Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
B
...
(I SEM), 2020-21
MATHEMATICS-I (KAS-103T)
MODULE 3(Differential Calculus II)

Syllabus: Partial derivatives, Total derivative, Euler’s Theorem for homogeneous functions,
Taylor, and Maclaurin’s Theorem for function of one and two variables, Maxima and Minima of
functions of several variables, Lagrange’s Method of Multipliers, Jacobians, Approximation of
errors
Course Outcomes:

S
...


CO 3

Course Outcome
Associate the concept of partial differentiation for determining
maxima, minima, expansion of series and Jacobians

BL

2,3

CONTENT

S
...


TOPIC

PAGE NO
...
1

Partial Derivatives and its applications

2

3
...
1

First order partial derivatives

2

3
...
2

Second and higher order partial derivatives

2

3
...
2
...
2
...
3

Euler’s Theorem for Homogeneous function

15

3
...
1

Homogeneous Function

15

3
...
2

Euler’s Theorem for Homogeneous Function

15

3
...
3

Deduction on Euler’s Theorem for homogeneous function

19

3
...
4
...
4
...
1

Taylor’s theorem for function of one variable

23

3
...
1
...
4
...
1

Taylor’s Theorem for function of two variables

26

3
...
2
...
5

Maxima and minima of function of two variables

32

1

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
3
...
1

Rule to find the extreme values of a function z = f(x, y)

32

3
...
2

Condition or f(x, y, z) to be maximum or minimum

32

3
...
6
...
7

44

3
...
1

Properties of Jacobians

44

3
...
2

Functional Relationship

45

3
...
9

E- Link for more understanding

57

3
...
Partial Derivatives and its Applications: The concept of partial differentiation is applied for function of two
or more independent variables
...
Partial derivatives are very useful in determining Maxima and Minima, Jacobian, Series expansion
of function of several variables, Error and Approximation, In solution of wave and heat equations etc
...
1
...
First order partial derivatives:
If𝑢 = 𝑓(𝑥, 𝑦) be any function of two independent variables 𝑥, 𝑦hen first order partial derivative of u with respect to ′𝑥′
is obtained by differentiating u with respect to ‘x’ treating y and function of ‘y’ as constant and is denoted by

𝜕𝑢
𝜕𝑥

Similarly fist order partial derivative of u with respect to ‘y’ is obtained by differentiating u with respect to ‘y’ treating
𝜕𝑢

‘x’ and functions of ‘x’ as constant and is denoted by𝜕𝑦
...


Solution: Here we have given 𝑢 = 𝑥 2 + 2𝑥𝑦 − 𝑦 2 + 2𝑥 − 3𝑦 + 5
𝜕𝑢

Therefore, to get value of 𝜕𝑥 we shall differentiate u with respect to ‘x’ treating ‘y’ as constant
𝝏𝒖
= 𝟐𝒙 + 𝟐𝒚 + 𝟐
𝝏𝒙
Also, to find
𝝏𝒖
𝝏𝒚

𝜕𝑢
𝜕𝑦

we shall differentiate u with respect to ‘y’ treating ‘x’ as constant

= 𝟐𝒙 − 𝟐𝒚 − 𝟑
...
1
...
Second and higher order partial derivatives:
If u  f ( x, y ) be any function of two independent variables x, y then second order partial derivative of u with respect
to ′𝑥′ is obtained by differentiating u with respect to ‘x’ twice in succession treating ‘y’ and function of ‘y’ as constant
𝜕2 𝑢

each time and is denoted by 𝜕𝑥 2
...

𝜕𝑢

Here if we differentiate 𝜕𝑦 with respect to ‘x’ treating ‘y’ as constant then we get second order derivative of u denoted
𝜕2 𝑢

𝜕𝑢

as𝜕𝑥𝜕𝑦
...
It is interesting to see that

𝜕2 𝑢
𝜕𝑥𝜕𝑦

𝜕2 𝑢

= 𝜕𝑦𝜕𝑥
...

Example-2: Find all first and second order partial derivatives of𝑢(𝑥, 𝑦) = 𝑥 𝑦
...
(1)
Differentiating equation (1) partially with respect to ‘x’ and ‘y’ we get
𝜕𝑢
= 𝑦𝑥 𝑦−1 … … … … … … … … … …
...
(3)
𝜕𝑦
Differentiating equation (2) partially with respect to ‘x’ and ‘y’ we get
𝜕2𝑢
= 𝑦(𝑦 − 1)𝑥 𝑦−2 … … … … … … … … … …
...
(5)
𝝏𝒚𝝏𝒙
Now differentiating equation (3) partially with respect to ‘y’ and ‘x’ we get
𝜕2𝑢
= 𝑥 𝑦 (𝑙𝑜𝑔𝑥)2 … … … … … … … … … …
...
(7)
𝝏𝒙𝝏𝒚
From equations (5) & (7), it is clear that

𝜕2 𝑢
𝜕𝑥𝜕𝑦

=

𝜕2 𝑢
𝜕𝑦𝜕𝑥


...
(2)
=
𝜕𝑥 𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛𝑦 + 𝑡𝑎𝑛𝑧
3

𝜕𝑢
𝜕𝑧

=2
...


KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
Similarly differentiating (1) partially with respect to ‘y’ and ‘z’ we get
𝜕𝑢
1
(𝑠𝑒𝑐 2 𝑦) … … … … … … … …
...
(4)
=
𝜕𝑧 𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛𝑦 + 𝑡𝑎𝑛𝑧
Multiplying equations (2), (3) and (4) with 𝑠𝑖𝑛2𝑥, 𝑠𝑖𝑛2𝑦 𝑎𝑛𝑑 𝑠𝑖𝑛2𝑧 respectively and then adding, we get
𝑠𝑖𝑛2𝑥

𝜕𝑢
𝜕𝑥

𝜕𝑢

+sin2y𝜕𝑦 + 𝑠𝑖𝑛2𝑧

𝜕𝑢
𝜕𝑧

1

= 𝑠𝑖𝑛2𝑥 {𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 (𝑠𝑒𝑐 2 𝑥)} +

1

1

𝑠𝑖𝑛2𝑦 {𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 (𝑠𝑒𝑐 2 𝑦)}+ 𝑠𝑖𝑛2𝑧 {𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 (𝑠𝑒𝑐 2 𝑧)}
𝑠𝑖𝑛2𝑥
𝒔𝒊𝒏𝟐𝒙

𝜕𝑢
𝜕𝑥

+sin2y

𝜕𝑢
𝜕𝑦

+ 𝑠𝑖𝑛2𝑧

𝜕𝑢
𝜕𝑧

=

2𝑡𝑎𝑛𝑥
𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧

+

2𝑡𝑎𝑛𝑦
𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧

+

2𝑡𝑎𝑛𝑧
𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧

+

𝝏𝒖
𝝏𝒖
𝝏𝒖
+ 𝐬𝐢𝐧𝟐𝐲
+ 𝒔𝒊𝒏𝟐𝒛
= 𝟐
...

𝜃

𝑟

𝑦

𝑥

Solution: Here we have
𝑥 = 𝑟𝑐𝑜𝑠𝜃 … … … … … … … …
...
(2)
Differentiating equation (1) partially with respect to ‘r’ treating ‘𝜃′ as constant, we get
(

𝝏𝒙
) = 𝒄𝒐𝒔𝜽
𝝏𝒓 𝜽

Similarly differentiating (2) partially with respect to ‘𝜃′ treating ‘r’ as constant we get
(

𝝏𝒚
) = 𝒓𝒄𝒐𝒔𝜽
𝝏𝜽 𝒓

Again, from equations (1) and (2), we get
𝑟 2 = 𝑥 2 + 𝑦 2 … … … … … … … … … … … … … … (3)
𝑦

and, 𝜃 = tan−1 (𝑥 ) … … … … … … … … … … … … (4)
Differentiating equation (3) partially with respect to ‘x’ treating ‘y’ as constant, we get
𝜕𝑟

2𝑟 (𝜕𝑥 ) = 2𝑥
𝑦

𝝏𝒓

𝒙

𝒐𝒓 (𝝏𝒙) = 𝒓
𝒚

Similarly Differentiating equation (3) partially with respect to ‘y’ treating ‘x’ as constant, we get

4

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD

(

𝜕𝜃
) =
𝜕𝑦 𝑥
𝝏𝜽

1

1

( )
𝑦 2 𝑥

1 + (𝑥 )
𝒙

( ) = 𝟐 𝟐
𝝏𝒚
𝒙 +𝒚
𝒙


...


𝜕𝑢
𝜕𝑥

𝜕𝑢

𝜕𝑢

3

𝜕

𝜕

𝜕 2

9

+ 𝜕𝑦 + 𝜕𝑧 = 𝑥+𝑦+𝑧 , and b) (𝜕𝑥 + 𝜕𝑦 + 𝜕𝑧 ) 𝑢 = − (𝑥+𝑦+𝑧)2
...
(2)
= 3
3
(
𝜕𝑥
𝑥 + 𝑦 + 𝑧 3 − 3𝑥𝑦𝑧)
Similarly differentiating equation (1) partially with respect to ‘y’ and ‘z’ we get
𝜕𝑢
1
(3𝑦 2 − 3𝑧𝑥 ) … … … … … … … …
...

𝝏𝒙

5

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD

Example-6:Find

𝜕𝑢
𝜕𝑟

𝑎𝑛𝑑

𝜕𝑢
𝜕𝜃

if 𝑢 = 𝑒 𝑟𝑐𝑜𝑠𝜃
...
cos(𝑟𝑠𝑖𝑛𝜃) … … … … … … … … … …
...
cos(𝑟𝑠𝑖𝑛𝜃) + 𝑒 𝑟𝑐𝑜𝑠𝜃
...
𝑠𝑖𝑛𝜃
𝜕𝑟
𝜕𝑢
= 𝑒 𝑟𝑐𝑜𝑠𝜃 {𝑐𝑜𝑠𝜃
...
sin(𝑟𝑠𝑖𝑛𝜃}
𝜕𝑟
𝜕𝑢
= 𝑒 𝑟𝑐𝑜𝑠𝜃
...
cos(𝑟𝑠𝑖𝑛𝜃) + 𝑒 𝑟𝑐𝑜𝑠𝜃
...
𝑟𝑐𝑜𝑠𝜃
𝜕𝜃
𝜕𝑢
= −𝑟𝑒 𝑟𝑐𝑜𝑠𝜃 {𝑠𝑖𝑛𝜃
...
𝑐𝑜𝑠𝜃}
𝜕𝜃
𝜕𝑢
= −𝑟𝑒 𝑟𝑐𝑜𝑠𝜃
...
(3)
𝜕𝜃

3
...
Total Derivative:
If 𝑢 = 𝑓(𝑥, 𝑦) be any function in two variables x, y and if 𝛿𝑥, 𝛿𝑦 are small increment in values of variables x and y
respectively
...

Similarly if 𝑢 = 𝑓(𝑥, 𝑦, 𝑧)
...
2
...


𝜕𝑢
𝜕𝑥

𝑑𝑥 +

𝜕𝑢
𝜕𝑦

𝑑𝑦 +

𝜕𝑢
𝜕𝑧

𝑑𝑧

Composite function and its Derivative: If𝑢 = 𝑓(𝑥, 𝑦), where 𝑥 = 𝜑(𝑡), 𝑦 = 𝜔(𝑡) then u is called a
composite function of single variable t,then the value of total derivative of u with respect to t is given by

𝑑𝑢 𝜕𝑢 𝑑𝑥 𝜕𝑢 𝑑𝑦
=

...

𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡
Also if 𝑧 = 𝑓(𝑥, 𝑦), where 𝑥 = 𝜑(𝑢, 𝑣), 𝑦 = 𝜔(𝑢, 𝑣) then z is called a composite function of two variables u , v and
the value of total derivative of z with respect to u, v are given by
6

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
𝝏𝒛
𝝏𝒖

𝝏𝒛 𝝏𝒙

𝝏𝒛 𝝏𝒚

= 𝝏𝒙
...
𝝏𝒗respectively
...
Find the value of

𝑑𝑢
𝑑𝑡

Solution: Here we have, 𝑢 = 𝑥 3 + 𝑦 3 … … … … … … … … … …
...
So we shall have
𝑑𝑢 𝜕𝑢 𝑑𝑥 𝜕𝑢 𝑑𝑦
=

...

… … … … … … … … … … … … … … (3)
𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡
Now from equations (1) & (2)
𝜕𝑢
𝜕𝑢
= 3𝑥 2 &
= 3𝑦 2 … … … … … … … … … … … … …
...
𝑐𝑜𝑠𝑡 − 𝑐𝑜𝑠 2 𝑡
...
(−𝑠𝑖𝑛𝑡) + (3𝑠𝑖𝑛𝑡 2 𝑡)
...
𝑐𝑜𝑠𝑡 − 𝑐𝑜𝑠 2 𝑡
...

Solution: Here we have, 𝑢 = 𝑥 2 − 𝑦 2 + 𝑠𝑖𝑛𝑦𝑧 … … … … … … … … (1)
Where 𝑦 = 𝑒 𝑥 , 𝑧 = 𝑙𝑜𝑔𝑥 … … … … … … … … … … … … … … …
...
So we shall have

7

and hence verify the result
...

+

...

… … … … … … … … … … … … … … (3)
𝑑𝑥 𝜕𝑥 𝑑𝑥 𝜕𝑦 𝑑𝑥 𝜕𝑧 𝑑𝑥
Now, differentiating equations (1) and (2), we get
𝜕𝑢
𝜕𝑢
𝜕𝑢
= 2𝑥,
= −2𝑦 + 𝑧𝑐𝑜𝑠𝑦𝑧 &
= 𝑦𝑐𝑜𝑠𝑦𝑧 … … … … … … … (4)
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝑑𝑦
𝑑𝑧 1
= 𝑒𝑥&
=
𝑑𝑥
𝑑𝑥 𝑥

… … … … … … … … … … … … … … … … … … … … (5)

Using equations (4) and (5) in equation (3), we get
𝑑𝑢
1
= 2𝑥 + (−2𝑦 + 𝑧𝑐𝑜𝑠𝑦𝑧)𝑒 𝑥 + 𝑦𝑐𝑜𝑠𝑦𝑧 ( )
𝑑𝑥
𝑥
Putting values of y & z from equation (2), we get
𝑑𝑢
1
= 2𝑥 + (−2𝑒 𝑥 + 𝑙𝑜𝑔𝑥
...

Solution: Given that 𝑣 = 𝑓 (2𝑥 − 3𝑦, 3𝑦 − 4𝑧, 4𝑧 − 2𝑥)
Let 𝑣 = 𝑓(𝑟, 𝑠, 𝑡) … … … … … … … … … … … … … … … … … … …
...
(2)
So, from equation (2), we get
𝜕𝑟
𝜕𝑟
𝜕𝑟
= 2,
= −3, = 0
𝜕𝑥
𝜕𝑦
𝜕𝑧

… … … … … … … … … … … … … … …
...
(4)

𝜕𝑡
𝜕𝑡
𝜕𝑡
= −2,
= 0, = 4
𝜕𝑥
𝜕𝑦
𝜕𝑧

… … … … … … … … … … … … … … …
...
𝜕𝑥
𝜕𝑓

𝜕𝑓

𝜕𝑓

𝜕𝑓

𝜕𝑓

𝑣𝑥 = 𝜕𝑟 (2) + 𝜕𝑠 (0) + 𝜕𝑡 (−2) Or 𝑣𝑥 = 2 ( 𝜕𝑟 − 𝜕𝑡 ) … … … … … … …
...

+

...

𝜕𝑦 𝜕𝑟 𝜕𝑦 𝜕𝑠 𝜕𝑦 𝜕𝑡 𝜕𝑦
𝜕𝑓
𝜕𝑟

(−3) +

𝜕𝑓
𝜕𝑠

(3) +

𝜕𝑓
𝜕𝑡

(0) Or 𝑣𝑦 = 3 (−

𝜕𝑓
𝜕𝑟

+

𝜕𝑓
𝜕𝑠

) … … … … … …
...
+

...

𝜕𝑧 𝜕𝑟 𝜕𝑧 𝜕𝑠 𝜕𝑧 𝜕𝑡 𝜕𝑧
𝜕𝑓

𝜕𝑓

𝜕𝑓

𝜕𝑓

𝜕𝑓

𝑣𝑧 = 𝜕𝑟 (0) + 𝜕𝑠 (−4) + 𝜕𝑡 (4) Or 𝑣𝑧 = 4 (− 𝜕𝑠 + 𝜕𝑡 ) … … … … … …
...
(7) and (8), we get
6𝑣𝑥 + 4𝑣𝑦 + 3𝑣𝑧 == 12 {(

𝜕𝑓
𝜕𝑟

−

𝑦−𝑥 𝑧−𝑥

Example-10:If 𝑢 = 𝑢 ( 𝑥𝑦 ,

𝑥𝑧

𝜕𝑓
𝜕𝑡

𝜕𝑓

) + (−

𝜕𝑟

𝜕𝑓

+

𝜕𝑠

) + (−

), then show that 𝑥 2
𝑦−𝑥 𝑧−𝑥

Solution: Here we have 𝑢 = 𝑢 ( 𝑥𝑦 ,

𝑥𝑧

𝜕𝑢
𝜕𝑥

𝜕𝑓
𝜕𝑠

+

𝜕𝑓
𝜕𝑡

)}or 6𝑣𝑥 + 4𝑣𝑦 + 3𝑣𝑧 = 0
...
(1)

Let 𝑢 = 𝑢 (𝑟, 𝑠) … … … … … … … … … … … … … … … … … … … (2)
𝑤ℎ𝑒𝑟𝑒 𝑟 =

𝑦−𝑥

1

1

or 𝑟 = 𝑥 − 𝑦 & s=

𝑥𝑦

𝜕𝑟

𝑧−𝑥

1

1

1

or 𝑠 = 𝑥 − 𝑧 …

𝑥𝑧
𝜕𝑟

1

𝜕𝑟

(3)
𝜕𝑠

1

𝜕𝑠

𝜕𝑠

1

From equation(3), we get 𝜕𝑥 = − 𝑥 2 , 𝜕𝑦 = 𝑦2 , 𝜕𝑧 = 0 and 𝜕𝑥 = − 𝑥 2 , 𝜕𝑦 = 0, 𝜕𝑧 = 𝑧 2
𝜕𝑢

Also, 𝜕𝑥 =
𝜕𝑢
𝜕𝑥

=

𝜕𝑢
𝜕𝑟

𝜕𝑢 𝜕𝑟


...
𝜕𝑥

1

𝜕𝑢

𝑥

𝜕𝑠

(− 2) +

1

𝜕𝑢

𝑥

𝜕𝑥

(− 2) or 𝑥 2

=−

𝜕𝑢
𝜕𝑟

−

𝜕𝑢
𝜕𝑠

… … … … … … … … … …
...

+

...
(5)

𝜕𝑟

𝜕𝑢 𝜕𝑢 𝜕𝑟 𝜕𝑢 𝜕𝑠
=

...

𝜕𝑧 𝜕𝑟 𝜕𝑧 𝜕𝑠 𝜕𝑧
𝜕𝑢
𝜕𝑧

=

𝜕𝑢
𝜕𝑟

𝜕𝑢

1

𝜕𝑢

or 𝑧 2 𝜕𝑧 =

(0) + ( 2 )
𝜕𝑠 𝑧

𝜕𝑢

… … … … … … … … … …
...

Example-11: If 𝑢 = 𝑓(𝑟), 𝑤ℎ𝑒𝑟𝑒 𝑥 = 𝑟𝑐𝑜𝑠𝜃, 𝑦 = 𝑟𝑠𝑖𝑛𝜃
...
(1)
Also, since 𝑥 = 𝑟𝑐𝑜𝑠𝜃, 𝑦 = 𝑟𝑠𝑖𝑛𝜃
Therefore 𝑟 2 = 𝑥 2 + 𝑦 2

… … … … … … … … … … … … (2)

Differentiating equation (2) partially with respect to ‘x’ and ‘y’ we get
𝜕𝑟
𝜕𝑥

𝑥 𝜕𝑟

𝑦

= 𝑟 , 𝜕𝑦 = 𝑟
...
𝜕𝑥

𝜕𝑢

𝑑𝑓

𝑥

or𝜕𝑥 = 𝑑𝑟
...


KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
and
𝜕𝑢
𝜕𝑦

=

𝑑𝑓 𝜕𝑟


...
( ) … … … … … … … … … … … … (4)
𝑟

Differentiating equation (3) partially with respect to ‘x’ we get
𝜕2 𝑢
𝜕𝑥 2
𝜕2 𝑢
𝜕𝑥 2

𝜕

𝑑𝑓

𝜕2 𝑢

𝑥

= 𝜕𝑥 { 𝑑𝑟
...
1−𝑥
...


𝑑𝑓 𝜕

𝑥

𝜕

𝑑𝑓

𝑥

= 𝑑𝑟 𝜕𝑥 {(𝑟 )} + 𝜕𝑥 ( 𝑑𝑟 )
...
1−𝑥
...
( 𝑟 )

𝜕 2 𝑢 𝑑𝑓 𝑟 2 − 𝑥 2
𝑑2 𝑓 𝑥2
=

...
(5)
𝜕𝑥 2 𝑑𝑟
𝑟3
𝑑𝑟 2 𝑟 2
Similarly differentiating equation (4) partially with respect to ‘y’ we get
𝜕 2 𝑢 𝑑𝑓 𝑟 2 − 𝑦 2
𝑑2 𝑓 𝑦 2
=

...
(6)
𝜕𝑦 2 𝑑𝑟
𝑟3
𝑑𝑟 2 𝑟 2
Adding equations (5) and (6) we get
𝜕 2 𝑢 𝜕 2 𝑢 𝑑𝑓 𝑟 2 − (𝑥 2 + 𝑦 2 )
𝑑2 𝑓 𝑥2 + 𝑦 2
{
}
+
=

...

(
)
+
𝜕𝑥 2 𝜕𝑦 2 𝑑𝑟 𝑟
𝑑𝑟 2
𝜕2 𝑢

𝜕2 𝑢

1

𝑑2 𝑓

𝑑𝑓

Or 𝜕𝑥 2 + 𝜕𝑦2 = 𝑓 ′′ (𝑟) + 𝑟 𝑓 ′ (𝑟)
...
Prove that( ) + ( ) = 𝑒 −2𝑢 {( ) + ( ) }
...
(1)
∴

𝜕𝑧
𝜕𝑥

𝜕𝑓

𝜕𝑧

𝜕𝑓

= 𝜕𝑥 &𝜕𝑦 = 𝜕𝑦

Also we have where 𝑥 = 𝑒 𝑢 𝑐𝑜𝑠𝑣, 𝑦 = 𝑒 𝑢 𝑠𝑖𝑛𝑣 … … … … … … … … … (2)
From equation (2) we get
𝜕𝑥
𝜕𝑢

𝜕𝑥

𝜕𝑦

𝜕𝑦

= 𝑒 𝑢 𝑐𝑜𝑠𝑣 = 𝑥 & 𝜕𝑣 = −𝑒 𝑢 𝑠𝑖𝑛𝑣 = −𝑦 Also 𝜕𝑢 = 𝑒 𝑢 𝑠𝑖𝑛𝑣 = 𝑦 & 𝜕𝑣 = 𝑒 𝑢 𝑐𝑜𝑠𝑣 = 𝑥

Now from equations (1) & (2), we have
𝜕𝑓
𝜕𝑢
𝜕𝑓
𝜕𝑣

𝜕𝑓 𝜕𝑥

𝜕𝑓 𝜕𝑦

= 𝜕𝑥 𝜕𝑢 + 𝜕𝑦 𝜕𝑢 or
𝜕𝑓 𝜕𝑥

𝜕𝑓 𝜕𝑦

= 𝜕𝑥 𝜕𝑣 + 𝜕𝑦 𝜕𝑣 or

𝜕𝑓
𝜕𝑢
𝜕𝑓
𝜕𝑣

𝜕𝑓

𝜕𝑓

= 𝑥 𝜕𝑥 + 𝑦 𝜕𝑦 … … … … … … … … … … … … … (3)
𝜕𝑓

𝜕𝑓

= −𝑦 𝜕𝑥 + 𝑥 𝜕𝑦 … … … … … … … … … … … … (4)

Now making square of equations (3) & (4) and then adding we get
10

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
So, differentiating equation (3) partially with respect to ‘x’ we get
𝜕𝑓 2

𝜕𝑓 2

𝜕𝑓

𝜕𝑢

𝜕𝑣

𝜕𝑥

𝜕𝑓 2

𝜕𝑓 2

𝜕𝑓 2

𝜕𝑓 2

𝜕𝑓 2

𝜕𝑓 2

𝜕𝑥

𝜕𝑦

𝜕𝑢

𝜕𝑣

( ) + ( ) = (𝑥

+𝑦

𝜕𝑓 2

𝜕𝑓

𝜕𝑦

𝜕𝑥

) + (−𝑦
𝜕𝑓 2

+𝑥

𝜕𝑓 2

𝜕𝑓 2

𝜕𝑓 2

𝜕𝑓 2

𝜕𝑓 2

𝜕𝑦

𝜕𝑢

𝜕𝑣

𝜕𝑥

𝜕𝑦

𝜕𝑓 2

) Or ( ) + ( ) = (𝑥 2 + 𝑦 2 ) ( ) + (𝑦 2 + 𝑥 2 ) ( )
𝜕𝑓 2

𝜕𝑓 2

𝜕𝑓 2

𝜕𝑓 2

( ) + ( ) = (𝑥 2 + 𝑦 2 ) {( ) + ( ) } Or ( ) + ( ) = 𝑒 2𝑢 {( ) + ( ) }
𝜕𝑢
𝜕𝑣
𝜕𝑥
𝜕𝑦
𝜕𝑢
𝜕𝑣
𝜕𝑥
𝜕𝑦
( ) + ( ) = 𝑒 −2𝑢 {( ) + ( ) }
...
Show that𝜕𝑥 2 + 𝜕𝑦2 = (𝑙 2 + 𝑚 2 ) (𝜕𝑢2 + 𝜕𝑣 2)
...
(1)

Where 𝑢 = 𝑙𝑥 + 𝑚𝑦, 𝑣 = 𝑙𝑦 − 𝑚𝑥 … … … … … … … …
...
𝜕𝑥
=

𝜕𝑧 𝜕𝑢


...
(3)

𝜕𝑧 𝜕𝑣


...
+𝑙

𝜕𝑧
𝜕𝑣

) … … … … … …
...
𝜕𝑢 − 𝑚 𝜕𝑣&𝜕𝑦 ≡ 𝑚 𝜕𝑢 + 𝑙 𝜕𝑣 … … … … … … … (5)

𝜕2 𝑧
𝜕𝑥 2

𝜕

𝜕𝑧

𝜕𝑧

= 𝜕𝑥 (𝑙
...
+𝑙

𝜕𝑧
𝜕𝑣

)

or

𝜕2 𝑧

𝜕𝑦 2

𝜕

𝜕𝑧

𝜕𝑧

= (𝑙 𝜕𝑢 − 𝑚 𝜕𝑣 ) (𝑙 𝜕𝑢 − 𝑚 𝜕𝑣)

= (𝑚

𝜕2𝑧
𝜕 2𝑧
𝜕2𝑧
𝜕2𝑧
2
2
=
𝑚
+
2𝑙𝑚
+
𝑙
𝜕𝑦 2
𝜕𝑢2
𝜕𝑢𝜕𝑣
𝜕𝑣 2

… … … … … … … … …
...
+𝑙

𝜕𝑧
𝜕𝑣

)

… … … … … … … … …
...


11

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
Differentiation of an implicit function:
For any implicitly defined function 𝑓(𝑥, 𝑦) = 0
𝜕𝑓

𝜕𝑓

We have, 𝜕𝑥 𝑑𝑥 + 𝜕𝑦 𝑑𝑦 = 0
𝑑𝑦
𝑑𝑥

𝜕𝑓

𝑑𝑦

= − 𝜕𝑥
𝜕𝑓

or

𝑑𝑥

𝜕𝑦

=−

𝑓𝑥
𝑓𝑥

𝑑2 𝑦

2

Also second order derivative is given by 𝑑𝑥 2 =

(𝑓𝑦 ) 𝑓𝑥𝑥 −2𝑓𝑥 𝑓𝑦 𝑓𝑥𝑦 +𝑓𝑦𝑦 (𝑓𝑥)2
(𝑓𝑦 )

Example-13: If 𝑥 3 + 3𝑥 2 𝑦 + 6𝑥𝑦 2 + 𝑦 3 = 1, 𝐹𝑖𝑛𝑑

𝑑𝑦
𝑑𝑥

3


...
(1)
𝜕𝑓

𝜕𝑓

Therefore from equation (1), 𝑓𝑥 = 𝜕𝑥 = 3𝑥 2 + 6𝑥𝑦 + 6𝑦 2 &𝑓𝑦 = 𝜕𝑦 = 3𝑥 2 + 12𝑥𝑦 + 3𝑦 2
∴
Or

𝑑𝑦

=−

𝑑𝑥
𝑑𝑦
𝑑𝑥

𝑓𝑥
𝑓𝑥

=−

will implies that

𝑥 2+2𝑥𝑦+ 2𝑦 2
𝑥 2+4𝑥𝑦+𝑦 2

Example-14: Find

𝑑𝑢
𝑑𝑥

𝑑𝑦
𝑑𝑥

=−

3𝑥 2+6𝑥𝑦+6𝑦 2
3𝑥 2+12𝑥𝑦+3𝑦 2


...


Solution: Here we have 𝑢 = 𝑥𝑙𝑜𝑔𝑥𝑦 … … … … … … … … …
...

12

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
𝜕𝑦

𝜕𝑧

𝜕𝑥

Example-15:If 𝜑(𝑥, 𝑦, 𝑧) = 0, show that( 𝜕𝑧 )
...

𝑥

𝑦

𝑧

Solution: Here we have 𝑓 (𝑥, 𝑦, 𝑧) = 0 … … … … … … … … … …
...
(4)
𝜕𝑦 𝑧
𝜕𝑥

From equations (2), (3) & (4), we get
𝜕𝜑

𝜕𝜑

𝜕𝜑

𝜕𝑦

𝜕𝑧

𝜕𝑥

𝜕𝑦
𝜕𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥
( )
...
(− 𝜕𝜑
)
...


13

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
Practice Exercise:
𝑥

Q-1: Find all first order derivatives of function 𝑢 = sin−1
...

𝜕2 𝑧

𝜕2 𝑧

Q-2: If𝑧 = 𝑓(𝑥 + 𝑐𝑡) + 𝜑(𝑥 − 𝑐𝑡), show that 𝜕𝑡 2 = 𝑐 2 𝜕𝑥 2
...
cos(𝑟𝑠𝑖𝑛𝜃) , 𝑦 = 𝑒 𝑟𝑐𝑜𝑠𝜃
...


𝜕𝑢

Q-4: If 𝑢 = 𝑢(𝑦 − 𝑧, 𝑧 − 𝑥, 𝑥 − 𝑦) show that 𝜕𝑥 + 𝜕𝑦 + 𝜕𝑧 = 0
...
Show that𝑥 𝜕𝑥 + 𝑦 𝜕𝑦 + 𝑧 𝜕𝑧 = 0
...

Q-7: If 𝑧 = 𝑢2 + 𝑣 2 , 𝑢 = 𝑟𝑐𝑜𝑠𝜃, 𝑣 = 𝑟𝑠𝑖𝑛𝜃
...


= 𝟎
...
Then show that
𝜕𝑃

𝜕𝑇

𝜕𝑉

( ) ( ) ( ) = −1
...
Show that𝜕𝑦
...


KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
3
...
Euler’s

Theorem for Homogeneous function:

3
...
1
...
A function 𝑢 = 𝑓(𝑥, 𝑦) is said to be a homogeneous function in variables x, y of degree n if it is expressible in any
𝑦
𝑥
one of the following form 𝑢 = 𝑥 𝑛 𝜑 ( ) 𝑜𝑟 𝑢 = 𝑦 𝑛 𝜔 ( )
...
Any polynomial function in two variables x and y is said to be homogeneous
If all its terms are of the same degree
...

𝑦 2

𝑦

𝑦 3

𝑦

Also, it can be expressed in form 𝑓(𝑥, 𝑦) = 𝑥 3 {1 − 5 (𝑥 ) + 3 (𝑥 ) + (𝑥 ) }or𝑓(𝑥, 𝑦) = 𝑥 3 𝜑 (𝑥 )
2
...

Ex: Function 𝑓(𝑥, 𝑦) = cos (
2 2

𝑥 2+𝑦 2

2𝑥𝑦
2 2

) will be an homogeneous function in two variables x, y of degree 0 as we have

𝜇 𝑥 +𝜇 𝑦
𝑥2 + 𝑦 2
0
(
)
𝑓 (𝜇𝑥, 𝜇𝑦) = cos (
)
𝑜𝑟
𝑓
𝜇𝑥,
𝜇𝑦
=
𝜇
𝑐𝑜𝑠
(
)
2𝜇2 𝑥𝑦
2𝑥𝑦
𝑦

For example the function 𝑓(𝑥, 𝑦) = x 2 tan−1 𝑥 will be a homogeneous function of degree 2 in variables x, y as it is
𝑦

𝑥2

𝑥

𝑦

identical with form 𝑓(𝑥, 𝑦) = 𝑥 𝑛 𝜑 ( ) but the function cos −1

is not homogeneous function of x, y
...
If 𝑢 = 𝑓(𝑥, 𝑦, 𝑧) is a homogeneous function in three variables x, y, z of degree n then it can be expressed in any of
𝑦 𝑧

𝑥 𝑧

𝑥 𝑦

𝑥 𝑥

𝑦 𝑦

𝑧 𝑧

these three forms, 𝑢 = 𝑥 𝑛 𝜑 ( , ) 𝑜𝑟 𝑢 = 𝑦 𝑛 𝜑 ( , ) 𝑜𝑟 𝑢 = 𝑧 𝑛 𝜑 ( , )

3
...
2
...

Proof: Since 𝑢 = 𝑓 (𝑥, 𝑦) is homogeneous function in two variables x, y of degree n therefore it can be expressed as
𝑦
𝑢(𝑥, 𝑦) = 𝑥 𝑛 𝜑 ( ) … … … … … … … … … … … … … … … …
...
( 2 ) … … … … … … … … … … … (2)
𝜕𝑥
𝑥
𝑥
𝑥
𝜕𝑢
𝑦
1
= 𝑥 𝑛 𝜑′ ( )
...

Solution: Here we have 𝑢 = (√𝑥 + √𝑦)5 … … … … … … … … … … … … … … … … … … (1)
𝑦

From (1) we have 𝑢 = (√𝑥)5 (1 + √𝑥 )

5

5

𝑦

5

or 𝑢 = 𝑥 2 (1 + √𝑥 )

𝑦

This is of form 𝑢 = 𝑥 𝑛 𝜑 (𝑥 ), so given function u is a homogeneous function in variables x, y of degree 𝑛 =
5


...

𝑦

𝑦

𝑥

𝑥

Solution: Here 𝑢 = 𝑓 ( ), so it is expressible as 𝑢 = 𝑥 0 𝑓 ( )
Therefore, u is a homogeneous function in variables x, y of degree 𝑛 = 0
...

1

4

𝑦

Solution: Here we have 𝑢 = 𝑥 3 𝑦 −3 tan−1 (𝑥 ) … … … … … … … … … … … … (1)
Replacing 𝑥 𝑏𝑦 𝜇𝑥 & 𝑦 𝑏𝑦 𝜇𝑦 in equation (1) we get
1 4
1
4
𝑦
𝑢 = 𝜇 (3−3) 𝑥 3 𝑦 −3 tan−1 ( )
𝑥

1
4
𝑦
𝑜𝑟 𝑢 = 𝜇 (−1) 𝑥 3 𝑦 −3 tan−1 ( )
𝑥

Clearly u is a homogeneous function in variables x, y of degree -1
...
(4)
𝑥

𝑥

Multiplying equations (3) & (4) by x & y respectively and then adding we get
𝑥

1
4
𝜕𝑢
𝜕𝑢 1 1 −4
𝑦
+𝑦
= 𝑥 3 𝑦 3 tan−1 ( ) + 𝑥 3 𝑦 −3
𝜕𝑥
𝜕𝑦 3
𝑥

1
𝑦 2

{1 + ( ) }

(

1
4
−𝑦
4 1 4
𝑦
) − 𝑥 3 𝑦 −3 tan−1 ( ) + 𝑥 3 𝑦 −3
𝑥
3
𝑥

𝑥

𝑥

1
4
𝜕𝑢
𝜕𝑢
𝑦
+𝑦
= (−1)𝑥 3 𝑦 −3 tan−1
𝜕𝑥
𝜕𝑦
𝑥

𝑜𝑟 𝒙

𝝏𝒖
𝝏𝒖
+𝒚
= −𝒖 … … … … … …
𝝏𝒙
𝝏𝒚

(5)

Example-4: If u be a homogeneous function in two variables x, y of degree n, then show that
𝜕2 𝑢

𝜕2 𝑢

𝜕𝑢

𝑥 𝜕𝑥 2 + 𝑦 𝜕𝑥𝜕𝑦 = (𝑛 − 1) 𝜕𝑥

ii)

𝑥

iii)

𝑥

𝜕2 𝑢

𝜕𝑥𝜕𝑦
2
2𝜕 𝑢
𝜕𝑥 2

+𝑦

𝜕2 𝑢

= (𝑛 − 1)

𝜕𝑦 2
𝜕2 𝑢

+ 2𝑥𝑦

𝜕𝑥𝜕𝑦

+𝑦

𝜕𝑢

𝜕𝑦
2𝑢
𝜕
2
𝜕𝑦 2

= 𝑛(𝑛 − 1)𝑢

Solution: Here u is a homogenous function in two variables x, y of degree n
∴By Euler’s theorem for homogeneous function, we shall have
𝑥

𝜕𝑢
𝜕𝑢
+𝑦
= 𝑛 𝑢 … … … … … … … … …
...
(2)
𝜕𝑥
𝜕𝑥𝜕𝑦
𝜕𝑥

Again, differentiating equation (1) partially with respect to y, we get
𝑥

𝑦 2

{1 + ( ) }
𝑥

From equations (2) & (5) Euler theorem is verified
...
(3)
𝜕𝑥𝜕𝑦
𝜕𝑦
𝜕𝑦
𝜕𝑦
𝜕𝑥𝜕𝑦
𝜕𝑦
𝜕𝑦

Multiplying equation (2) and (3) by x & y respectively and then adding, we get
𝑥2

𝜕2𝑢
𝜕2𝑢
𝜕2𝑢
𝜕𝑢
𝜕𝑢
2
+
2𝑥𝑦
+
𝑦
= (𝑛 − 1)𝑥
+ (𝑛 − 1)𝑦
2
2
𝜕𝑥
𝜕𝑥𝜕𝑦
𝜕𝑦
𝜕𝑥
𝜕𝑦

𝑥2

𝜕2𝑢
𝜕2𝑢
𝜕2𝑢
𝜕𝑢
𝜕𝑢
2
+
2𝑥𝑦
+
𝑦
= (𝑛 − 1) {𝑥
+ 𝑦 } … … … … … … … … … … … … … … … (4)
2
2
𝜕𝑥
𝜕𝑥𝜕𝑦
𝜕𝑦
𝜕𝑥
𝜕𝑦

Using equation (1) in equation (4), we get
17

𝑦
( )
𝑥

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
𝝏𝟐 𝒖

𝝏𝟐 𝒖

𝝏𝟐 𝒖

𝒙𝟐 𝝏𝒙𝟐 + 𝟐𝒙𝒚 𝝏𝒙𝝏𝒚 + 𝒚𝟐 𝝏𝒚𝟐 = 𝒏(𝒏 − 𝟏)𝒖
...

𝑦

𝑥

Example-5:If 𝑢 = 𝑥 𝑛 𝑓 ( ) + 𝑦 −𝑛 𝜑 ( ), then show that
𝜕2 𝑢
𝑥 2 𝜕𝑥 2

𝜕2 𝑢

+ 2𝑥𝑦 𝜕𝑥𝜕𝑦 + 𝑦

𝑥
2
2𝜕 𝑢
𝜕𝑦 2

𝑦

𝜕𝑢

𝜕𝑢

+ 𝑥 𝜕𝑥 + 𝑦 𝜕𝑦 = 𝑛2 𝑢
...
(1)
𝑥
𝑢2 = 𝑦 −𝑛 𝜑 ( ) … … … … … … … … … … … … …
...


(3)

From equation (1) it is clear that 𝑢1 is a homogenous function in x, y of degree n, therefore from Euler’s theorem
𝑥

𝜕𝑢1
𝜕𝑢1
+𝑦
= 𝑛𝑢1 … … … … … … … … … … … … (4)
𝜕𝑥
𝜕𝑦

Also 𝑥 2

𝜕2 𝑢1
𝜕𝑥 2

𝜕2 𝑢

+ 2𝑥𝑦 𝜕𝑥𝜕𝑦1 + 𝑦 2

𝜕2 𝑢1
𝜕𝑦 2

= 𝑛 (𝑛 − 1)𝑢1 … … …
...
(7)

Adding equations (4) and (6) we get
𝑥

𝜕
𝜕
(𝑢 + 𝑢2 ) + 𝑦 (𝑢1 + 𝑢2 ) = 𝑛(𝑢1 − 𝑢2 )
𝜕𝑥 1
𝜕𝑦

0𝑟 𝑥

𝜕𝑢
𝜕𝑢
+𝑦
= 𝑛(𝑢1 − 𝑢2 ) … …
...
(9)

At last adding equations (8) & (9), we get

18

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
𝝏𝟐 𝒖

𝝏𝟐 𝒖

𝝏𝟐 𝒖

𝝏𝒖

𝝏𝒖

𝒙𝟐 𝝏𝒙𝟐 + 𝟐𝒙𝒚 𝝏𝒙𝝏𝒚 + 𝒚𝟐 𝝏𝒚𝟐 + 𝒙 𝝏𝒙 + 𝒚 𝝏𝒚 = 𝒏𝟐 𝒖
...


3
...
3
...


Proof: Since 𝑣 = 𝑓(𝑢) is given to be homogeneous function in variables x, y of degree n,
Therefore, by Euler’s theorem
𝑥

𝜕𝑣
𝜕𝑥

+𝑦

𝑓 ′ (𝑢) {𝑥

𝜕𝑣
𝜕𝑦

= 𝑛𝑣

Or 𝑥

𝜕
𝜕𝑥

𝑓(𝑢) + 𝑦

𝜕
𝜕𝑦

𝑓(𝑢) = 𝑛𝑓(𝑢) {since𝑣 = 𝑓(𝑢)

𝜕𝑢
𝜕𝑢
𝝏𝒖
𝝏𝒖
𝒇(𝒖)
+ 𝑦 } = 𝑛𝑓 (𝑢) 𝑂𝑟 𝒙
+𝒚
=𝒏 ′
𝜕𝑥
𝜕𝑦
𝝏𝒙
𝝏𝒚
𝒇 (𝒖)

Note: (i)If𝑓 (𝑢) be homogeneous function in three variables x, y, z
...

𝜕2 𝑢

𝜕2 𝑢

Example-6:If 𝑓(𝑢) be the homogenous function in variables x, y of degree n, then show that 𝑥 2 𝜕𝑥 2 + 2𝑥𝑦 𝜕𝑥𝜕𝑦 +
𝜕2 𝑢

𝑦 2 𝜕𝑦2 = 𝑔(𝑢){𝑔′ (𝑢) − 1}
...
(2)
𝜕𝑥
𝜕𝑦

Differentiating equation (2) partially with respect to x we get,
𝑥

𝜕 2 𝑢 𝜕𝑢
𝜕2𝑢
𝜕𝑢
𝜕2𝑢
𝜕2𝑢
𝜕𝑢
′( )
+
+
𝑦
=
𝑔
𝑢
𝑜𝑟
𝑥
+
𝑦
= {𝑔′ (𝑢) − 1}
… … … … … … … … …
...
(4)
𝜕𝑦𝜕𝑥 𝜕𝑦
𝜕𝑦
𝜕𝑦
𝜕𝑦𝜕𝑥
𝜕𝑦
𝜕𝑦

Multiplying equation (3) & (4) by x & y respectively, we get

19

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD

𝑥2

𝜕2𝑢
𝜕2𝑢
𝜕2𝑢
𝜕𝑢
𝜕𝑢
2
+
2𝑥𝑦
+
𝑦
= {𝑔′ (𝑢) − 1} (𝑥
+ 𝑦 ) … … … … … … … … … … … … (5)
2
2
𝜕𝑥
𝜕𝑥𝜕𝑦
𝜕𝑦
𝜕𝑥
𝜕𝑦

Using equation (2) in equation (5) we get
𝝏𝟐 𝒖

𝝏𝟐 𝒖

𝝏𝟐 𝒖

𝒙𝟐 𝝏𝒙𝟐 + 𝟐𝒙𝒚 𝝏𝒙𝝏𝒚 + 𝒚𝟐 𝝏𝒚𝟐 = 𝒈(𝒖){𝒈′ (𝒖) − 𝟏}
...


), then u is not homogenous function but

… … … … … … … … … … … …
...
So, from deduction to Euler’s theorem
𝜕𝑢

𝜕𝑢

𝑥 𝜕𝑥 + 𝑦 𝜕𝑦 = 𝑛
Therefore 𝑥

𝜕𝑢
𝜕𝑥

𝑓(𝑢)
𝑓 ′ (𝑢)

+𝑦

, Now since 𝑛 = 3 & 𝑓(𝑢) = 𝑒 𝑢

𝜕𝑢
𝜕𝑦

𝑒𝑢

=3

𝑜𝑟 𝒙

𝑒𝑢

𝝏𝒖
𝝏𝒙

+𝒚

𝝏𝒖
𝝏𝒚

𝑥 2+𝑦 2 +𝑧 2

= 𝟑
...

𝑥 2+𝑦 2 +𝑧 2

Solution: We have 𝑢 = sin−1 (𝑎𝑥+𝑏𝑦+𝑐𝑧 ), clearly u is not homogeneous function
But sin 𝑢 = (

𝑥 2+𝑦 2 +𝑧 2
𝑎𝑥+𝑏𝑦+𝑐𝑧

) … … … … … … … … … … … … … … … …
...
Therefore buy deduction to Euler’s theorem we
have 𝑥

𝜕𝑢
𝜕𝑥

+𝑦

𝜕𝑢
𝜕𝑦

+𝑧

𝜕𝑢
𝜕𝑧

=𝑛

𝑓(𝑢)
𝑓 ′ (𝑢)

… … … … … … … … …
...

𝜕𝑥
𝜕𝑦
𝜕𝑧
𝑐𝑜𝑠𝑢
𝝏𝒖

𝝏𝒖

𝝏𝒖

𝒙 𝝏𝒙 + 𝒚 𝝏𝒚 + 𝒛 𝝏𝒛 = 𝒕𝒂𝒏𝒖
...
(1)
20

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
1

Here u is not homogeneous but 𝑓 (𝑢) = 𝑠𝑖𝑛𝑢 will be homogeneous function in x, y of degree 𝑛 = − 12
...


), is not homogenous but 𝑓 (𝑢) = 𝑡𝑎𝑛𝑢 will be homogeneous function in x, y of

degree 𝑛 = 2
...
𝑐𝑜𝑠𝑢𝑢
𝜕𝑥
𝜕𝑦
𝑐𝑜𝑠𝑢
𝜕𝑥
𝜕𝑦
𝑥

𝜕𝑢
𝜕𝑢
+𝑦
= 𝑠𝑖𝑛2𝑢 … … … … … … … … …
...

1

Q-2: Verify Euler’s theorem for 𝑧 =

1

𝑥 3+𝑦 3
1

1


...
show that 𝑥 𝜕𝑥 + 𝑦 𝜕𝑦 + 𝑧 𝜕𝑧 + 3 𝑡𝑎𝑛𝑢 = 0
...


Ans: (i) 3u and (ii) 6u
...

1

Q-6: If 𝑢 = csc

1

−1 √𝑥 2 + 𝑦 2
1
1
𝑥3 + 𝑦3

, then evaluate 𝑥

𝜕𝑢
𝜕𝑥

+𝑦

𝜕𝑢
𝜕𝑦


...

Q-7: If 𝑢 = tan−1

𝑦2
𝑥

𝜕2 𝑢

𝜕2 𝑢

𝜕2 𝑢

, show that 𝑥 2 𝜕𝑥 2 + 2𝑥𝑦 𝜕𝑥𝜕𝑦 + 𝑦 2 𝜕𝑦2 = −𝑠𝑖𝑛2𝑢 𝑠𝑖𝑛2 𝑢
...
4
...
4
...
Taylor’s and Maclaurin’s theorem for function in one variable:
3
...
1
...
Taylor’s theorem for function of one variable:
If 𝑦 = 𝑓(𝑥) be a function in single variable x and all its derivatives up to desired order be finite and
continuous for value of x lying in 𝑎 ≤ 𝑥 ≤ 𝑎 + ℎ
...
𝑓 ′ (𝑎) + ℎ 2 𝑓 ′′ (𝑎) + ℎ 3 𝑓 ′′′ (𝑎) + ⋯ … … … … …
...
Taking 𝑎 + ℎ = 𝑥 𝑜𝑟 ℎ =
𝑥 − 𝑎 in equation (1) we get
𝑓 (𝑥) = 𝑓 (𝑎) +

1
1
1
(𝑥 − 𝑎)𝑓 ′ (𝑎) + (𝑥 − 𝑎)2 𝑓 ′′ (𝑎) + (𝑥 − 𝑎)3 𝑓 ′′′ (𝑎) + … (2)
1!
2!
3!

Here equation (2) represents Taylor’s series expansion of 𝑦 = 𝑓(𝑥) about the point 𝑥 = 𝑎 or in powers
of (𝑥 − 𝑎)
...
𝑓 ′ (𝑥) + ℎ 2 𝑓 ′′ (𝑥) + ℎ 3 𝑓 ′′′ (𝑥) + ⋯ (3)
1!
2!
3!
Equation (3) represent Taylor’s series expansion of 𝑓 (𝑥 + ℎ ) in powers of h
...
4
...
2
...
𝑓 ′ (0) + 𝑥 2
...
𝑓 ′′′ (0) + … (1)
1!
2!
3!
This is known as Maclaurin’s Theorem and series expansion represented by equation (1) is known as
Maclaurin’s series expansion of 𝑦 = 𝑓(𝑥) in powers of x
...

Example-1: Expand 𝑠𝑖𝑛𝑥 in powers of x up to fifth derivative terms
...
𝑓 ′ (0) + 𝑥 2
...
𝑓 ′′′ (0) + … … … … … …
...
(1) + 𝑥 2
...
(−1) + 𝑥 4
...
(1) + ⋯ …
...

Solution: we know that Taylor’s series expansion of a function about any point 𝑥 = 𝑎
is given by
1
1
1
(𝑥 − 𝑎)𝑓 ′ (𝑎) + (𝑥 − 𝑎)2 𝑓 ′′ (𝑎) + (𝑥 − 𝑎)3 𝑓 ′′′ (𝑎) + … … (1)
1!
2!
3!
𝜋
So, we take𝑎 = in this question
...
(2)

𝜋
1
𝑓 ′ (𝑥) = −𝑠𝑖𝑛𝑥 𝑜𝑟 𝑓 ′ ( ) = −
4
√2
𝜋
1
𝑓 ′′ (𝑥) = −𝑐𝑜𝑠𝑥 𝑜𝑟 𝑓 ′′ ( ) = −
4
√2
𝜋
1
𝑓 ′ ′′(𝑥) = 𝑠𝑖𝑛𝑥 𝑜𝑟 𝑓 ′′′ ( ) =
4
√2
𝜋
1
𝑓 ′𝑣 (𝑥) = 𝑐𝑜𝑠𝑥 𝑜𝑟 𝑓 ′𝑣 ( ) =
4
√2
𝜋
1
𝑓 𝑣 (𝑥) = −𝑠𝑖𝑛𝑥 𝑜𝑟 𝑓 𝑣 ( ) = −
4
√2
Putting all these values in equation (1) we get
1
1
𝜋
1
1
𝜋 2
1
1
𝜋 3 1
1
𝜋 4 1
𝑐𝑜𝑠𝑥 =
+ (𝑥 − ) (− ) + (𝑥 − ) (− ) + (𝑥 − ) ( ) + (𝑥 − ) ( )
4
2!
4
3!
4
4!
4
√2 1!
√2
√2
√2
√2
1
𝜋 5
1
+ (𝑥 − ) (− ) + ⋯
5!
4
√2
𝒄𝒐𝒔𝒙 =

𝟏
√𝟐

⌈𝟏 −

𝟏
𝝅
𝟏
𝝅 𝟐 𝟏
𝝅 𝟑 𝟏
𝝅 𝟒 𝟏
𝝅 𝟓
(𝒙 − ) − (𝒙 − ) + (𝒙 − ) + (𝒙 − ) − (𝒙 − ) + ⋯ … … … … … … ⌉
𝟏!
𝟒
𝟐!
𝟒
𝟑!
𝟒
𝟒!
𝟒
𝟓!
𝟒

Example-3: Expand 𝑓(𝑥) = log(1 + 𝑥) in powers of (𝑥 − 1)
...

Here we have𝑓(𝑥) = 𝑙𝑜𝑔𝑒 (1 + 𝑥) … … … … … … … … … … … … … … … … … … … … (2)
Now 𝑓(𝑥) = 𝑙𝑜𝑔𝑒 (1 + 𝑥) 𝑜𝑟 𝑓 (1) = 𝑙𝑜𝑔𝑒 2
Also differentiating equation (2) with respect to x we get
𝑓 (𝑥) = 𝑓 (𝑎) +

𝑓 ′ (𝑥 ) =

1
(1 + 𝑥)

𝑓 ′′ (𝑥) = −

𝑜𝑟 𝑓 ′ (1) =

1
2

1
1
𝑜𝑟 𝑓 ′′ (1) = −
2
(1 + 𝑥)
4
24

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
2
1
𝑜𝑟 𝑓 ′′′ (1) =
3
(1 + 𝑥)
4
Putting above values in equation (2), we get
𝑓 ′′′ (𝑥) =

𝑙𝑜𝑔𝑒 (1 + 𝑥) = 𝑙𝑜𝑔𝑒 2 +

1
1
1
1
1
1
(𝑥 − 1) ( ) + (𝑥 − 1)2 (− ) + (𝑥 − 1)3 ( ) + …
1!
2
2!
4
3!
4

𝟏
𝟏
𝟏
(𝒙 − 𝟏)𝟑 + …
𝒍𝒐𝒈𝒆 (𝟏 + 𝒙) = 𝒍𝒐𝒈𝒆 𝟐 + (𝒙 − 𝟏) − (𝒙 − 𝟏)𝟐 +
𝟐
𝟖
𝟐𝟒
𝜋
Example-4: Expand 𝑠𝑖𝑛 ( 4 + ℎ) in powers of h
...
𝑓 ′ (𝑎) + ℎ 2 𝑓 ′′ (𝑎) + ℎ 3 𝑓 ′′′ (𝑎) + ⋯ … … … … …
...
𝑓 ′ ( ) + ℎ 2 𝑓 ′′ ( ) + ℎ 3 𝑓 ′′′ ( ) + ⋯ … … … … …
...
𝑓 ′ (0) + 𝑥 2
...
𝑓 ′′′ (0) + … … … … … …
...
𝑡𝑎𝑛𝑥 𝑜𝑟 𝑓 ′ (𝑥) = 𝑡𝑎𝑛𝑥 𝑜𝑟 𝑓 ′ (0) = 𝑡𝑎𝑛0 = 0
𝑓 ′′ (𝑥) = 𝑠𝑒𝑐 2 𝑥 𝑜𝑟 𝑓 ′′ (0) = 1
𝑓 ′′′ (𝑥) = 2𝑠𝑒𝑐𝑥
...
𝑡𝑎𝑛𝑥 = 2𝑓 ′′ (𝑥)
...
𝑓 ′′′ (𝑥) + 2𝑓 𝑖𝑣 (𝑥)
...
𝑓 ′′ (𝑥) 𝑜𝑟 𝑓 𝑣 (𝑥) = 6𝑓 ′′ (𝑥)
...
𝑓 ′ (𝑥)
𝑓 𝑣 (0) = 0
2

𝑓 𝑣𝑖 (𝑥) = 6𝑓 ′′ (𝑥)
...
𝑓 ′′ (𝑥)
25

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
2

𝑓 𝑣𝑖 (𝑥) = 8𝑓 ′′ (𝑥)
...
𝑓 𝑖𝑣 (0) + 6(𝑓 ′′′ (0)) + 2𝑓 𝑣 (0)𝑓 ′ (0) 𝑜𝑟 𝑓 𝑣𝑖 (0) = 16
Therefore, from equation (1),
1
1
1
1
1
1
log 𝑠𝑒𝑐𝑥 = 0 + 𝑥
...


3
...
2
...
Taylor’s Theorem for function of two variables: If 𝑓(𝑥, 𝑦) be any function in two variables x
and y such that f(x, y) and all its partial derivatives up to desired order are finite and continuous for any point
(x, y) subject to 𝑎 ≤ 𝑥 ≤ 𝑎 + ℎ, 𝑏 ≤ 𝑦 ≤ 𝑏 + 𝑘 , then
1 𝜕𝑓
𝜕𝑓
1
𝜕2𝑓
𝜕2𝑓
𝜕 2𝑓
𝑓(𝑎 + ℎ, 𝑏 + 𝑘) = 𝑓 (𝑎, 𝑏) + {ℎ
+ 𝑘 } + {ℎ 2 2 + 2ℎ𝑘
+ 𝑘2 2 }
1! 𝜕𝑥
𝜕𝑦
2!
𝜕𝑥
𝜕𝑥𝜕𝑦
𝜕𝑦
3
3
3
3
1
𝜕 𝑓
𝜕 𝑓
𝜕 𝑓
𝜕 𝑓
+ {ℎ 3 3 + +3ℎ 2 𝑘 2 + 3ℎ𝑘 2
+ +𝑘 3 3 } + … … … … …
...
Also, in this
series values of all order partial derivatives are determined at point (a, b)
...
(2)

Here equation (2) represents Taylor’s series expansion of a function 𝑓 (𝑥, 𝑦) about point (a, b)
...

Note: For simplicity sometime we represent first order derivatives
𝜕2 𝑓

𝜕2 𝑓

𝜕𝑓
𝜕𝑥

𝜕𝑓

𝑏𝑦 𝑓𝑥 , 𝜕𝑦 𝑏𝑦 𝑓𝑦 and second order

𝜕2 𝑓

derivatives 𝜕𝑥 2 𝑏𝑦 𝑓𝑥 2 , 𝜕𝑥𝜕𝑦 𝑏𝑦 𝑓𝑥𝑦 , 𝜕𝑦2 𝑏𝑦 𝑓𝑦2 and so on other higher order derivatives terms are
represented
...
(3)
2
3!
𝜕𝑥
𝜕𝑥 𝜕𝑦
𝜕𝑥𝜕𝑦
𝜕𝑦

𝑓 (𝑥 + ℎ, 𝑦 + 𝑘) = 𝑓(𝑥, 𝑦) +

Note: In Taylor’s series expansion represented by equation (3), values of all order partial derivatives are
determined at point (x, y)
...
4
...
2
...

26

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
The above series is known as Maclaurin’s series expansion of given function 𝑓 (𝑥, 𝑦)
...

In other words, we can say that Maclaurin’s series is nothing but Taylor’s series expansion about the point
(0,0)
...

Solution: We know that Taylor’s series expansion of any function 𝑓(𝑥, 𝑦) in powers of
(𝑥 − 𝑎)& (𝑦 − 𝑏) is given by
𝑓 (𝑥, 𝑦) = 𝑓(𝑎, 𝑏) +
𝜕2 𝑓
𝑏)2 𝜕𝑦2 }

1
1!

{(𝑥 − 𝑎)

𝜕𝑓

𝜕𝑓

𝜕3 𝑓

1

𝜕2 𝑓

1

𝜕2 𝑓

+ (𝑦 − 𝑏) 𝜕𝑦 } + 2! {(𝑥 − 𝑎)2 𝜕𝑥 2 + 2(𝑥 − 𝑎)(𝑦 − 𝑏) 𝜕𝑥𝜕𝑦 + (𝑦 −
𝜕𝑥
𝜕3 𝑓

𝜕3 𝑓

𝜕3 𝑓

+ 3! {(𝑥 − 𝑎)3 𝜕𝑥 3 + +3(𝑥 − 𝑎)2 (𝑦 − 𝑏) 𝜕𝑥 2𝜕𝑦 + 3(𝑥 − 𝑎)(𝑦 − 𝑏)2 𝜕𝑥𝜕𝑦2 + +(𝑦 − 𝑏)3 𝜕𝑦3 }

+
...
Now from
equation (2),
𝑓 (𝑥, 𝑦) = 𝑥 2 𝑦 + 3𝑦 − 2 𝑜𝑟 𝑓(1, −2) = −10
𝜕𝑓
𝜕𝑥
𝜕𝑓
𝜕𝑦

= 2𝑥𝑦 𝑜𝑟

𝜕𝑓
𝜕𝑥

= 𝑥 2 + 3 𝑜𝑟

(1, −2) = −4
𝜕𝑓
𝜕𝑦

(1, −2) = 4

𝜕2𝑓
𝜕2𝑓
𝜕2𝑓
𝜕2𝑓
𝜕2𝑓
𝜕 2𝑓
(
)
= 2𝑦 𝑜𝑟
1, −2 = −4,
= 2𝑥 𝑜𝑟
= 2, 2 = 0 𝑜𝑟 2 (1, −2) = 0
𝜕𝑥 2
𝜕𝑥 2
𝜕𝑥𝜕𝑦
𝜕𝑥𝜕𝑦
𝜕𝑦
𝜕𝑦
𝜕3𝑓
𝜕3𝑓
𝜕3𝑓
𝜕3𝑓
𝜕2𝑓
(
)
(
)
(1, −2) = 0
(1,
−2)
=
0,
=
2
𝑜𝑟
1,
−2
=
2,
1,
−2
=
0,
𝜕𝑥 3
𝜕𝑥 2 𝜕𝑦
𝜕𝑥 2 𝜕𝑦
𝜕𝑥𝜕𝑦 2
𝜕𝑦 3
Putting all these values in equation (1), we get
1

1

𝑥 2 𝑦 + 3𝑦 − 2 = −10 + 1! {(𝑥 − 1)(−4) + (𝑦 + 2)(4)}+ 2! {(𝑥 − 1)2 (−4) + 2(𝑥 − 1)(𝑦 + 2)(2) +
1

(𝑦 + 2)2 (0)} + {(𝑥 − 1)3 (0) + 3(𝑥 − 1)2 (𝑦 + 2)(2) + 3(𝑥 − 1)(𝑦 + 2)2 (0) + (𝑦 + 2)3 (0)}
3!
𝒙𝟐 𝒚 + 𝟑𝒚 − 𝟐 = −𝟏𝟎 − 𝟒(𝒙 − 𝟏) + 𝟒(𝒚 + 𝟐) − 𝟐(𝒙 − 𝟏)𝟐 + 𝟐(𝒙 − 𝟏)(𝒚 + 𝟐) + (𝒙 − 𝟏)𝟐 (𝒚 + 𝟐)
...

Solution: By Taylor’s series we have
𝑓 (𝑥, 𝑦) = 𝑓(𝑎, 𝑏) +
𝜕2 𝑓
𝑏)2 𝜕𝑦2 }

1

1
1!

{(𝑥 − 𝑎)

𝜕𝑓

𝜕𝑓

1

𝜕2 𝑓

𝜕2 𝑓

+ (𝑦 − 𝑏) 𝜕𝑦 } + 2! {(𝑥 − 𝑎)2 𝜕𝑥 2 + 2(𝑥 − 𝑎)(𝑦 − 𝑏) 𝜕𝑥𝜕𝑦 + (𝑦 −
𝜕𝑥

𝜕3 𝑓

𝜕3 𝑓

𝜕3 𝑓

𝜕3 𝑓

+ 3! {(𝑥 − 𝑎)3 𝜕𝑥 3 + +3(𝑥 − 𝑎)2 (𝑦 − 𝑏) 𝜕𝑥 2𝜕𝑦 + 3(𝑥 − 𝑎)(𝑦 − 𝑏)2 𝜕𝑥𝜕𝑦2 + +(𝑦 − 𝑏)3 𝜕𝑦3 } +
...
(2),
27

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
To get required series expansion, we take a=b=1and find all values at (1, 1)
Now from equation (2), 𝑓(𝑥, 𝑦) = 𝑥 𝑦 𝑜𝑟 𝑓(1,1) = 1
...

𝑥
𝑜𝑟
𝜕𝑥𝜕𝑦 2
𝜕𝑥𝜕𝑦 2
𝜕3𝑓
𝜕3𝑓
𝑦(
3
)
(1,1) = 0
=
𝑥
𝑙𝑜𝑔𝑥
𝑜𝑟
𝜕𝑦 3
𝜕𝑦 3
Using all these values in equation (1) we get
𝟏
𝒙𝒚 = 𝟏 + (𝒙 − 𝟏) + (𝒙 − 𝟏)(𝒚 − 𝟏) + (𝒙 − 𝟏)𝟐 (𝒚 − 𝟏) + ⋯ … … … … … …
...
Hence, computef(1
...
9)
...
(1)
Therefore, in this problem we select 𝑎 = 1, 𝑏 = 1
...
1 − 𝑥
...

2
𝒚

𝝅

𝟏

𝟏

𝟏

𝟏

𝐟(𝐱, 𝐲) = 𝐭𝐚𝐧−𝟏 (𝒙) = 𝟒 − 𝟐 (𝒙 − 𝟏) + 𝟐 (𝒚 − 𝟏) + 𝟒 (𝒙 − 𝟏)𝟐 − 𝟒 (𝒚 − 𝟏)𝟐 +
...
1 & 𝑦 = 0
...
1, 0
...
1 − 1) + (0
...
1 − 1)2 − (0
...

Example-9: Expand

(𝑥+ℎ)(𝑦+𝑘)
(𝑥+ℎ)+(𝑦+𝑘)

in powers of h and k up to and inclusive second-degree terms
...
(1)
1! 𝜕𝑥
𝜕𝑦
2!
𝜕𝑥
𝜕𝑥𝜕𝑦
𝜕𝑦
(𝑥+ℎ)(𝑦+𝑘)

𝑥𝑦

Here let us take 𝑓(𝑥 + ℎ, 𝑦 + 𝑘) = (𝑥+ℎ)+(𝑦+𝑘)such that 𝑓 (𝑥, 𝑦) = 𝑥+𝑦 … … … (2)
Differentiating equation (2) partially with respect to x and y we get
𝜕𝑓 (𝑥 + 𝑦)𝑦 − 𝑥𝑦
...
1
=
(𝑥 + 𝑦)2
𝜕𝑦

𝑜𝑟

𝜕𝑓
𝑥2
=
𝜕𝑦 (𝑥 + 𝑦)2

𝜕2 𝑓

𝜕2 𝑓

𝜕𝑥 2

and

𝑦 2 {−2}

2𝑦 2

= (𝑥+𝑦)3 = − (𝑥+𝑦)3 ,
𝜕2 𝑓
𝜕𝑥𝜕𝑦

=

𝜕𝑦 2

𝑥 2{−2}

(𝑥+𝑦)2 (2𝑥)−𝑥 2 {2(𝑥+𝑦)}
(𝑥+𝑦)4

2𝑥 2

= (𝑥+𝑦)3 = − (𝑥+𝑦)3
𝑜𝑟

𝜕2 𝑓
𝜕𝑥𝜕𝑦

=

2𝑥𝑦
(𝑥+𝑦)3

Putting all the values in equation (1), we get
(𝑥 + ℎ )(𝑦 + 𝑘)
𝑥𝑦
1
𝑦2
𝑥2
}
=
+ {ℎ
+
𝑘
(𝑥 + ℎ ) + (𝑦 + 𝑘) 𝑥 + 𝑦 1! (𝑥 + 𝑦)2
(𝑥 + 𝑦)2
1 2 −2𝑦 2
2𝑥𝑦
−2𝑥 2
2
{ℎ
} + … … … … … …
...

+ 2ℎ𝑘
+𝑘
(𝑥 + 𝑦)3
(𝑥 + 𝑦)3
(𝑥 + 𝑦)3
2!
29

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD

(𝑥 + ℎ )(𝑦 + 𝑘)
𝑥𝑦
𝑦2
𝑥2
𝑦2
2𝑥𝑦
𝑥2
2
=
+
ℎ+
𝑘−
ℎ +
ℎ𝑘 −
𝑘2 + ⋯
(𝑥 + ℎ ) + (𝑦 + 𝑘) 𝑥 + 𝑦 (𝑥 + 𝑦)2
(𝑥 + 𝑦)2
(𝑥 + 𝑦)3
(𝑥 + 𝑦)3
(𝑥 + 𝑦)3
Example-10: Expand 𝑒 𝑎𝑥 sin 𝑏𝑦 in powers of x and y as far as terms of third degree
...


𝜕2 𝑓
𝜕𝑥 2

+ 2𝑥𝑦

𝜕2 𝑓
𝜕𝑥𝜕𝑦

+ 𝑦2

𝜕2 𝑓
𝜕𝑦 2

}+

1
3!

{𝑥 3

𝜕3 𝑓
𝜕𝑥 3

+ +3𝑥 2 𝑦

(1)
...

So, let 𝑓(𝑥, 𝑦) = 𝑒 𝑎𝑥 sin 𝑏𝑦

𝑜𝑟 𝑓(0, 0) = 0

On differentiating 𝑓 (𝑥, 𝑦) partially with respect to x and y we get
𝜕𝑓
= 𝑎𝑒 𝑎𝑥 sin 𝑏𝑦 𝑜𝑟 𝑓 (0,0) = 0
𝜕𝑥
𝜕𝑓
𝜕𝑓
(0,0) = 𝑏
= 𝑏𝑒 𝑎𝑥 cos 𝑏𝑦 𝑜𝑟
𝜕𝑦
𝜕𝑦
𝜕2𝑓
𝜕 2𝑓
2 𝑎𝑥
(0,0) = 0
=
𝑎
𝑒
sin
𝑏𝑦
𝑜𝑟
𝜕𝑥 2
𝜕𝑥 2
𝜕2𝑓
𝜕2𝑓
2 𝑎𝑥
(0,0) = 0
=
−𝑏
𝑒
sin
𝑏𝑦
𝑜𝑟
𝜕𝑦 2
𝜕𝑦 2
𝜕2𝑓
𝜕2𝑓
= 𝑎𝑏𝑒 𝑎𝑥 cos 𝑏𝑦 𝑜𝑟
= 𝑎𝑏
𝜕𝑥𝜕𝑦
𝜕𝑥𝜕𝑦
𝜕3𝑓
𝜕 3𝑓
3 𝑎𝑥
(0,0) = 0
=
𝑎
𝑒
sin
𝑏𝑦
𝑜𝑟
𝜕𝑥 3
𝜕𝑥 3
𝜕3𝑓
𝜕3𝑓
3 𝑎𝑥
(0,0) = −𝑏3
=
−𝑏
𝑒
cos
𝑏𝑦
𝑜𝑟
𝜕𝑦 3
𝜕𝑦 3
𝜕3𝑓
𝜕3𝑓
2
𝑎𝑥
(0,0) = 𝑎2 𝑏
=
𝑎
𝑏𝑒
cos
𝑏𝑦
𝑜𝑟
𝜕𝑥 2 𝜕𝑦
𝜕𝑥 2 𝜕𝑦
𝜕3𝑓
𝜕3𝑓
2 𝑎𝑥
(0,0) = 0
=
−𝑎𝑏
𝑒
sin
𝑏𝑦
𝑜𝑟
𝜕𝑥𝜕𝑦 2
𝜕𝑥𝜕𝑦 2
Putting all these values in equation (1), we get
𝑒

𝑎𝑥

1
1
1
𝜕3𝑓
2
2
{0
{0
{0
sin 𝑏𝑦 = 0 +
+ 𝑏𝑦} +
+ 2𝑎𝑏𝑥𝑦 + 0)} +
+ +3𝑎 𝑏
...
𝑥 𝑦 − 𝑦3 + ⋯
2
6
30

𝜕3 𝑓
𝜕𝑥 2𝜕𝑦

+

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD

Practice Exercise:
Q-1: State Taylor’s series in powers of (𝑥 − 𝑎) for function of one variable
...

𝜋

Q-3: Obtain Taylor’s series expansion of 𝑠𝑖𝑛𝑥 in powers of (𝑥 − 2 )
...

Ans: 𝒆𝒙 𝒄𝒐𝒔𝒚 =

𝒆
√𝟐

𝝅

(𝒙−𝟏)𝟐

𝟒

𝟐

[𝟏 + (𝒙 − 𝟏) − (𝒚 − ) +

𝝅 𝟐

𝝅

− (𝒙 − 𝟏) (𝒚 − 𝟒 ) − (𝒚 − 𝟒 ) + … ]

Q-7: Expand 𝑦 𝑥 about (1, 1) up to second degree terms and hence evaluate(1
...
03
...
&(1
...
03 = 1
...

Ans: 𝒄𝒐𝒔𝒙 𝒄𝒐𝒔𝒚 = 𝟏 −

𝒙𝟐
𝟐

−

𝒚𝟐
𝟐

𝒙𝟒

+ 𝟐𝟒 +

𝒙𝟐 𝒚 𝟐
𝟒

+

𝒚𝟒
𝟐𝟒

+ ………

Q-9: Obtain linearized form 𝑇(𝑥, 𝑦) of the function 𝑓(𝑥, 𝑦) = 𝑥 2 − 𝑥𝑦 +

𝑦2
2

+ 3 at the point

(3, 2)using Taylor’s series expansion
...

Q-10: Expand 𝑒 −(𝑥
𝟐 +𝒚𝟐 )

Ans: 𝒆−(𝒙

2 +𝑦 2 )

cos 𝑥𝑦 about the point (0, 0) up to second order derivative terms
...


31

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
3
...


Maxima and minima of function of two variables

A function 𝑓 (𝑥, 𝑦) is said to have a maximum value at 𝑥 = 𝑎, 𝑦 = 𝑏 if
𝑓(𝑎, 𝑏) > 𝑓(𝑎 + ℎ, 𝑏 + 𝑘), for small and independent values of ℎ and 𝑘, positive or negative
...

Thus, f (x, y) has a maximum or minimum value at a point (𝑎, 𝑏) according as
𝐹 = 𝑓(𝑎 + ℎ, 𝑏 + 𝑘) − 𝑓(𝑎, 𝑏) < 𝑜𝑟 > 0
The maximum and minimum value of a function is called its extreme value
...
5
...
Rule to find the extreme values of a function 𝒛 = 𝒇(𝒙, 𝒚)
𝜕𝑧

𝜕𝑧

1
...
Solve 𝜕𝑥 = 0 by 𝜕𝑦 = 0 simultanously
Let (𝑎, 𝑏), (𝑐, 𝑑) …
...
For each solution in step 2 find𝑟 = 𝜕𝑥 2, 𝑠 = 𝜕𝑥𝜕𝑦 𝑡 = 𝜕𝑦 2
4
...

(b) If 𝑟𝑡 − 𝑠 2 > 0 and 𝑟 > 0 for a particular (𝑎, 𝑏) of step 2, then 𝑧 has minimum value at (𝑎, 𝑏)
(c) If 𝑟𝑡 − 𝑠 2 < 0 for a particular (𝑎, 𝑏) of step 2, then 𝑧 has no extreme value at (𝑎, 𝑏)
3
...
2
...
Find 𝜕𝑥 , 𝜕𝑦 and
2
...


𝜕𝑧

𝜕𝑧 2

,

𝜕2𝑢

,

𝜕2𝑢

,

𝜕2𝑢

𝜕𝑦𝜕𝑧 𝜕𝑧𝜕𝑥 𝜕𝑥𝜕𝑦


...


𝐴 𝐻 𝐺
𝐻
| = 𝐷1 and find |𝐻 𝐵 𝐹 | = 𝐷2
𝐵
𝐺 𝐹 𝐶
The given function u will be minimum if 𝐴 > 0, 𝐷1 > 0, 𝐷2 > 0
And will have a maximum if 𝐴 < 0, 𝐷1 > 0, 𝐷2 < 0
...
If above conditions are not satisfied, we have neither maximum nor minimum
...
Also find 𝐴𝐵 − 𝐻 = |
𝐻
2

32

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
Example 1
...
Let 𝑓(𝑥, 𝑦) = 𝑥 2 + 𝑦 2 + 6𝑥 + 12
By taking the partial derivative with respect to x and y respectively
𝜕𝑓
𝜕𝑥

𝜕 2𝑓

𝜕𝑓

𝜕 2𝑓

𝜕2𝑓

= 2𝑥 + 6, 𝜕𝑦 = 2𝑦, 𝑟 = 𝜕2 𝑥 = 2, 𝑠 = 𝜕𝑥𝜕𝑦 = 0 and 𝑡 = 𝜕2𝑦 = 2

By equating the first order partial derivative to 0, we get
2𝑥 + 6 = 0 … … … … … … … … … … … (1) and
2𝑦 = 0 … … … … … … … … … … … …
...

Solution: Here 𝑓(𝑥, 𝑦) = 𝑥 3 + 𝑦 3 − 3𝑎𝑥𝑦
𝑓𝑥 = 3𝑥 2 − 3𝑎𝑦, 𝑓𝑦 = 3𝑦 2 − 3𝑎𝑥, 𝑟 = 𝑓𝑥𝑥 = 6𝑥, 𝑠 = 𝑓𝑥𝑦 = −3𝑎, 𝑡 = 𝑓𝑦𝑦 = 6𝑦
By equating the first order partial derivative to zero
𝑥 2 − 𝑎𝑦 = 0 … … … … … … …
...
(2)
From equation (1) 𝑦 =

𝑥2
𝑎

Put the value of y in equation (2)

𝑥4
𝑎2

− 𝑎𝑥 = 0 𝑜𝑟 𝑥(𝑥 3 − 𝑎3 ) 𝑜𝑟 𝑥 = 0, 𝑎

When 𝑥 = 0, 𝑦 = 0; when 𝑥 = 𝑎, 𝑦 = 𝑎
There are two stationary points (0, 0) and (𝑎, 𝑎)
Now, 𝑟𝑡 − 𝑠 2 = 36𝑥𝑦 − 9𝑎2
At (𝟎, 𝟎)𝑟𝑡 − 𝑠 2 = −9𝑎2 < 0
It means 𝑓(𝑥, 𝑦) has no extreme value at (0, 0)
33

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
At (𝒂, 𝒂)𝑟𝑡 − 𝑠 2 = 27𝑎2 > 0
𝑓(𝑥, 𝑦) has extreme value at (0, 0)
𝑟 = 6𝑎
if 𝑎 > 0, 𝑟 > 0 so that 𝑓(𝑥, 𝑦)has a minimum value at (𝑎, 𝑎) and minimum value = −𝑎3
if 𝑎 > 0, 𝑟 < 0 so that 𝑓(𝑥, 𝑦)has a maximum value at (𝑎, 𝑎)and maximum value = 𝑎3
Example 3:Examine for minimum and maximum values 𝑠𝑖𝑛 𝑥 + 𝑠𝑖𝑛 𝑦 + 𝑠𝑖𝑛 (𝑥 + 𝑦)
Solution: Here 𝑓(𝑥, 𝑦) = 𝑠𝑖𝑛 𝑥 + 𝑠𝑖𝑛 𝑦 + 𝑠𝑖𝑛 (𝑥 + 𝑦)
𝑓𝑥 = cos 𝑥 + cos(𝑥 + 𝑦)
𝑓𝑦 = cos 𝑦 + cos(𝑥 + 𝑦)
𝑟 = 𝑓𝑥𝑥 = −sin 𝑥 − sin(𝑥 + 𝑦)
𝑠 = 𝑓𝑥𝑦 = cos 𝑥 − sin(𝑥 + 𝑦)
𝑡 = 𝑓𝑦𝑦 = −sin 𝑦 − sin(𝑥 + 𝑦)
By equating the first order partial derivative to zero
cos 𝑥 + cos(𝑥 + 𝑦) = 0 …
...
(2)
By subtracting equation (2) from (1)
𝑐𝑜𝑠 𝑥 = 𝑐𝑜𝑠 𝑦 𝑜𝑟 𝑥 = 𝑦
put the value in equ (1) 𝑐𝑜𝑠 2𝑥 = −𝑐𝑜𝑠𝑥 = 𝑐𝑜𝑠(𝜋 − 𝑥 )
𝑥=

𝜋
=𝑦
3
𝜋 𝜋

The stationary point is ( 3 , 3 )
𝜋 𝜋

At ( 3 , 3 ) 𝑟 = −√3, 𝑠 =
𝑟𝑡 − 𝑠 2 =

√3
,𝑡
2

= −√3

9
> 0 𝑎𝑛𝑑 𝑟 < 0
4
𝜋 𝜋

So 𝑓(𝑥, 𝑦) has a maximum value at ( 3 , 3 ) and maximum value is

34

3 √3
2


...
In a plane triangle ABC, find the maximum value of 𝑐𝑜𝑠𝐴 𝑐𝑜𝑠𝐵 𝑐𝑜𝑠𝐶
...
In a triangle 𝐴 + 𝐵 + 𝐶 = 𝜋
Convert the given function into two variables A and B
So cos 𝐴 cos 𝐵 cos 𝐶 = −𝑐𝑜𝑠𝐴 cos 𝐵 𝑐𝑜𝑠 (𝜋 − 𝐴 − 𝐵) = 𝑓(𝐴, 𝐵)
Taking the partial derivative of 𝑓
𝜕𝑓
= − cos 𝐵 [− sin 𝐴 cos(𝐴 + 𝐵) − cos 𝐴 𝑠𝑖𝑛(𝐴 + 𝐵)] = cos 𝐵 sin(2𝐴 + 𝐵)]
𝜕𝐴
𝜕𝑓
= − cos 𝐴[− sin 𝐵 cos(𝐴 + 𝐵) − cos 𝐵 𝑠𝑖𝑛(𝐴 + 𝐵)] = cos 𝐴 sin(𝐴 + 2𝐵)]
𝜕𝐵
𝑟 = 2 cos 𝐵 cos(2𝐴 + 𝐵), 𝑠 =

cos(2𝐴 + 2𝐵), 𝑡 = 2 cos 𝐴 cos(𝐴 + 2𝐵)

By equating the first order partial derivative to zero
𝜕𝑓
𝜕𝐴

= 0 and

𝜕𝑓
𝜕𝐵

=0

𝑐𝑜𝑠 𝐵 𝑠𝑖𝑛(2𝐴 + 𝐵) = 0 … … … … … … … … …
...
(3)
𝑠𝑖𝑛(𝐴 + 2𝐵) = 0 𝑜𝑟 𝐴 + 2𝐵 = 𝜋…(4)
solving equation (3) and (4) 𝐴 = 𝐵 =
𝜋

𝜋

𝜋

𝜋

𝜋
3

so ( 3 , 3 ) is the stationary point
At ( 3 , 3 )
1
𝑟 = −1, 𝑠 = − , 𝑡 = −1
2

35

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD

𝑟𝑡 − 𝑠 2 =

3
> 0 𝑎𝑛𝑑 𝑟 = −1 < 0
4
𝜋

So 𝑓(𝐴, 𝐵) is maximum at 𝐴 = 𝐵 =

1

and maximum value is 8
...
A rectangular box, open at the top, is to have a given capacity
...

Solution
...
Let 𝑉 be the
given capacity and 𝑆 is surface
Capacity is given so V is constant
𝑉

𝑉 = 𝑥𝑦𝑧 or 𝑧 = 𝑥𝑦
𝑆 = 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 = 𝑥𝑦 +

2𝑉 2𝑉
+
= 𝑓(𝑥, 𝑦)
𝑦
𝑥

By taking the partial derivative
2𝑉

2𝑉

𝑓𝑥 = 𝑦 − 𝑥 2 , 𝑓𝑦 = 𝑥 − 𝑦 2 , 𝑟 = 𝑓𝑥𝑥 =

4𝑉
𝑥3

, 𝑠 = 𝑓𝑥𝑦 = 1, 𝑡 = 𝑓𝑦𝑦 =

4𝑉
𝑦3

By equating the first order partial derivative to zero
2𝑉

and

𝑓𝑥 = 𝑦 − 𝑥 2 = 0…
...
(2)
From equation (1) 𝑦 =

2𝑉
𝑥2
𝑥4

Put the value of y in equation (2) we get 𝑥 − 2𝑉 4𝑉 2 = 0
𝑥3
𝑥 (1 − ) = 0
2𝑉
𝑥 = (2𝑉 )

1⁄
3

and 𝑦 =

Hence ((2𝑉 )
At ((2𝑉 )

2𝑉
𝑥2

so 𝑦 = (2𝑉 )

1⁄
1
3 , (2𝑉 ) ⁄3 )

1⁄
3

is the stationary point

1⁄
1
3 , (2𝑉 ) ⁄3 )

𝑟 = 2, 𝑠 = 1 and 𝑡 = 2
𝑟𝑡 − 𝑠 2 = 3 > 0 𝑎𝑛𝑑 𝑟 = 2 > 0
So 𝑓(𝑥, 𝑦) has minimum value at 𝑥 = 𝑦 = (2𝑉 )

1⁄
3 and𝑧

36

𝑉

𝑦

= 𝑥𝑦 = 2

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD

Example 6
...

𝑥−3
1

=

𝑦−5
−2

=

𝑧−7
1

= 𝛿 and

𝑥+1
7

=

𝑦+1
−6

=

𝑧+1
1

=𝜇

So 𝑃 (𝛿 + 3, 5 − 2𝛿, 7 + 𝛿 ), 𝑄(7𝜇 − 1, −1 − 6𝜇, 𝜇 − 1)are two points on the first and second line
respectively
...

𝑓(𝛿, 𝜇) = 𝑃𝑄2 = 6𝛿 2 + 86𝜇2 − 40𝛿𝜇 + 116
Taking the partial derivatives
𝜕𝑓
= 12𝛿 − 40𝜇
𝜕𝛿
𝜕𝑓
= 172𝜇 − 40𝛿
𝜕𝜇
𝑟=

𝜕2𝑓
= 12
𝜕𝛿 2

𝑠=

𝜕2𝑓
= −40
𝜕𝛿𝜕𝜇

𝑡=

𝜕2𝑓
= 172
𝜕𝜇2

Equate the first order partial derivative to zero
𝜕𝑓
= 12𝛿 − 40𝜇 = 0 … …
...
6
...
This method cannot
determine the extreme value
...
6
...
Working procedure of Lagrange’s method of undetermined multipliers
Let 𝑓(𝑥, 𝑦, 𝑧) be a function to be examined for extreme value and let the variables x, y and z are connected
by the relation 𝜑(𝑥, 𝑦, 𝑧) = 0 … (1)
I
...


Consider the Lagrange’s function 𝑔(𝑥, 𝑦, 𝑧) = 𝑓 (𝑥, 𝑦, 𝑧) + 𝜇 𝜑(𝑥, 𝑦, 𝑧)
To find the stationary value of 𝑔(𝑥, 𝑦, 𝑧), we take 𝑑𝑔 = 0
𝜕𝑓

𝜕𝜑

𝜕𝑓

𝜕𝜑

𝜕𝑓

𝜕𝜑

Such that (𝜕𝑥 + 𝜇 𝜕𝑥 ) 𝑑𝑥 + (𝜕𝑦 + 𝜇 𝜕𝑦 ) 𝑑𝑦 + ( 𝜕𝑧 + 𝜇 𝜕𝑧 ) 𝑑𝑧 = 0
The above equation holds well if
(

𝜕𝑓
𝜕𝜑
𝜕𝑓
𝜕𝜑
𝜕𝑓
𝜕𝜑
+ 𝜇 ) 𝑑𝑥 = 0, ( + 𝜇 ) 𝑑𝑦 = 0, ( + 𝜇 ) 𝑑𝑧 = 0
𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧

These equations with equation (1) give the values of 𝑥, 𝑦, 𝑧 and 𝜇 for extreme value
...
Prove that their product is maximum when they
are equal
...
(1)
and 𝑓 = 𝑥𝑦𝑧 …
...
(3)
To find the stationary value differentiates equation (3) w
...
t
...
…(4)
𝑧𝑥 + 𝜇 = 0…
...
(6)
Multiply (4), (5) and (6) by 𝑥, 𝑦 𝑎𝑛𝑑 𝑧 respectively and adding
3𝑥𝑦𝑧 + 𝜇 (𝑥 + 𝑦 + 𝑧) = 0
38

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
From equation (1) and (2), we get
3𝑓 + 𝜇𝑘 = 0
𝜇=−

3𝑓
𝑘

Substitute the value of 𝜇 in equ
...

Let 𝑧 be a function of 𝑥, 𝑦
...

𝑓 = 𝑥𝑦(𝑘 − 𝑥 − 𝑦)
𝜕𝑓
= 𝑘𝑦 − 2𝑥𝑦 − 𝑦 2
𝜕𝑥
𝜕𝑓
= 𝑘𝑥 − 2𝑥𝑦 − 𝑥 2
𝜕𝑦
𝜕2𝑓
𝑟 = 2 = −2𝑦
𝜕𝑥
𝑠=

𝜕2𝑓
= 𝑘 − 2𝑥 − 2𝑦
𝜕𝑥𝜕𝑦

𝜕2𝑓
𝑡 = 2 = −2𝑥
𝜕𝑦
𝑘 𝑘 𝑘

At stationary point (3 , 3 , 3 )
𝑟𝑡 − 𝑠 2 =

𝑘2
2𝑘
> 0 𝑎𝑛𝑑 𝑟 = −
<0
3
3

So 𝑓(𝑥, 𝑦) has maximum value

𝑘3
27

𝑘 𝑘 𝑘

at stationary point(3 , 3 , 3 )
...

Solution: Let 2𝑥, 2𝑦 and 2𝑧 be the length, breadth, and height of the rectangular solid and 𝑟 be the radius of
the sphere
...
(1)
And 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑟 2 …
...
(3)
to find the stationary point, 𝑑𝑔 = 0
(8𝑦𝑧 + 𝜇2𝑥 )𝑑𝑥 + (8𝑧𝑥 + 𝜇2𝑦)𝑑𝑦 + (8𝑥𝑦 + 𝜇2𝑧)𝑑𝑧 = 0
8𝑦𝑧 + 𝜇2𝑥 = 0…
...
(5)
8𝑥𝑦 + 𝜇2𝑧 = 0…
...


Example 9: The temperature T at any point (𝑥, 𝑦, 𝑧) in space is 𝑇 = 400𝑥𝑦𝑧 2
...

Solution: Temperature T at any point (𝑥, 𝑦, 𝑧) in space is 𝑇 = 400𝑥𝑦𝑧 2 …
...
(2)

the Lagrange’s function 𝑔(𝑥, 𝑦, 𝑧) = 400𝑥𝑦𝑧 2 + 𝜇(𝑥 2 + 𝑦 2 + 𝑧 2 − 1) …
...
(4)
400𝑥𝑧 2 + μ2y = 0…
...
(6)
Multiply (4), (5) and (6) by 𝑥, 𝑦 and 𝑧 respectively and adding, we get
40

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
1600𝑥𝑦𝑧 2 + μ2(𝑥 2 + 𝑦 2 + 𝑧 2 ) = 0
By using equation (2) μ = −800𝑥𝑦𝑧 2
Substitute the value of 𝜇 in equ
...
Find the maximum and minimum distances of the point (3, 4, 12) from the sphere 𝑥 2 + 𝑦 2 +
𝑧2 = 1
Solution: Let 𝑃(𝑥, 𝑦, 𝑧) be any point on the sphere
...

Let

𝑓 (𝑥, 𝑦, 𝑧) = (𝑥 − 3)2 + (𝑦 − 4)2 + (𝑧 − 12)2 …
...
(2)

Let the Lagrange’s function
𝑔(𝑥, 𝑦, 𝑧) = (𝑥 − 3)2 + (𝑦 − 4)2 + (𝑧 − 12)2 + 𝜇(𝑥 2 + 𝑦 2 + 𝑧 2 − 1)…(3)
For stationary value, 𝑑𝑔 = 0
(2(𝑥 − 3) + 2𝜇𝑥 )𝑑𝑥 + (2(𝑦 − 4) + 2𝜇𝑦)𝑑𝑦 + (2(𝑧 − 12) + 2𝜇𝑧)𝑑𝑧 = 0
2(𝑥 − 3) + 2𝜇𝑥 = 0
...
… (5)
2(𝑧 − 12) + 2𝜇𝑧 = 0 …
...
(7)
3

4

12

From equations (4), (5) and (6) 𝑥 = 1+𝜇 , 𝑦 = 1+𝜇 , 𝑧 = 1+𝜇
Substitute these values in equation (7), we get (1 + 𝜇)2 = 169 𝑜𝑟 1 + 𝜇 = ±13
Or 𝜇 = 12 𝑜𝑟 − 14

41

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
3

4

12

3

4

12

For 𝜇 = 12, 𝑥 = 13 , 𝑦 = 13 , 𝑧 = 13 the point is 𝑄 (13 , 13 , 13)
3

4

12

3

4

12

For 𝜇 = −14, 𝑥 = − 13 , 𝑦 = − 13 , 𝑧 = − 13 the point is 𝑅 (− 13 , − 13 , − 13)
By using the distance formula between two points
𝑃𝑄 = 12 and 𝑃𝑅 = 14
So 𝑄 (

3

4

12

3

4

12

, , ) is at minimum distance from P and the distance is 12
...


42

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
Practice Exercise:
Q-1: Examine for extreme values 𝑓 (𝑥, 𝑦) = 𝑥 3 + 𝑦 3 − 3𝑥𝑦Ans: Min value = -1 at (𝟏, 𝟏)
Q-2: Examine for extreme values 𝑓 (𝑥, 𝑦) = 3𝑥 2 − 𝑦 2 + 𝑥 3 Ans: Max value = 4 at (−𝟐, 𝟎)
Q-3: Examine for extreme values 𝑓 (𝑥, 𝑦) = 𝑥 3 𝑦 2 (1 − 𝑥 − 𝑦)Ans: Max value =

𝟏

𝟏 𝟏

at (𝟐 , 𝟑)
𝟒𝟑𝟐

Q-4: Determine the points where the function 𝑓 (𝑥, 𝑦) = 𝑥 2 + 𝑦 2 + 6𝑥 + 12 has a maximum or minimum
...
Ans:1
Q-6: Find the shortest distances from the point (1, 2, 2) from the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 36Ans: 3
Q-7: Find the dimension of a rectangular box of maximum capacity whose surface area is given when

(a) Box is open at top
𝑺

𝟏

𝑺

𝑺

(b) Box is closed
...
distance =21
...
distance = 4
...
7
...

𝜕(𝑢,𝑣)

(1) If 𝑢(𝑥, 𝑦) and 𝑣(𝑥, 𝑦) are two functions then the jacobian of u and v is denoted by 𝐽(𝑢, 𝑣) or 𝜕(𝑥,𝑦)
and its value is

𝜕𝑢

𝜕𝑢

𝜕𝑥
|𝜕𝑣

𝜕𝑦
|
𝜕𝑣

𝜕𝑥

𝜕𝑦
𝜕𝑢1

(2) If 𝑢1 , 𝑢2 , 𝑢3 , … , 𝑢𝑛 are function 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛 then

𝜕(𝑢1,𝑢2 ,𝑢3,…,𝑢𝑛 )
𝜕(𝑥1,𝑥2,𝑥3,…,𝑥𝑛 )

𝜕𝑥1

is | …

𝜕𝑢𝑛
𝜕𝑥1

…
…
…

𝜕𝑢1
𝜕𝑥𝑛

…|

𝜕𝑢𝑛
𝜕𝑥𝑛

3
...
1
...
Chain Rule Property
If 𝑢(𝑟, 𝑠) and 𝑣(𝑟, 𝑠) are two functions i
...
u and v are two function of 𝑟 𝑎𝑛𝑑 𝑠 and 𝑟(𝑥, 𝑦) and
𝑠(𝑥, 𝑦) are two functions i
...
𝑟 𝑎𝑛𝑑 𝑠 are two function of 𝑥 𝑎𝑛𝑑 𝑦 then𝑢 and 𝑣 becomes the
functions of 𝑥 and 𝑦
𝜕(𝑢, 𝑣) 𝜕(𝑢, 𝑣) 𝜕(𝑟, 𝑠)
=
∗
𝜕(𝑥, 𝑦) 𝜕(𝑟, 𝑠) 𝜕(𝑥, 𝑦)
𝜕(𝑢,𝑣)

2
...
If 𝑢𝑖 and 𝑥𝑖 are related as following
𝑢1 = 𝑓(𝑥1 )
𝑢2 = 𝑓(𝑥1 , 𝑥2 )
… … … … … …
...
𝜕𝑥2 … 𝜕𝑥𝑛
1

2

𝑛

𝜕(𝑢,𝑣)

4
...
If 𝑢1 , 𝑢2 , 𝑢3 are implicit function 𝑥1 , 𝑥2 , 𝑥3 i
...

𝐹1 (If 𝑢1 , 𝑢2 , 𝑢3 , 𝑥1 , 𝑥2 , 𝑥3 ) = 0
𝐹2 (If 𝑢1 , 𝑢2 , 𝑢3 , 𝑥1 , 𝑥2 , 𝑥3 ) = 0
𝐹3 (If 𝑢1 , 𝑢2 , 𝑢3 , 𝑥1 , 𝑥2 , 𝑥3 ) = 0
𝜕(𝐹1 ,𝐹2 ,𝐹3 )

𝜕(𝑢1 , 𝑢2 , 𝑢3 )
)
= (−1)3 [𝜕(𝑥1,𝑥2,𝑥3⁄
𝜕(𝐹1 ,𝐹2 ,𝐹3 ) ]
𝜕 (𝑥1 , 𝑥2 , 𝑥3 )
𝜕(𝑢1 ,𝑢2 ,𝑢3 )

44

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
3
...
2
...
Then the necessary condition for the existence of a relation
of the form 𝐹(If 𝑢1 , 𝑢2 , … , 𝑢𝑛 ) = 0 is that

𝜕(𝑢1 ,𝑢2 ,𝑢3 ,…,𝑢𝑛 )

=0

𝜕(𝑥1 ,𝑥2 ,𝑥3 ,…,𝑥𝑛 )

Example 1: If 𝑢 = 𝑥 + 2𝑦 + 𝑧, 𝑣 = 𝑥 + 2𝑦 + 3𝑧 and 𝑤 = 2𝑥 + 3𝑦 + 5𝑧 then find the Jacobian

Solution: According to the definition of Jacobian

𝜕(𝑢,𝑣,𝑤)
𝜕(𝑥,𝑦,𝑧)

=

𝜕𝑢

𝜕𝑢

𝜕𝑢

𝜕𝑥
| 𝜕𝑣

𝜕𝑦
𝜕𝑣

𝜕𝑧
𝜕𝑣 |

| 𝜕𝑥
𝜕𝑤

𝜕𝑦
𝜕𝑤

1
= |1
𝜕𝑧 |
2
𝜕𝑤

𝜕𝑥

𝜕𝑦

𝜕𝑧

Solution: According to the definition of jacobian

𝜕(𝑢,𝑣,𝑤)
𝜕(𝑥,𝑦,𝑧)

𝜕𝑢

𝜕𝑢

𝜕𝑥

𝜕𝑦
𝜕𝑣

𝜕𝑧
𝜕𝑣 |

𝜕𝑤

𝜕𝑦
𝜕𝑤

𝑦𝑧
𝑦
+𝑧
|
=
𝜕𝑧 |
1
𝜕𝑤

𝜕𝑥

𝜕𝑦

𝜕𝑧

| 𝜕𝑣
= 𝜕𝑥
|

𝜕(𝑥,𝑦,𝑧)

2 1
2 3|=2
3 5

Example 2: If 𝑢 = 𝑥𝑦𝑧, 𝑣 = 𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥 𝑎𝑛𝑑 𝑤 = 𝑥 + 𝑦 + 𝑧 then find the jacobian
𝜕𝑢

𝜕(𝑢,𝑣,𝑤)

𝜕(𝑢,𝑣,𝑤)
𝜕(𝑥,𝑦,𝑧)

𝑧𝑥
𝑧+𝑥
1

𝑥𝑦
𝑥 + 𝑦|
1

To solve the above determinant, we want to reduce the matrix in simplest form so we will use the property
(elementary operation) of matrix
...
If 𝑢 = 𝑟𝑐𝑜𝑠𝜃, 𝑣 = 𝑟𝑠𝑖𝑛𝜃 find 𝜕(𝑟,𝜃) and 𝜕(𝑢,𝑣)
...
Therefore, chain rule is verified
...

By applying the elementary operation𝑅1 → 𝑅1 + 𝑅2 + 𝑅3 , we get
1
0
|𝑣(1 − 𝑤) 𝑢(1 − 𝑤)
𝑣𝑤
𝑢𝑤

0
−𝑢𝑣|= 𝑢2 𝑣
𝑢𝑣
46

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD

Example 6: If u, v and w are the roots of the cubic equation (𝑥 − 𝑎)3 + (𝑦 − 𝑏)3 + (𝑧 − 𝑐 )3 = 0 then find
the value of

𝜕(𝑢,𝑣,𝑤)
𝜕(𝑎,𝑏,𝑐)

Solution: The cubic equation(𝑥 − 𝑎)3 + (𝑦 − 𝑏)3 + (𝑧 − 𝑐 )3 = 0
3𝑥 3 − 3𝑥 2 (𝑎 + 𝑏 + 𝑐 ) + 3𝑥(𝑎2 + 𝑏2 + 𝑐 2 ) − (𝑎3 + 𝑏3 + 𝑐 3 ) = 0
Given that 𝑢, 𝑣 𝑎𝑛𝑑 𝑤 are the roots of the cubic equation
...
so use the property no
...

𝐹1 = 𝑢 + 𝑣 + 𝑤 − 𝑎 − 𝑏 − 𝑐 = 0
𝐹2 = 𝑢𝑣 + 𝑣𝑤 + 𝑤𝑢 − 𝑎2 − 𝑏2 − 𝑐 2 = 0
𝐹3 = 𝑢𝑣𝑤 −

𝑎3 + 𝑏3 + 𝑐 3
=0
3
𝜕(𝐹1 ,𝐹2 ,𝐹3 )

𝜕 (𝑢, 𝑣, 𝑤)
= (−1)3 [
𝜕(𝑎, 𝑏, 𝑐 )

𝜕(𝑎,𝑏,𝑐)

−1
𝜕(𝐹1 , 𝐹2 , 𝐹3 )
= |−2𝑎
𝜕(𝑎, 𝑏, 𝑐 )
−𝑎2

⁄𝜕(𝐹1 ,𝐹2 ,𝐹3 )]
𝜕(𝑢,𝑣,𝑤)

−1
−2𝑏
−𝑏2

−1
−2𝑐 |
−𝑐 2

By applying the elementary operation 𝐶2 → 𝐶2 − 𝐶1 and 𝐶3 → 𝐶3 − 𝐶2 , we get
−1
|−2𝑎
−𝑎2

0
−2𝑏 + 2𝑎
−𝑏2 +𝑎2

0
−2𝑐 + 2𝑎|
−𝑐 2 +𝑎2

= −2(𝑎 − 𝑏)(𝑏 − 𝑐)(𝑐 − 𝑎)
1
𝜕(𝐹1 , 𝐹2 , 𝐹3 )
|
= 𝑣+𝑤
𝜕 (𝑢, 𝑣, 𝑤)
𝑣𝑤

1
𝑢+𝑤
𝑢𝑤

1
𝑢 + 𝑣|
𝑢𝑣

By applying the elementary operation 𝐶2 → 𝐶2 − 𝐶1 and 𝐶3 → 𝐶3 − 𝐶2 , we get
1
|𝑣 + 𝑤
𝑣𝑤

0
𝑢−𝑣
𝑤(𝑢 − 𝑣)
47

0
𝑢−𝑤 |
𝑣(𝑢 − 𝑤)

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
= −(𝑢 − 𝑣)(𝑣 − 𝑤)(𝑤 − 𝑢)
So
𝜕(𝐹1 ,𝐹2 ,𝐹3 )

𝜕(𝑢, 𝑣, 𝑤)
2(a − b)(b − c)(c − a)
𝜕(𝑎,𝑏,𝑐)
= (−1)3 𝜕(𝐹 ,𝐹 ,𝐹 ) = −
1 2 3
𝜕(𝑎, 𝑏, 𝑐 )
(u − v)(v − w)(w − u)
𝜕(𝑢,𝑣,𝑤)

Example 7: Prove that 𝑢 = 𝑥 + 2𝑦 + 𝑧, 𝑣 = 𝑥 − 2𝑦 + 3𝑧 𝑎𝑛𝑑 𝑤 = 2𝑥𝑦 − 𝑥𝑧 + 4𝑦𝑧 − 2𝑧 2 are not
independent and find the relation between them
...
e
...

𝜕𝑢
| 𝜕𝑥
𝜕(𝑢, 𝑣, 𝑤)
𝜕𝑣
=
𝜕 (𝑥, 𝑦, 𝑧)
𝜕𝑥
|𝜕𝑤
𝜕𝑥

𝜕𝑢
𝜕𝑦
𝜕𝑣
𝜕𝑦
𝜕𝑤
𝜕𝑦

𝜕𝑢
𝜕𝑧 |
1
𝜕𝑣
=| 1
𝜕𝑧
2𝑦 − 𝑧
𝜕𝑤|
𝜕𝑧

2
1
−2
3
|
2𝑥 + 4𝑧 −𝑥 + 4𝑦 − 4𝑧

By applying the elementary operation 𝐶2 → 𝐶2 − 2𝐶1 and 𝐶3 → 𝐶3 − 𝐶2 , we get
|

1
1
2𝑦 − 𝑧

i
...


0
−4
2𝑥 + 6𝑧 − 4𝑦

𝜕(𝑢,𝑣,𝑤)
𝜕(𝑥,𝑦,𝑧)

0
2
|=0
−𝑥 + 2𝑦 − 3𝑧

=0

so the function 𝑢, 𝑣 and 𝑤 are not independent i
...
dependent
...

𝑢 + 𝑣 = 2𝑥 + 4𝑧 …(1)
𝑢 − 𝑣 = 4𝑦 − 2𝑧 …(2)
By multiplying both equations, we get
𝑢2 − 𝑣 2 = 4(2𝑥𝑦 − 𝑥𝑧 + 4𝑦𝑧 − 2𝑧 2 )
𝑢2 − 𝑣 2 = 4𝑤
is the relation between 𝑢, 𝑣 and 𝑤
...
Find the relation
between them
...
e
...

48

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD

𝜕(𝑢,𝑣)

1+𝑦 2

𝜕𝑢

𝜕𝑢

𝜕𝑥

𝜕𝑦
(1−𝑥𝑦)2
|=| 1
𝜕𝑣

𝜕𝑥

𝜕𝑦

𝐽(𝑢, 𝑣) = 𝜕(𝑥,𝑦)=|𝜕𝑣

1+𝑥 2

1+𝑥 2
(1−𝑥𝑦)2
|
1

=0

1+𝑦 2

So 𝑢 𝑎𝑛𝑑 𝑣 are functionally related
𝑡𝑎𝑛−1 𝑥 + 𝑡𝑎𝑛−1 𝑦 = 𝑡𝑎𝑛−1

𝑥+𝑦
1 − 𝑥𝑦

𝑣 = 𝑡𝑎𝑛−1 𝑢
So 𝑢 𝑎𝑛𝑑 𝑣 are functionally related with the relation 𝑣 = 𝑡𝑎𝑛−1 𝑢
...


Q-2:If 𝑢3 + 𝑣 3 = 𝑥 + 𝑦, 𝑢2 + 𝑣 2 = 𝑥 3 + 𝑦 3 then show that

𝑦 2−𝑥 2

𝜕(𝑢,𝑣)

= 2𝑢𝑣(𝑢−𝑣)
...


𝜕(𝑢,𝑣,𝑤) 𝜕(𝑥,𝑦,𝑧)

=1

Q-5:Use the Jacobian to prove that the function 𝑢 = 𝑥 + 𝑦 − 𝑧, 𝑣 = 𝑥 − 𝑦 + 𝑧 and𝑤 = 𝑥 2 + 𝑦 2 + 𝑧 2 −

2𝑦𝑧 are not independent of one another
...

Q-7:If 𝑢 = 𝑥 + 𝑦 + 𝑧, 𝑣 = 𝑥 2 + 𝑦 2 + 𝑧 2 and 𝑤 = 𝑥 3 + 𝑦 3 + 𝑧 3 − 3𝑥𝑦𝑧, prove that 𝑢, 𝑣 𝑎𝑛𝑑 𝑤 are not

independent and hence find the relation between them
...
Find the relation between them
...
Ans: 𝒗 = 𝐬𝐢𝐧 𝒖

50

𝒖(𝒖𝟐+𝟑𝒗)
𝟒

= 4𝑟 3
...
Prove that 𝑢 𝑎𝑛𝑑 𝑣

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
3
...


Approximation of Errors

Let 𝑧 is a function of 𝑥 𝑎𝑛𝑑 𝑦
...
(1)

If 𝛿𝑥 and 𝛿𝑦are the small changes (or error) in 𝑥, 𝑦, the 𝛿𝑧 is corresponding small changes (or error) in 𝑧
𝑧 + 𝛿𝑧 = 𝑓(𝑥 + 𝛿𝑥, 𝑦 + 𝛿𝑦)
...
100

Example 1:Find the percentage error in the area of an ellipse when an error of +1 amount is made in
measuring the major and minor axis
...

Area A of an ellipse
𝐴 = 𝜋𝑥𝑦
Taking log on both sides
𝑙𝑜𝑔𝐴 = 𝑙𝑜𝑔𝜋 + 𝑙𝑜𝑔𝑥 + 𝑙𝑜𝑔𝑦
After differentiating, we get

𝛿𝐴
𝐴

=0+

𝛿𝑥
𝑥

+

𝛿𝑦
𝑦

For the percentage error

51

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
𝛿𝐴
𝛿𝑥
𝛿𝑦

...
100 +
...
so
𝛿𝐴

...
Find the maximum percentage error in T due to
the possible errors upto 1%in 𝑥 and 2
...

Solution: Given that 𝑇 = 2𝜋√

𝑥
𝑦

Taking log on both sides
1
1
𝑙𝑜𝑔𝑇 = 𝑙𝑜𝑔2𝜋 + 𝑙𝑜𝑔𝑥 + 𝑙𝑜𝑔𝑦
2
2
Differentiating

𝛿𝑇
𝑇

1 𝛿𝑥

=2

𝑥

1 𝛿𝑦

−2

𝑦

For the percentage error
𝛿𝑇
1 𝛿𝑥
1 𝛿𝑦
1

...
100 −

...
5)
𝑇
2 𝑥
2 𝑦
2
1

For maximum percentage error in 𝑇 =

2

(1 + 2
...
75%

Example 3: If the base radius and height of a cone are measured as 4 and 8 inches with a possible error of

...
08 inches respectively, calculate the percentage error in calculating volume of the cone
...
04 and 𝛿ℎ =
...
04

...
03
𝑉
4
8
Percentage error in volume=
...
100 = 3%

52

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
Example 4: A balloon is in the form of right circular cylinder of radius 1
...
If the radius in increased by
...
05m
...

Solution: Given that 𝑟 = 1
...
001 and 𝛿ℎ =
...
2𝑟𝛿𝑟
...
3𝑟 2 𝛿𝑟
3
𝛿𝑉 = 𝜋𝑟(2ℎ𝛿𝑟 + 𝑟𝛿ℎ + 4𝑟𝛿𝑟)
𝛿𝑉 𝜋𝑟(2ℎ𝛿𝑟 + 𝑟𝛿ℎ + 4𝑟𝛿𝑟)
=
4
𝑉
𝜋𝑟 2 ℎ + 𝜋𝑟 3
3

Substituting the values of 𝑟, ℎ, 𝛿𝑟, 𝛿ℎ, we get
𝛿𝑉
...
100 =


...
100 = 2
...
100) (100 𝑙𝑜𝑔10 𝑒) = 100 𝑙𝑜𝑔10 𝑒
𝛿𝑦 =


...
0043429
100

Example 6: In estimating the number of bricks in a pile which is measured to be (5𝑚
...
5𝑚), the count
of bricks is taken as 100 bricks per 𝑚3
...
The cost of bricks is 2000 rs
...

Solution: Let 𝑥, 𝑦 and 𝑧 be the length, breadth, and height of the pile
...
10𝑚
...
100 =

...
100 +
...
𝑉 =

...
100 = 1500
2000

Error in the cost = 1500
...

Example 7: Find the possible percentage error in computing the parallel resistance r of three resistances𝑟1 ,
1

1

𝑟

𝑟1

𝑟2 and 𝑟3 from the formula =
1

1

+

𝑟2

1

1

1

1

2

3

+

1
𝑟3

if 𝑟1 , 𝑟2 and 𝑟3 are each in error by 1
...


Solution: Given that 𝑟 = 𝑟 + 𝑟 + 𝑟
1

1

1

1

1

2

3

Differentiating − 𝑟 2 𝛿𝑟 = − 𝑟 2 𝛿𝑟1 − 𝑟 2 𝛿𝑟2 − 𝑟 2 𝛿𝑟3
For the percentage error
1 1
1 1
1 1
1 1
(
...
100) + (
...
100)
𝑟 𝑟
𝑟1 𝑟1
𝑟2 𝑟2
𝑟3 𝑟3
1 1
1
1
1
(
...
2) + (1
...
2)
𝑟 𝑟
𝑟1
𝑟2
𝑟3
1 1
1 1 1
(
...
2) ( + + )
𝑟 𝑟
𝑟1 𝑟2 𝑟3
1 1
1
(
...
2
...
100) = 1
...
2%

Example 8: Find approximate value of(1
...
01
...
So, differentiate 𝑓 w
...
t
...
04)3
...
04, 𝑦 = 3 𝑎𝑛𝑑 𝑑𝑦 =
...
12
𝑓(𝑥, 𝑦) = 1
(1
...
01 = 𝑓(𝑥, 𝑦) + 𝑑𝑓 = 1 + 0
...
12

55

KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
Practice Exercise:
Q-1: Find approximate value of[(0
...
01)2 + (1
...
Ans:

2
...
99 𝑎𝑛𝑑 𝑦 = 3
...
Ans: 4
...
Ans:3%
Q-5:Find the possible percentage error in computing the parallel resistance r of two resistances 𝑟1 and 𝑟2
1

1

1

1

2

from the formula 𝑟 = 𝑟 + 𝑟 if 𝑟1 and 𝑟2 are each in error by +2%
...
𝟐%
𝑙

Q-6: The period of a simple pendulum is 𝑇 = 2𝜋√
...
Ans: 3
...
5 and the velocity 𝑣 changes from 1600 to 1590
...

Q-8: In estimating the number of bricks in a pile which is measured as(6𝑚
...
4𝑚), the tape is stretched

1% beyond its standard length
...
per
thousand bricks
...
Ans:𝟒𝟑
...

Q-10:Find the percentage error in calculated volume of a right circular cone whose altitude is same as the

base radius and is measured as 5 𝑐𝑚 with a possible error of 0
...
Ans: 𝟏
...

2
...

4
...

6
...

8
...

10
...

12
...

14
...

16
...

18
...

20
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube
...
youtube

---