Partial differentiation, total differentiation, Jacobian, Taylor's expansion, stationary points,maxima & minima (Extreme values),constraint maxima & minima ( Lagrangian multiplier), differentiation of implicit functions.
2. A function of several variables is a function where the domain is a
subset of π π and range is π .
A real valued function of πβvariables is a function
π βΆ π· β R, where the domain D is a subset of π π.
So: for each (π₯1, π₯2, . . . , π₯ π) in D, the value of f is a real number
π(π₯1, π₯2, . . . , π₯ π ).
For example,
1.π(π₯, π¦) = π₯ β π¦
(a function of 2 variables defined for all (π₯, π¦) β π 2)
2. If π is a function defined by
π(π₯, π¦) = 9 β cos(π₯) + sin(π₯2 + π¦2),
(a function of 2 variables defined for all (π₯, π¦) β π 2
)
3. 2. π π₯, π¦, π§ =
1
π₯2+π¦2+π§2
Then f is a function of 3 variables, defined whenever π₯2 + π¦2 + π§2 β
0
This is all (π₯, π¦, π§) β π 3 except for (π₯, π¦, π§) = (0, 0, 0).
4. Partial Derivative
If π(π₯, π¦) is a function of two variables π₯ and π¦,
the partial derivative of π with respect to π₯, is given by ππ₯ π₯, π¦ =
lim
ββ0
π π₯+β,π¦ βπ π₯,π¦
β
The partial derivative of π with respect to π¦, is given by ππ¦ π₯, π¦ =
lim
ββ0
π π₯,π¦+β βπ π₯,π¦
β
5. Note
β’ If π(π₯, π¦) is a function of two variables π₯ and π¦, then
Partial derivative of π(π₯, π¦) with respect to π₯ is
ππ
ππ₯
, it is
denoted by ππ₯
β’ Partial derivative of π(π₯, π¦) with respect to π¦ is
ππ
ππ¦
, it is denoted by
ππ¦
β’ Partial second derivative of π(π₯, π¦) with respect to π₯ is
π2 π
ππ₯2 , it is
denoted by ππ₯π₯
β’ Partial second derivative of π(π₯, π¦) with respect to π¦ is
π2 π
ππ¦2 , it is
denoted by ππ¦π¦
β’ Partial derivative of
ππ
ππ₯
with respect to π¦ is
π2 π
ππ¦ππ₯
, it is denoted by ππ₯π¦
and so on
12. Homogeneous Function:
A function π(π₯, π¦) is called a homogeneous function of the degree β²πβ² if
the following relationship is valid for all π‘ > 0: π π‘π₯, π‘π¦ = π‘ π
π π₯, π¦ .
Example:
Consider π π₯, π¦ =
π₯2+π¦2
π₯π¦
π π‘π₯, π‘π¦ =
π‘π₯ 2+ π‘π¦ 2
π‘π₯ π‘π¦
=
π‘2 π₯2+π‘2 π¦2
π‘2 π₯π¦
=
π‘2 π₯2+π¦2
π‘2 π₯π¦
=
π₯2+π¦2
π₯π¦
= π‘0 π(π₯, π¦)
Hence f(x, y ) is homogeneous of order zero
13. Eulerβs Theorem:
If π’ = π π₯, π¦ is a homogeneous function of degree βnβ, then π₯
ππ’
ππ₯
+ π¦
ππ’
ππ¦
= ππ’
(OR) π₯
ππ
ππ₯
+ π¦
ππ
ππ¦
= ππ
Problem 1:
If π = π
π
π
, prove that π
ππ
ππ
+ π
ππ
ππ
= π
Solution:
Given π’ π₯, π¦ = π
π₯
π¦
For any π‘ > 0,
π’ π‘π₯, π‘π¦ = π
π‘π₯
π‘π¦
= π
π₯
π¦
= π‘0
π
π₯
π¦
= π‘0
π’(π₯, π¦)
Hence π’ = π
π₯
π¦
is a homogeneous function of order zero
π. π π = 0
By Eulerβs Theorem , π₯
ππ’
ππ₯
+ π¦
ππ’
ππ¦
= ππ’
βΉ π₯
ππ’
ππ₯
+ π¦
ππ’
ππ¦
= 0 u = 0
Hence proved.
.
14. Problem 2:
If π = π¬π’π§βπ π
π
+ πππ§βπ π
π
, prove that π
ππ
ππ
+ π
ππ
ππ
= π.
Solution:
Given π’(π₯, π¦) = sinβ1 π₯
π¦
+ tanβ1 π¦
π₯
For any π‘ > 0,
π’(π‘π₯, π‘π¦) = sinβ1 π‘π₯
π‘π¦
+ tanβ1 π‘π¦
π‘π₯
= sinβ1 π₯
π¦
+ tanβ1 π¦
π₯
= π‘0
( sinβ1 π₯
π¦
+ tanβ1 π¦
π₯
)
Hence π’(π₯, π¦) is a homogeneous function of order zero
π. π π = 0
By Eulerβs Theorem,
π₯
ππ’
ππ₯
+ π¦
ππ’
ππ¦
= ππ’ = 0 π’ = 0
15. Total differential coefficient
If π’ = π π₯, π¦ , the total differential of π’ is given by
ππ’ = π’ π₯ ππ₯ + π’ π¦ ππ¦ (OR) ππ = ππ₯ ππ₯ + ππ¦ ππ¦
If π’ = π π₯, π¦ , and x = g(t) , y = h(t) , the total differential of π’ is given
by
ππ’
ππ‘
=
ππ’
ππ₯
ππ₯
ππ‘
+
ππ’
ππ¦
ππ¦
ππ‘
= π’ π₯
ππ₯
ππ‘
+ π’ π¦
ππ¦
ππ‘
(OR)
ππ
ππ‘
=
ππ
ππ₯
ππ₯
ππ‘
+
ππ
ππ¦
ππ¦
ππ‘
= ππ₯
ππ₯
ππ‘
+ ππ¦
ππ¦
ππ‘
33. Necessary Conditions
The necessary conditions for π(π₯, π¦) to have a maximum or minimum at a point
(π, π) are that ππ₯ π, π = 0 πππ ππ¦(π, π) = 0 ππ‘ (π, π)
Sufficient Conditions
The sufficient conditions for π(π₯, π¦) to have a maximum or minimum at a point
(π, π) are as follows
Let π = ππ₯π₯ π, π , π = ππ₯π¦ π, π πππ π‘ = ππ¦π¦ π, π
The function π(π₯, π¦) has maximum or minimum (extreme values) at ( a, b) if
1. ππ₯ π, π = 0 πππ ππ¦(π, π) = 0
2. ππ‘ β π 2
> 0
3. π(π₯, π¦) has maximum or minimum at (π, π) according as π < 0 or π > 0
34. Procedure to find maximum or minimum (extreme values):
Step 1. Find ππ₯ πππ ππ¦
Step 2. Put ππ₯ = 0 πππ ππ¦ = 0
Step 3. Solve the simultaneous equations in a and y, find the values of x and y
Say (π, π), (π, π) β¦ . ( these points are called stationary / critical points)
Step 4.Find π = ππ₯π₯ , π‘ = ππ¦π¦ πππ π = ππ₯π¦ (at each pair (π, π) , (π, π) β¦ . )
Step 5. Find ππ‘ β π 2 (at each pair (π, π) , (π, π) β¦ . )
Step 6.
(i) If rπ‘ β π 2
> 0 πππ π < 0 ππ‘ (π, π) , then π(π₯, π¦) has maximum at (π, π)
And the maximum value is π(π, π)
(ii) If rπ‘ β π 2 > 0 πππ π > 0 ππ‘ (π, π) , then π(π₯, π¦) has minimum at (π, π)
And the minimum value is π(π, π)
(iii) If rπ‘ β π 2 < 0 ππ‘ (π, π) , then π(π₯, π¦) has neither maximum nor minimum at
(π, π) (no extreme values at (π, π)) and the point (π, π) is called saddle point.
(iv) If ππ‘ β π 2
= 0 ππ‘ (π, π), the case is doubtful and need further investigation.
Step 7. Step 6 to be repeated for other pair of values (c, d) β¦ to examine extreme values
40. To find the point at which π(π, π) has maximum/minimum:
ππ‘ β π 2 = 6 + 6π₯ β2 β 0 = β12 β 12π₯
At the Point (0,0):
ππ‘ β π 2
ππ‘ 0,0 = β12 β 12 0 = β12 < π
Hence the point (0,0) is the saddle point of π(π₯, π¦)
At the Point (-2,0):
ππ‘ β π 2
ππ‘ β2,0 = β12 β 12 β2 = β12 + 24 = 12 > π
πat β2,0 = 6 + 6 β2 = 6 β 12 = β6 < π
Hence at the point β2,0 the function π π₯, π¦ has maximum.
To find the maximum value:
Put π₯, π¦ = (β2,0) in (1), π β2,0 = 3(β2)2 β 0 2 + β2 3 = 12 β 8 = 4
41. Constrained Maxima and Minima β Lagrangian Multiplier
If π(π₯, π¦, π§) is a function of three variables π₯, π¦, π§ , we will find the extreme values
(maximum or minimum) of π(π₯, π¦, π§) with respect to a constraint β π₯, π¦, π§ = 0
Procedure
Step 1. Identify the constraint equation β π₯, π¦, π§ = 0
Step 2. Identify the main function for which we have to find the extreme value, let it be
π(π₯, π¦, π§)
Step 3. Form the equation πΉ = π + πβ
Step 4. Find πΉπ₯ , πΉπ¦, πΉπ§
Step 5. Put πΉπ₯ = 0 , πΉπ¦ = 0, πΉπ§ = 0 and solve all the equations including β π₯, π¦, π§ = 0
Step 6. Find the values of π₯, π¦, π§ and π
Step 7. The values of π₯, π¦, π§ gives the extreme values of π(π₯, π¦, π§)
42. Note :
Distance of a point π₯1, π¦1, π§1 ππππ π₯, π¦, π§ is given by
π = π₯ β π₯1
2 + π¦ β π¦1
2 + π§ β π§1
2
Square of the distance is π2
= π₯ β π₯1
2
+ π¦ β π¦1
2
+ π§ β π§1
2
43. Problem 1:
Find the length of the shortest line form the point (π, π,
ππ
π
) to the surface
π = ππ
Solution:
The square of the distance from the point (0,0,
25
9
) to (π₯, π¦, π§) is (π2)
π π₯, π¦, π§ = π₯ β 0 2 + π¦ β 0 2 + π§ β
25
9
2
π π₯, π¦, π§ = π₯2 + π¦2 + π§ β
25
9
2
------- (1)
Subject to ( to the surface) β π₯, π¦π§ = π§ β π₯π¦ = 0 ------------(2)
(β΅ π§ = π₯π¦ βΉ π§ β π₯π¦ = 0)
Consider the Lagrangian function F = π π₯, π¦, π§ + ππ π₯, π¦, π§
πΉ = π₯2
+ π¦2
+ π§ β
25
9
2
+ π(π§ β π₯π¦)
πF
ππ±
= 2x β Ξ»y
πF
πy
= 2y β Ξ»x
πF
πz
= 2 z β
25
9
+ Ξ»
45. Hence π₯ = π¦ = Β±
4
3
& π§ = π₯π¦ = π¦2 =
4
3
2
=
16
9
From (1) square distance is π2 = π₯2 + π¦2 + π§ β
25
9
2
=
4
3
2
+
4
3
2
+
16
9
β
25
9
2
=
16
9
+
16
9
+ β
9
9
2
= 2
16
9
+ β1 2
=
32
9
+ 1 =
32+9
9
=
41
9
The required minimum distance is (d) =
41
9
=
41
3
units
Problem 2:
A rectangular box open at the top is to have a volume of 32 cc .Find the dimensions of the
box that requires the least material for its construction
Solution:
Given a rectangular open box with volume 32cc
Let us take the length, width and height of the box be x, y and z respectively.
46. Hence the volume of the box is π₯π¦π§ = 32 , which is the given constrain
(condition).
Let β π₯, π¦, π§ = π₯π¦π§ β 32 ------ (1)
Requirement of least material to construct the open top box is the total least surface
area of the box.
Total surface area of the open rectangular box is π₯π¦ + 2π₯π§ + 2π¦π§
Let π π₯, π¦, π§ = π₯π¦ + 2π₯π§ + 2π¦π§ ------ (2)
53. Problem 3:
Find the dimensions of the rectangular box, open at the top, of maximum capacity
whose surface area is 432 square meter.
Solution:
Given a rectangular open box with surface area 432 sq.m
Let us take the length, width and height of the box be x, y and z respectively.
Hence the surface area π₯π¦ + 2π₯π§ + 2π¦π§ = 432, which is the given constrain(condition)
Let β π₯, π¦, π§ = π₯π¦ + 2π₯π§ + 2π¦π§ β 432 ------ (1)
Requirement of a open rectangular box with maximum capacity (volume)
Total volume of the open rectangular box is π₯π¦π§
Let π π₯, π¦, π§ = π₯π¦π§ ------ (2)
60. Jacobian
β’ If π’ = π(π₯, π¦) and π£ = π(π₯, π¦) are two continuous functions of two
independent variables x and y then the functional determinant
π½ =
ππ’
ππ₯
ππ’
ππ¦
ππ£
ππ₯
ππ£
ππ¦
=
π’ π₯ π’ π¦
π£ π₯ π£ π¦
is called Jacobian of π’ , π£ with respect to π₯, π¦ and it is
denoted by
π π’,π£
π π₯,π¦
β’ If u , v, w are functions of x ,y , z then jacobian of u , v , w with respect to x , y , z
is given by
π π’,π£,π€
π π₯,π¦,π§
=
ππ’
ππ₯
ππ’
ππ¦
ππ’
ππ§
ππ£
ππ₯
ππ£
ππ¦
ππ£
ππ§
ππ€
ππ₯
ππ€
ππ¦
ππ€
ππ§
=
π’ π₯ π’ π¦ π’ π§
π£ π₯ π£ π¦ π£π§
π€ π₯ π€ π¦ π€π§
61. Two important Properties of Jacobian
1. If π’, π£ are functions of π₯, π¦ and π₯, π¦ are the function of π, π then
π π’,π£
π π,π
=
π π’,π£
π π₯,π¦
π π₯,π¦
π π,π
2. If π’, π£ are the functions of π₯, π¦ then
π π₯,π¦
π π’,π£
=
1
π π’,π£
π π₯,π¦
Note :
Two functions π’(π₯, π¦) & π£(π₯, π¦) are functionally depended if
π π’,π£
π π₯,π¦
=0
62. Problem 1:
If π =
ππ
π
, π =
ππ
π
, π =
ππ
π
find
π π,π,π
π π,π,π
Solution:
Wkt, the jacobian of π’, π£, π€ with respect to π₯, π¦, π§ is given by
π π’,π£,π€
π π₯,π¦,π§
=
π’ π₯ π’ π¦ π’ π§
π£ π₯ π£ π¦ π£π§
π€ π₯ π€ π¦ π€π§
Given
π’ =
π¦π§
π₯
, π£ =
π§π₯
π¦
, π€ =
π₯π¦
π§
72. π π₯,π¦
π π’,π£
=
π₯ π’ π₯ π£
π¦π’ π¦π£
=
1 β π£ βπ’
π£ π’
= 1 β π£ π’ + π’π£ = π’ β π’π£ + π’π£ = π’
Note :
Implicit vs Explicit
Explicit: "y = some function of x". When we know x we can calculate y directly.
An explicit function is one which is given in terms of the independent variable.
Example , consider π¦ = π₯2 + 3π₯ β 8
here y is the dependent variable and is given in terms of the independent variable x.
More Examples : π¦ = π₯ + 3 , π¦ = π₯2
β π2
ππ‘π.,
Implicit: "some function of y and x equals something else".
Implicit functions, on the other hand, are usually given in terms of both dependent and
independent variables.
Example, consider π¦ + π₯2 β 3π₯ + 8 = 0
More Examples: π₯2
+ π¦2
= π2
, π₯3
+ π₯π¦2
+ 4π₯ = 5, ππ‘π. ,
73. Differentiation of implicit functions
If π(π₯, π¦ ) is the given implicit function , then
ππ¦
ππ₯
= β
ππ
ππ₯
ππ
ππ¦
Producer to find the differentiation:
(i) Take π(π₯, π¦)
(ii) Find
ππ
ππ₯
&
ππ
ππ¦
(iii) Find
ππ¦
ππ₯
= β
ππ
ππ₯
ππ
ππ¦
Problem 1:
If π π
+ π π
= π, then find
π π
π π
Solution:
Given π₯ π¦
+ π¦ π₯
= π
βΉ π₯ π¦
+ π¦ π₯
β π = 0
Take π π₯, π¦ = π₯ π¦
+ π¦ π₯
β π