 = 1, we have
1 ==
or
1±
Hence,
169
√
(1 ± f )2
p
f = ±13
p
f = 12, 14.
(124)
√
√
[We have left out the negative values of f because f = AP +ve ]
Hence, maximum AP = 14 and minimum AP = 12.
To understand the following example, please watch the video https://youtu.be/6Fgpspatibs.
Example 19.5 Show that the rectangular solid of maximum volume that can be inscribed in a
sphere is a cube.
Solution: Let 2x, 2y, 2z be the length, breadth and height of the rectangular solid so that its
volume
V = 8xyz
(125)
Let R be the radius of the sphere so that
x2 + y 2 + z 2 = R 2
Then
and
(126)
F (x, y, z) = 8xyz + A(x2 + y 2 + z 2 − R2 )
∂F
∂F
∂F
= 0,
= 0 and
= 0 give
∂x
∂y
∂z
or
M
8yz + 2xλ = 0, 8zx + 2yλ = 0, 8xy + 2zλ = 0
2x2 λ = −8xyz = 2y 2 λ = 2z 2 λ
Thus for a maximum volume x = y = z. That means, the rectangular solid is a cube.
SP
Exercise 19.6 1. A tent on a square base of side x, has its sides vertical of height y and the
top is a regular pyramid of height h. Find x and y in terms of h, if the canvas required for its
construction is to be minimum for the tent to have a given capacity.
2. Find the volume of the greatest rectangular parallelopiped that can be inscribed in the ellipsoid
x2 y 2 z 2
+ 2 + 2 = 1.
a2
b
c
3. Find the dimensions of the rectangular box, open at the top, of maximum capacity whose
surface is 432 sq. cm.
4. Find the minimum value of x2 + y 2 + z 2 subject to the condition ax + by + cz − d = 0, a2 + b2 + c2 6= 0,
d 6= 0.
5. Find the point within a triangle such that the sum of squares of its distances from the side of
the triangle is a minimum.
6. Prove that the extreme value of u = ax2 + by 2 + cz 2 subject to the conditions x2 + y 2 + z 2 =
l2
m2
u2
1, lx + my + nz = 0 is given by 2
+ 2
+ 2
= 0.
a −u b −u c −u
7. Show that of all triangle, having given perimeter the largest is an equilateral triangle.
8. Find the stationary values of f (x, y, z) = sin xsiny sin z, where x, y, z are angle of triangle. (that
is x + y + z = π).
9. Find the point on the surface z 2 = x2 + y 2 that are closed to points (1, 1, 0).
10. Divide ’a’ into two parts such that the product of ’n’ power of one number and ’m’ power of
other number is least.
20
Taylor’s Theorem of two variables
To understand this topic, please watch the video https://youtu.be/OQJukemDIlU.
Taylor’s theorem for one variables, we have learn in the previous studies. Here, we are establish
the Taylor’s series and Maclaurin’s Series for function of two variables, to do these we required
following lemma.
Lemma 20.1 Let z = f (x, y) be real valued function of two variables x and y defined on domain
D. Let (a, b) ∈ D. Let x = a + ht and y = b + kt, for all t ∈ R. Then prove that
i
∂
∂ iz
∂
+k
z, for all i ∈ N.
= h
∂ti
∂x
∂y
Proof: We shall prove this result using principle of mathematical induction. We note that z =
f (x, y), x = a + ht, y = b + kt, we have
Therefore,
dz
=
dt
M
dz
∂z dx ∂z dy
=
+
dt
∂x dt
∂y dt
∂z
∂z
= h
+k .
∂x
∂y
∂
∂
h
+k
∂x
∂y
z
SP
Hence, the result is true for i = 1. Now,
d dz
d2 z
d ∂z dx ∂z dy
=
+
=
dt2
dt dt
dt ∂x dt
∂y dt
d
∂z
∂z
=
h
+k
dt
∂x
∂y
∂
∂z
∂z ∂x
∂
∂z
∂z ∂y
=
h
+k
+
h
+k
∂x
∂x
∂y ∂t
∂y
∂x
∂y ∂t
2
2
2
2
∂ z
∂ z
∂ z
∂ z
= h2 2 + hk
+ hk
+ k2 2
∂x
∂y∂x
∂x∂y
∂y
2
∂
∂
+k
z.
=
h
∂x
∂y
Hence, the result is true for i = 2.
Assume that, the result is true for i = k, for k < n, k ∈ N. That means
k
dk z
∂
∂
= h
+k
z.
dtk
∂x
∂y
Now,
dk+1 z
d
=
k+1
dt
dt
=
=
=
=
=
dk z
dtk
k !
d
∂
∂
h
+k
z
dt
∂x
∂y
k !
k !
∂
∂
∂
dx
∂
∂
∂
dy
h
+k
z
+
h
+k
z
∂x
∂x
∂y
dt
∂y
∂x
∂y
dt
k !
k !
∂
∂
∂
∂
∂
∂
h
+k
z +k
h
+k
z
h
∂x
∂x
∂y
∂y
∂x
∂y
k ∂
∂
∂
∂
h
+k
z h
+k
z
∂x
∂y
∂x
∂y
k+1
∂
∂
h
+k
z.
∂x
∂y
The result is true for i = k + 1. Therefore, by principle of mathematical induction the result is
true for any i ∈ N.
where
SP
M
Theorem 20.2 (Taylor’s theorem) Let z = f (x, y) be function of two variables, whose all partial
derivatives exist in neighborhood N of (a, b) ∈ D. Also, all partial derivatives are continuous on N .
Suppose that (a + h, b + k) ∈ D with h > 0, k > 0. Then prove that there exist θ ∈ (0, 1) such that
2
∂
∂
1
∂
∂
f (a, b) + · · ·
f (a + h, b + k) = f (a, b) + h
+k
f (a, b) +
h
+k
∂x
∂y
2!
∂x
∂y
n−1
1
∂
∂
+
h
+k
f (a, b) + Rn ,
(n − 1)!
∂x
∂y
1
Rn =
n!
n
∂
∂
+k
h
f (a + θh, b + θk).
∂x
∂y
Proof: Define g : [0, 1] → R as
g(t) = f (a + ht, b + kt)
for every
t ∈ [0, 1].
Let x = a + ht and y = b + kt. Now, t = 0 then x = a and y = b. So,
g(0) = f (a, b) = f (x, y) = z.
Since, all partial derivatives of f exists and x and y are differentiable function of t. Therefore, g(t)
is also n times differentiable function on [0, 1]. Thus, the function g is satisfies all the condition
of Manchurian theorem on [0, 1].
Therefore, these exist θ ∈ (0, 1) such that
g(1) = g(0) + g 0 (0) +
1 00
1
1
g (0) + g 000 (0) + · · · +
g (n−1) (0) + Rn ,
2!
3!
(n − 1)!
1 (n)
g (θ).
n!
Here g(1) = f (a + h, b + h) and g(0) = f (a, b).
Also, using
i
∂ iz
∂
∂
= h
+k
z,
∂ti
∂x
∂y
where Rn =
we have
2
∂
∂
1
∂
∂
f (a + h, b + k) = f (a, b) + h
+k
f (a, b) +
h
+k
f (a, b) + · · ·
∂x
∂y
2!
∂x
∂y
n
1
∂
∂
+
h
+k
f (a, b) + Rn ,
(n − 1)!
∂x
∂y
where
1
Rn =
n!
n−1
∂
∂
h
+k
f (a + θh, b + θk).
∂x
∂y
which complete the proof of theorem.
Remark 20.3
1. The above Taylor’s series also can be written as
f (a + h, b + k) = f (a, b) + hfx (a, b) + kfy (a, b) +
1 2
h fxx (a, b) + 2hkfxy (a, b) + k 2 fyy (a, b) + . . .
2!
2. Putting a + h = x and b + k = y so that h = x − a , k = y − b, we get
M
f (x, y) = f (a, b) + [(x − a)fx (a, b) + (y − b)fy (a, b)]
1 +
(x − a)2 fxx (a, b) + 2(x − a)(y − b)fxy (a, b) + (y − b)2 fyy (a, b) + · · ·
2!
(127)
This is Taylor’s expansion of f (x, y) in powers of (x − a) and (y − b). It is used to expand f (x, y)
in the neighborhood of (a, b).
3. Putting a = 0, b = 0 in (127), we get
1 2
x fxx (0, 0) + 2xyfxy (0, 0) + y 2 fyy (0, 0) + · · ·
2!
SP
f (x, y) = f (0, 0) + [xfx (0, 0) + yfy (0, 0)] +
This is Maclaurin’s expansion of f (x, y).
Example 20.4 Expand ex log(1 + y) in powers of x and y up to terms of third degree.
Solution: Here Now Maclaurin’s expansion of f (x, y) gives
f (x, y) = ex log(1 + y)
fx (x, y) = ex log(1 + y)
1
fy (x, y) = ex
1+y
fxx (x, y) = ex log(1 + y)
1
fxy (x, y) = ex
1+y
fyy (x, y) = −ex (1 + y)−2
fxxx (x, y) = ex log(1 + y)
1
fxxy (x, y) = ex
1+y
fxyy (x, y) = −ex (1 + y)−2
fyyy (x, y) = −ex (1 + y)−2
f (0, 0) = 0
fx (0, 0) = 0
fy (0, 0) = 1
fxx (0, 0) = 0
fxy (0, 0) = 1
fyy (0, 0) = −1
fxxx (0, 0) = 0
fxxy (0, 0) = 1
fxyy (0, 0) = −1
fyyy (0, 0) = 2
f (x, y) = f (0, 0) + [xfx (0, 0) + yfy (0, 0)]
1 2
+
x fxx (0, 0) + 2xyfxy (0, 0) + y 2 fyy (0, 0)
2!
1 3
+
x fxxx (0, 0) + 3x2 yfxyy (0, 0) + 3xy 2 fxyy (0, 0) + y 3 fyyy (0, 0) + · · · .
3!
(128)
Therefore
ex log(1 + y) = 0 + [x(0) + y(1)]
1 2
+
x (0) + 2xyfxy (1) + y 2 (−1)
2!
1 3
+
x (0) + 3x2 y(1) + 3xy 2 (−1) + y 3 (2) + · · ·
3!
1
1
1
= y + xy − y 2 + (x2 y − xy 2 ) + y 3 + · · · .
2
2
3
To understand this topic, please watch the video https://youtu.be/T9lkQXwDja8.
Example 20.5 Expand x2 y + 3y − 2 in powers of (x − 1) and (y + 2) using Taylor’s Theorem.
Solution: Taylor’s expansion of f (x, y) in powers of (x − a) and (y − b) is given by
M
f (x, y) = f (a, b) + [(x − a)fx (a, b) + (y − b)fy (a, b)]
1 +
(x − a)2 fxx (a, b) + 2(x − a)(y − b)fxy (a, b) + (y − b)2 fyy (a, b)
2!
1 +
(x − a)3 fxxx (a, b) + 3(x − a)2 (y − b)fxxy (a, b)
3!
+3(x − a)(y − b)2 fxyy (a, b) + (y − b)3 fyyy (a, b) + · · · .
Hence, a = 1, b = −2 and
f (1, −2) = −10
fx (1, −2) = −4
fy (1, −2) = 4
fxx (1, −2) = −4
fxy (1, −2) = 2
fyy (1, −2) = 0
fxxx (1, −2) = 0
fxxy (1, −2) = 2
fxyy (1, −2) = 0
fyyy (1, −2) = 0.
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f (x, y) = x2 y + 3y − 2
fx (x, y) = 2xy
fy (x, y) = x2 + 3
fxx (x, y) = 2y
fxy (x, y) = 2x
fyy (x, y) = 0
fxxx (x, y) = 0
fxxy (x, y) = 2
fxyy (x, y) = 0
fyyy (x, y) = 0
(129)
All partial derivatives of higher order vanish.
Substituting these values in (129), we get
x2 y + 3y − 2 = −10 + [(x − 1)(−4) + (y + 2)4]
1
+ (x − 1)2 (−4) + 2(x − 1)(y + 2)(2) + (y + 2)2 (0)
2
1
+ (x − 1)3 (0) + 3(x − 1)2 (y + 2)(2) + 3(x − 1)2 (y + 2)(0) + (y + 2)3 (0) + · · · .
6
= −10 − 4(x − 1) + 4(y + 2) − 2(x − 1)2 + 2(x − 1)(y + 2) + (x − 1)2 (y + 2).
Example 20.6 Expand f (x, y) = tan−1
y
x
Hence compute f (1.1, 0.9) approximately.
Solution: Here a = 1, b = 1.
in powers of (x − 1) and (y − 1) upto third-degree terms.
y
f (x, y) = tan−1
x
−y
fx (x, y) = 2
x + y2
x
fy (x, y) = 2
x + y2
2xy
fxx (x, y) = 2
(x + y 2 )2
y 2 − x2
fxy (x, y) = 2
x + y2
−2xy
fyy (x, y) = 2
x + y2
2y 3 − 6x2 y
fxxx (x, y) =
(x2 + y 2 )3
2x3 − 6xy 2
fxxy (x, y) =
(x2 + y 2 )3
6x2 y − 2y 3
fxyy (x, y) =
(x2 + y 2 )3
6xy 2 − 2x3
fyyy (x, y) =
(x2 + y 2 )3
π
f (1, 1) = tan−1 (1) =
4
1
fx (1, 1) = −
2
1
fy (1, 1) =
2
1
fxx (1, 1) =
2
fxy (1, 1) = 0
1
2
1
fxxx (1, 1) = −
2
1
fxxy (1, 1) = −
2
1
fxyy (1, 1) =
2
1
fyyy (1, −2) = .
2
fyy (1, 1) = −
Taylor’s expansion of f (x, y) in powers of (x − 1) and (y − 1) is given by
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f (x, y) = f (1, 1) + [(x − 1)fx (1, 1) + (y − 1)fy (1, 1)]
1 +
(x − 1)2 fxx (1, 1) + 2(x − 1)(y − 1)fxy (1, 1) + (y − 1)2 fyy (1, 1)
2!
1 (x − 1)3 fxxx (1, 1) + 3(x − 1)2 (y − 1)fxxy (1, 1)
+
3!
+ 3(x − 1)(y − 1)2 fxyy (1, 1) + (y − 1)3 fyyy (1, 1) + · · · .
π
1
1
=
+ (x − 1) −
+ (y − 1)
4
2
2
1
1
1
2
2
+
(x − 1)
+ 2(x − 1)(y − 1) (0) + (y − 1) −
2!
2
2
1
1
1
1
1
3
2
2
3
(x − 1) −
+ 3(x − 1) (y − 1) −
+ 3(x − 1)(y − 1)
+ (y − 1)
+ ··· .
+
3!
2
2
2
2
1
π 1
− [(x − 1) − (y − 1)] + (x − 1)2 − (y − 1)2
=
4 2
4
1 (x − 1)3 + 3(x − 1)2 (y − 1) − 3(x − 1)(y − 1)2 − (y − 1)3 + · · · .
−
12
Putting x = 1.1 and y = 0.9, we get
π 1
1
1 − [(0.2)] + [0] −
(0.1)3 + 3(0.1)3 − 3(0.1)3 − (−0.1)3
4 2
4
12
= 0.7854 − 0.1000 + 0.0003 = 0.6857.
f (x, y) =
Exercise 20.7
1. Expand the following function as for as terms of third degree.
(a) sin x cos y
π
(b) ex sin y at −1,
4
π
2
(c) xy + cos(xy) about 1,
.
2
2. Expand f (x, y) = xy in powers of (x − 1) and (y − 1).
3. If f (x, y) = tan−1 (xy), compute f (0.9, −1.2) approximately.
21
Multiple points
To understand this topic, please watch the video https://youtu.be/ZbwvT6_EdcM.
Ordinary Point
Let f (x, y) = 0 be continuous function such that fx , fy exist. A point on the curve f (x, y) = 0 at
which fx and fy do not vanish simultaneously is called ordinary points.
Singular Point
Let f (x, y) = 0 be continuous function such that fx , fy exist. A point on the curve f (x, y) = 0 at
which fx and fy both vanish simultaneously is called singular points.
Multiple points
Let f (x, y) = 0 be curve in R2 and P be a point on the curve. If more than one branch of curve
passes through ’P ’ then ’P ’ is called multiple point.
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Double Point
If two branch of curve passes through ’P ’ then ’P ’ is double point.
Remark 21.1 1. If k number of branch of curve f (x, y) = 0 passing through ’P ’ is called multiple
point with order ’k’.
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2. If P (a, b) is a double point then we draw two tangent T1 and T2 at P at each branch and
according to its type the double points are classified as follows:
Node: If T1 and T2 are real and different then P is called node.
Cusp: If two tangent T1 and T2 are real and equal then P is called cusp.
Isolated: If T1 and T2 are imaginary then P is called isolated or conjugate point.
How to find double points of curve
Let f (x, y) = 0 be the equation of the curve in R2 . Suppose that fx and fy exists then apply
following steps.
1. Find fx and fy and solve fx = 0 and fy = 0. Suppose that point (a, b) is a solution.
2. Check if (a, b) lies on the curve, if yes then point (a, b) is a double point.
3. Find r = fxx (a, b), s = fxy (a, b), t = fyy (a, b).
4. (a) If s2 − rt > 0 then (a, b) is node.
(b) If s2 − rt < 0 then (a, b) is isolated.
(c) If s2 − rt = 0 then (a, b) is cusp.
Remark 21.2 If the origin (0, 0) is a double point of a given curve then by equating lowest degree
term equal to zero. Obtain equation of the tangent at origin. For example x3 + y 3 = 3axy. Here the
lowest degree terms is 3axy. Therefore, 3axy = 0.
Therefore, x = 0 or y = 0 are the tangent at origin. Hence, (0, 0) is a node.
Remark 21.3 If the origin is not a double point and say (a, b) 6= (0, 0) is a double point then in order
to find the equation of tangent X = x + a, Y = y + b in the given equation. Then X and Y are new
variables. Find the tangent by equating lowest degree terms equal to 0.
Example 21.4 Find the double points of y 3 + x3 = 3axy (a > 0).
Solution: Here f (x, y) = x3 + y 3 − 3axy. Now, fx = 0 gives 3x2 − 3ay = 0. Therefore, x2 = ay.
and fy = 0 gives 3y 2 − 3ax = 0. Therefore, y 2 = ax.
Solving x2 = ay and y 2 = ax, we get
y = 0 or y 3 − a3 = 0
Therefore, y = 0 and y = a. That gives x = 0 and x = a. Hence,(0, 0) and (a, a) are singular points.
Now, we note that
fxx (x, y) = 6x, fxy (x, y) = −3a, fyy (x, y) = 6y.
Point (a, b)
(0, 0)
(a, a)
r = fxx (a, b)
0
6a
s = fxy (a, b)
−3a
−3a
t = fyy (a, b)
0
6a
s2 − rt
9a2 > 0
2
9a − 36a2 < 0
double point or not
node
(a, a) not on the curve.
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From the above table, we say that (0, 0) is node for the curve y 3 + x3 = 3axy (a > 0)
Example 21.5 Find the double points and classified them for x(x2 + y 2 ) = 6y 2 .
Solution: Here f (x, y) = x3 + xy 2 − 6y 2 . Taking partial derivative of f with respect to x and y, we
get
fx (x, y) = 3x2 + y 2 ,
and fy (x, y) = 2xy − 12y.
Now, fy (x, y) = 0 gives 2y(x − 6) = 0. Hence, y = 0 and x = 6.
Using y = 0 in fx (x, y) = 0 gives x = 0 and x = 6 in fx (x, y) = 0 gives y 2 = −3(36), which is not
possible.
Therefore, (0, 0) is double point. Now, fxx (x, y) = 6x, fxy (x, y) = 2y and fyy = 2x − 12. Now,
r = fxx (0, 0) = 0,
s = fxy (0, 0) = 0,
t = fyy (0, 0) = −12.
Therefore, s2 − rt = 0. Hence, (0, 0) is cusp.
Example 21.6 Find the double point of the x4 − 2y 3 − 2x2 − 3y 2 + 1 = 0.
Solution: Here f (x, y) = x4 − 2y 3 − 2x2 − 3y 2 + 1. Taking partial derivative of f (x, y), we get
fx (x, y) = 4x3 − 4x and fy (x, y) = −6y 2 − 6y
Now, fx (x, y) = 0 gives x(x2 − 1) = 0. Hence x = 0, x = 1 or x = −1.
Now, using fy (x, y) = 0 gives y(y + 1) = 0. Hence, y = 0 or y = −1.
Therefore, the points (0, 0), (0, −1), (−1, 0), (−1, −1), (1, 0) or (1, −1) are possible double points.
But, only the points (0, −1), (−1, 0) and (1, 0) lies on the curve.
Now,
fx x(x, y) = 12x2 − 4, fxy (x, y) = 0, fyy (x, y) = −12y − 6.
Point (a, b)
(0, −1)
(−1, 0)
(1, 0)
r = fxx (a, b)
−4
8
8
s = fxy (a, b)
0
0
0
t = fyy (a, b)
6
−6
−6
s2 − rt
24 > 0
48 > 0
48 > 0
double point or not
(0, −1) is a node.
(−1, 0) is a node.
(−1, 0) is a node.
Therefore, (0, −1), (−1, 0) and (1, 0) are nodes for the given curve.
Example 21.7 Find the double point of the x3 + y 3 − 12x − 27y + 70 = 0.
Solution: Here f (x, y) = x3 + y 3 − 12x − 27y + 70. Taking partial derivative of f (x, y), we get
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fx (x, y) = 3x2 − 12 and fy (x, y) = 3y 2 − 27.
SP
Now, fx (x, y) = 0 gives x2 − 4 = 0. Hence x = 2, x = −2.
Now, using fy (x, y) = 0 gives y 2 − 9 = 0. Hence, y = 3 or y = −3.
Therefore, the points (2, 3), (2, −3), (−2, 3), (−2, −3) are possible double points. But, only the points
(2, 3) lies on the curve.
Now,
fxx (x, y) = 6x, fxy (x, y) = 0, fyy (x, y) = 6y.
Therefore, s2 − rt = 0 − 18(12) < 0. Hence, (2, 3) is isolated point.
Exercise 21.8 Find the double point of the following.
1. x3 + 3x2 − y 2 + 3x − 2y = 0
2. (x − 1)2 + (x − 1)5/2 − y + 3 = 0
3. xy 2 − (x + y)2 = 0
4. x3 + y 3 − 3x2 − 3xy + 3x + 3y + 1 = 0
5. (ax)2/3 + (by)2/3 = k 2/3 .
22
Curvature
To understand this topic, please watch the video https://youtu.be/GGymHxaaMCg.
Let P be any point on a given curve and Q a neighboring point. Let arc AP = s and arc P Q = δs.
Let the tangents at P and Q make angle ξ and ξ + δξ with the x−axis, so that the angle between
the tangents at P and Q = δξ(Fig. 21).
In moving from P to Q through a distance δs, the tangent has turned through the angle δξ. This
is called the total bending or total curvature of the arc P Q.
δξ
Therefore, the average curvature of arc P Q = .
δs
The limiting value of average curvature when Q approaches P (i.e., δs → 0) is defined as the
curvature of the curve at P .
dξ
Thus curvature K (at P ) = .
ds
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Figure 21
SP
Remark 22.1 1. Since δξ is measured in radians, the unit of curvature is radians per unit length
e.g., radians per centimeter.
2. Radius of curvature: The reciprocal of the curvature of a curve at any point P is called the
ds
radius of curvature at P and is denoted by ρ, so the ρ = .
dξ
3. Center of curvature: A point C on the normal at any point P of a curve distant ρ from it, is
called the center of curvature at P .
4. Circle of curvature: A circle with center C (center of curvature at P ) and radius ρ is called
the circle of curvature at P .
Theorem 22.2 (Radius of Curvature in Cartesian Coordinates) Prove that the radius of curvature of curve y = f (x) in R2 at any point P is given by
(1 + y12 )3/2
ρ=
,
y2
where y1 =
dy
d2 y
and y2 = 2 .
dx
dx
Proof: Here from the figure 21 the tangent at P1 at point P makes angle ξ at X−axis.
Thus, the slope of T1 at P =tan ξ
dy
.
But for the curve y = f (x), the slope is
dx
dy
= tan ξ
dx
(130)
Therefore,
1 + y12 = 1 + tan2 ξ
(131)
Differentiating (130) with respect to x, we get
d2 y
2 dξ
= sec2 ξ
=
y
=
sec
ξ
2
dx2
dx
Hence,
dξ ds
ds dx
.
ds
sec2 ξ
1
dx .
=
dξ
y2
ds
Finally, we have
(1 + tan2 ξ)
ρ=
but using
y2
ds p
= 1 + y12 , we have
dx
(1 + y12 )3/2
.
y2
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ρ=
ds
ds
(1 + y12 )
dx =
dx ,
y2
Example 22.3 Find the radius of curvature at the point
3a 3a
,
2 2
of the Folium x3 + y 3 = 3axy.
SP
Solution: Differentiating x3 + y 3 = 3axy with respect to x, we get
dy
2
2 dy
= 3a y + x
3x + 3y
dx
dx
or
(y 2 − ax)
Therefore,
dy
dx
dy
= ay − x2 .
dx
(132)

3a 3a = −1.


,
2 2

Differentiating equation (132), we get
dy
dy
dy
d2 y
2y
−a
+ (y 2 − ax) 2 = a − 2x.
dx
dx
dx
dx
Therefore,
d2 y
dx2
Hence,
"
ρ at
3a 3a
,
2 2
1+
=
dy
dx
2
dy
dx2
32

3a 3a = − 3a .


,
2 2

2 #3/2
=
[1 + (−1)2 ]
−32/3a
3/2
3a
= √ . ( in mgnitute).
8 2
Example 22.4 Find the radius of curvature at the point (a, 0) of the curve xy 2 = a3 − x3 .
Solution: We have xy 2 = a3 − x3 or
y 2 = a3 x−1 − x2 .
Therefore,
2y
or
At (a, 0),
dy
= −a3 x−2 − 2x
dx
dy
a3
x
=− 2 − .
dx
2x y y
dy
dx
→ ∞, so we find
from xy 2 = a3 − x3 Therefore,
dx
dy
x − 2y + y 2
or
dx
2xy
=− 2
dy
3x + y 2
dx
= 0.
dy (a,0)
or
dx
dx
(3x + y ) −2y
− 2x − (−2xy) 6x + 2y
d2 x
dy
dy
=
2
2
2
2
dy
(3x + y )
2 dx
3a
(3a2 + 0)(0 − 2a) − 0
=− .
=
2
2
2
dy (a,0)
(3a + 0)
2
or
Hence,
2
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2
M
Therefore,
dx
dx
= −3x2
dy
dy
"
1+
ρ(a,0) =
dx
dy
d2 x
dy 2
(a,0)
#3/2
=
3a
(1 + 0)3/2
=− .
(−2/3a)
2
(a,0)
Radius of Curvature at the origin
Corollary 22.5 If x-axis is tangent to a curve at the origin then ρ at (0, 0) is
2
x
ρ(0,0) = lim
.
x→0
2y
dy
Proof: Since x-axis is a tangent at (0, 0),
= 0 or (y1 )0 = 0. By L-Hospital Rule, we have
dx (0,0)


2
0
x
 2x 
lim
= lim 
as form

dy
x→0
x→0
2y
0
2
dx
1
0
= lim 2 as form
x→0 d y
0
2
dx
1
.
=
(y2 )0
Therefore,
ρ(0,0)
(1 + (y12 )0 )3/2
1
=
=
= lim
(y2 )0
(y2 )0 x→0
x2
2y
.
Corollary 22.6 If y-axis is tangent to a curve at the origin then ρ at (0, 0) is
2
y
ρ(0,0) = lim
.
y→0
2x
Example 22.7 Find ρ at the origin for the curves y 4 + x3 + a(x2 + y 2 ) − a2 y = 0.
solution: Equating to zero the lowest degree terms, we get y = 0.
Therefore, x-axis is the tangent at the origin. Dividing throughout by y, we have
2
x2
x
3
y +x·
+a
+ y − a2 = 0.
y
y
Let x → 0, so that
x2
= ρ.
x→0 2y
lim
Therefore,
M
0 + 0 · 2ρ + a(2ρ + 0) − a2 = 0
or
a
ρ= .
2
SP
Example 22.8 Find ρ at the origin for the curve y − x = x2 + 2xy + y 2 .
Solution: Equating to zero the lowest degree terms, we get y = x, as the tangent at the origin,
x2
which is neither of the coordinates axes. Putting y = y 0 (0)x + y 00 (0) + . . . in the given equation,
2
we get
2
x2
x2
x2
0
00
2
0
00
0
00
y (0)x + y (0) + . . . − x = x + 2x y (0)x + y (0) + . . . + y (0)x + y (0) + . . .
2
2
2
Equating coefficients of x and x2 ,
y 0 (0) − 1 = 0,
or
y 00 (0)
= 1 + 2y 0 (0) + (y 0 (0))2
2
y 0 (0) = 1, y 00 (0) = 2 + 4 · 1 + 2 · 12 = 8.
ρ(0,0) =
(1 + (y 0 (0))2 )3/2
(1 + 1)3/2
1
=
= √ .
00
y (0)
8
2 2
Theorem 22.9 (Radius of Curvature in Parametric Coordinates) Prove that the radius of Cur3/2
(f 02 + g 02 )
vature for the parametric curve x = f (t) and y = g(t) at the point P (t0 ) is ρ = 0 00
.
f g − g 0 f 00
Proof: Given that x = f (t) and y = g(t). Assume that y = F (x) is the same curve generated by
x = f (t) and y = g(t). At t = t0 , suppose that x = x0 and y = y0 . Now, the radius of curvature at
point P is
(1 + y1 )3/2
ρ=
.
(133)
y2
Now,
y1 =
y2 =
dy
dy/dt
g 0 (t)
=
= 0 .
dx
dx/dt
f (t)
dy1
f 0 (t)g 00 (t) − g 0 (t)f 00 (t) 1
=
.
dx
(f 0 (t))2
f 0 (t)
Hence,
y2 =
f 0 (t)g 00 (t) − g 0 (t)f 00 (t)
.
(f 0 (t))3
Putting value of y1 and y2 in equation (133), we get
0 2 !3/2
g
1+
3/2
f0
(f 02 + g 02 )
ρ= 0
= 0 00
.
f (t)g 00 (t) − g 0 (t)f 00 (t)
f g − g 0 f 00
(f 0 (t))3
Example 22.10 Show that the radius of curvature at any point of the cycloid x = a(θ + sin θ),
θ
y = a(1 − cos θ) is 4a cos .
2
Solution: We have
M
dx
dy
= a(1 + cos θ),
= a sin θ.
dθ
dθ
Therefore,
d
d2 y
=
2
dx
dθ
Hence,
"
1+
ρ=
SP
Now,
θ
dy
θ
cos
2
sin
dy
a sin θ
2
2 = tan θ .
= dθ =
=
θ
dx
dx
a(1 + cos θ)
2
2 cos2
dθ
2
dy
dx
dy
dx
·
dθ
1
1
1
θ
θ
= sec2 ·
= sec2 ·
dx
2
2 a(1 + cos θ)
2
2
2 #3/2
2
dy
dx2
θ
4a 1 + tan
2
=
θ
sec4
2
2
1
2a cos2
θ
2
=
1
θ
sec4 .
4a
2
3/2
θ
= 4a sec
2
2
3/2
cos4
θ
θ
= 4a cos .
2
2
Example 22.11 Prove that the radius of curvature at any point of the astroid x2/3 + y 2/3 = a2/3 , is
three times the length of the perpendicular from the origin to the tangent at that point.
Solution: The parametric equation of the curve is
x = a cos3 t, y = a sin3 t
Therefore,
x0 =
Also,
dx
dy
= −3a cos2 t sin t, and y 0 =
= 3a sin2 t cos t
dt
dt
d2 x
2
2
=
−3a
cos
t
−
2
cos
t
sin
t
= 3a cos t(2 sin2 t − cos2 t);
2
dt
d2 y
y 00 = 2 = 3a 2 sin t cos2 t − 2 sin3 t = 3a sin t(2 cos2 t − sin2 t).
dt
x00 =
Now,
x02 + y 02 = 9a2 cos4 t sin2 t + sin4 t cos2 t = 9a2 sin2 t cos2 t
and
x0 y 00 − y 0 x00 = −9a2 cos2 t sin2 t(2 cos2 t − sin2 t) − 9a2 cos2 t sin2 t(2 sin2 t − cos2 t) = −9a2 sin2 t cos2 t
Therefore,
3/2
ρ=
(f 02 + g 02 )
27a3 sin3 t cos3 t
= −3a sin t cos t.
=
f 0 g 00 − g 0 f 00
−9a2 sin2 t cos2 t
Since,
dy
y0
= 0 = − tan t,
dx
x
Therefore, equation of tangent l at (a cos3 t, a sin3 t) is y − a sin3 t = − tan t(x − a cos3 t). That is
x tan t + y − a sin t = 0.
0 + 0 − a sin t
The length of perpendicular (p) from (0, 0) on l = √ 2
= −a sin t cos t. Thus ρ = 3p.
tan t + 1
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Example 22.12 If ρ1 and ρ2 be the radii of curvature at the ends of focal chord of the parabola
y 2 = 4ax then show that (ρ1 )−2/3 + (ρ2 )−2/3 = (2a)−2/3 .
Solution: Given parabola is y 2 = 4ax or x = at2 , y = 2at. If dashes denote differentiation with
respect to t, then
x0 = 2at, y 0 = 2a, x00 = 2a, y 00 = 0
. Therefore,
(x02 + y 02 )3/2
(4at2 + 4a2 )3/2
= 0 00
=
= 2a(1 + t2 )3/2 .
0
00
2
xy −yx
0 − 4a
SP
ρ
(at2 ,2at)
If P (t1 ) and Q(t2 ) be the extremities of the focal chord of the parabola, then
t1 t2 = −1
Hence
Thus,
ρ1 = ρP (t1 ) = 2a(1 + t21 )3/2 ; ρ2 = ρQ(t2 ) = 2a(1 + t22 )3/2 ;
−2/3
ρ1
−2/3
+ ρ2
= (2a)−2/3 (1 + t21 )−1 + (1 + t22 )−1
1
1
−2/3
+
= (2a)
1 + t21 1 + t22
1
t1
−2/3
= (2a)
+
= (2a)−2/3 .
1 + t21 1 + t21
y2
CD3
x2
Example 22.13 Show that the radius of curvature of P on an ellipse 2 + 2 = 1 is
, where
a
b
ab
CD is the semi-diameter conjugate to CP .
Solution: Two diameters of an ellipse are said to be conjugate if each bisects chords parallel
to the
If CP and CD
are two semi-conjugate diameters and P is (a cos θ, b sin θ) then D is a
other.
π
π
a cos θ +
, b sin θ +
, that means (−a sin θ, b cos θ).
2
2
Also, C(0, 0) is the center of the ellipse.
Therefore,
p
CD = a2 sin2 θ + b2 cos2 θ.
At P , we have x = a cos θ and y = b sin θ. Therefore,
dy
dy
b cos θ
b
= dθ =
= − cot θ;
dx
dx
−a sin θ
a
dθ
d2 y
dθ
b
b
= − 2 cosec3 θ
= cosec2 θ
2
dx
a
dx
a
Now,
"
1+
dy
dx
2 #3/2
ρ =
M
d2 y
dx2
3/2
b2
2
1 + 2 cot θ
a
=
b
− 2 cosec3 θ
a
a2
(a2 sin2 θ + b2 cos2 θ)3/2
=
b cosec3 θ
a3 sin3 θ
(a2 sin2 θ + b2 cos2 θ)3/2
CD3
=
=
.
ab
ab
SP
Theorem 22.14 (Radius of curvature for polar curve) Prove that the radius of curvature for polar curve r = f (θ) is given by
(r2 + r12 )3/2
ρ= 2
.
r + 2r12 − rr2
Proof: First we shall show that, if φ be the angle between the radius vector and the tangent at
dθ
any point of the curve r = f (θ). Then tan θ = r .
dr
Let P (r, θ) and Q(r + δr, θ + δθ) be two neighboring points on the curve (See in Figure 22). Join
P Q and draw P M ⊥ OQ. Then from the angled 4OM P . M P = r sin δθ, OM = r cos δθ.
M Q = OQ − OM = r + δr − r cos δθ
= δr + r(1 − cos δθ) = δr + 2r sin2
If ∠M QP = α, then
tan α =
MP
=
MQ
r sin δθ
δθ
.
2
.
δθ
δr + 2r sin
2
In the limit as Q → P ( that means δθ → 0), the chord P Q turns about P and becomes the tangent
2
M
at P and α → φ.
SP
Figure 22
tan φ =
lim (tan α) = lim
r sin δθ
δθ
δr + 2r sin2
2
sin δθ
r
δθ
= lim


δθ
δ→0 δθ  sin 2 
δr
+ r sin 
δθ 
δθ
2
2
r·1
dθ
= =r .
dr
dr
+r·0·1
dθ
Q→P
δ→0
Hence, we have
tan φ = r
dθ
.
dr
We note that,
ξ = θ + φ.
Differentiating with respect to s, we have
1
dξ
dθ dφ
dθ dφ dθ
=
=
+
=
+
ρ
ds
ds ds
ds ds ds
dθ
dφ
=
1+
.
ds
dθ
(134)
Also we know that
tan θ = r
or
−1
φ = tan
r
r1
dθ
r
=
dr
r1
, where r1 =
dr
.
dθ
Differentiating with respect to θ, we have
dφ
=
dθ
1
r1 · r1 − rr2
r12 − rr2
= 2
.
2 ·
r12
r + r12
r
1+
r1
(135)
Also,
q
ds
= r2 + r12
dθ
Substituting the value from (135) and (136) in (134), we get
1
1
r12 − rr2
.
=p
· 1+ 2
ρ
r + r12
r2 + r12
Hence,
(136)
3/2
(r2 + r12 )
.
r2 + 2r12 − rr2
M
ρ=
Example
22.15 Show that the radius of curvature at any point of the cardioid r = a(1−cos θ) varies
√
as r.
SP
Solution: Differentiating r = a(1 − cos θ) with respect to θ, we get
r1 = a sin θ, r2 = a cos θ,
Therefore,
r2 + r12
and
3/2
3/2
= a2 (1 − cos θ)2 + a2 sin2 θ
= a3 [2(1 − cos θ)]3/2
r2 − rr2 + 2r12 = a2 (1 − cos θ)2 − a2 (1 − cos θ) cos θ + 2a2 sin2 θ = 3a2 (1 − cos θ).
Thus,
√
√ √
3/2
(r2 + r12 )
a3 2 2(1 − cos θ)3/2
2 2
2
2a r 1/2 √
1/2
=
=
a(1
−
cos
θ)
=
≈ r.
ρ= 2
r + 2r12 − rr2
3a2 (1 − cos θ)
3
3
a
Exercise 22.16
1. Find the radius of curvature at any point
(a) (at2 , 2at) of the parabola y 2 = 4ax.
x
(b) (0, c) of the catenary y = c cosh .
c
3
(c) (a, 0) of the curve y = x (x − a).
2. Show that for the rectangular hyperbola xy = c2 , ρ =
3. Find the radius of curvature at
(a) (a, 0) on the curve y 2 =
a2 (a − x)
.
x
(x2 + y 2 )3/2
.
2c2
a a
√
√
√
,
on the curve x + y = a.
4 4
π
(c) x = of the curve y = 4 sin x − sin 2x.
2
4. Find the radius of curvature at any point on the
(b)
(a) x = a cos θ, y = b sin θ
(b) x = a(θ − sin θ), y = a(1 − cos θ).
(c) x = a(cos t + t sin t), y = a(sin t − t cos t).
5. Find the radius of the curvature at the point (r, θ) on each of the curves
(a) r = a(1 − cos θ)
(b) rn = an cos nθ.
ρ2
is constant.
r
2a
7. Find the radius of curvature for the parabola
= 1 + cos θ.
r
SP
M
6. For the cardioid r = a(l + cosθ), show that