Functions of severable variables
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The document presents the concept of functions of several variables along with their domains and ranges, offering examples of real-valued functions. It discusses partial derivatives, including first and second derivatives, through solved problems, demonstrating how to compute them for given functions. Additionally, the document introduces homogeneous functions and Euler's theorem, along with total differential coefficients and several examples to illustrate these mathematical principles.
![Taylor’s expansion of 𝑓(𝑥, 𝑦) about the point (1,2) is given by
𝑓 𝑥, 𝑦 = 𝑓 1,2 + 𝑥 − 1 𝑓𝑥 1,2 + 𝑦 − 2 𝑓𝑦 1,2 +
1
2!
𝑥 − 1 2 𝑓𝑥𝑥 1,2 + 2 𝑥 − 1 𝑦 − 2 𝑓𝑥𝑦 1,2 + 𝑦 − 2 2 𝑓𝑦𝑦 1,2 +
1
3!
𝑥 − 1 3
𝑓𝑥𝑥𝑥 1,2 + 3 𝑥 − 1 2
𝑦 − 2 𝑓𝑥𝑥𝑦(1,2) + 3 𝑥 − 1 𝑦 − 2 2
𝑓𝑥𝑦𝑦 (1,2)
+ 𝑦 − 2 3 𝑓𝑦𝑦𝑦 1,2
+ ⋯
𝑓 𝑥, 𝑦 = 13 + 7 𝑥 − 1 + 16 𝑦 − 2
+
1
2!
[6 𝑥 − 1 2
+ 2 4 𝑥 − 1 𝑦 − 2 + 14 𝑦 − 2 2
]
+
1
3!
6 𝑥 − 1 3 + 3 0 𝑥 − 1 2 𝑦 − 2 + 3 2 𝑥 − 1 𝑦 − 2 2 + 6 𝑦 − 2 3
+ ⋯
= 13 + 7 𝑥 − 1 + 16 𝑦 − 2
+
1
2!
[6 𝑥 − 1 2
+ 8 𝑥 − 1 𝑦 − 2 + 14 𝑦 − 2 2
]
+
1
3!
6 𝑥 − 1 3 + 0 + 6 𝑥 − 1 𝑦 − 2 2 + 6 𝑦 − 2 3 + ⋯
= 13 + 7 𝑥 − 1 + 16 𝑦 − 2 +
6
2
𝑥 − 1 2
+
8
2
𝑥 − 1 𝑦 − 2 +
14
2
𝑦 − 2 2
+
6
6
𝑥 − 1 3
+
6
6
𝑥 − 1 𝑦 − 2 2
+
6
6
𝑦 − 2 3
+… , (2! = 1.2 = 2, 3! = 1.2.3 = 6)](https://image.slidesharecdn.com/functionsofseverablevariables-201226040640/85/Functions-of-severable-variables-25-320.jpg)
![𝜕 𝑢,𝑣,𝑤
𝜕 𝑥,𝑦,𝑧
=
𝑢 𝑥 𝑢 𝑦 𝑢 𝑧
𝑣 𝑥 𝑣 𝑦 𝑣𝑧
𝑤 𝑥 𝑤 𝑦 𝑤𝑧
=
−
𝑦𝑧
𝑥2
𝑧
𝑥
𝑦
𝑥
𝑧
𝑦
−
𝑧𝑥
𝑦2
𝑥
𝑦
𝑦
𝑧
𝑥
𝑧
−
𝑥𝑦
𝑧2
=
1
𝑥
1
𝑦
1
𝑧
−
𝑦𝑧
𝑥
𝑧 𝑦
𝑧 −
𝑧𝑥
𝑦
𝑥
𝑦 𝑥 −
𝑥𝑦
𝑧
=
1
𝑥𝑦𝑧
−
𝑦𝑧
𝑥
−
𝑧𝑥
𝑦
−
𝑥𝑦
𝑧
− 𝑥2 − 𝑧 𝑧 −
𝑥𝑦
𝑧
− 𝑦𝑥 + 𝑦 𝑧𝑥 − −
𝑧𝑥
𝑦
𝑦
=
1
𝑥𝑦𝑧
[ −
𝑦𝑧
𝑥
𝑥2 − 𝑥2 − 𝑧 −𝑥𝑦 − 𝑥𝑦 + 𝑦 𝑧𝑥 + 𝑧𝑥 ]
=
1
𝑥𝑦𝑧
−
𝑦𝑧
𝑥
0 − 𝑧 −2𝑥𝑦 + 𝑦 2𝑧𝑥 =
1
𝑥𝑦𝑧
2𝑥𝑦𝑧 + 2𝑥𝑦𝑧 =
4𝑥𝑦𝑧
𝑥𝑦𝑧
= 4](https://image.slidesharecdn.com/functionsofseverablevariables-201226040640/85/Functions-of-severable-variables-64-320.jpg)
Functions of severable variables
- 2. A function of several variables is a function where the domain is a subset of 𝑅 𝑛 and range is 𝑅. A real valued function of 𝑛–variables is a function 𝑓 ∶ 𝐷 → R, where the domain D is a subset of 𝑅 𝑛. So: for each (𝑥1, 𝑥2, . . . , 𝑥 𝑛) in D, the value of f is a real number 𝑓(𝑥1, 𝑥2, . . . , 𝑥 𝑛 ). For example, 1.𝑓(𝑥, 𝑦) = 𝑥 − 𝑦 (a function of 2 variables defined for all (𝑥, 𝑦) ∈ 𝑅2) 2. If 𝑓 is a function defined by 𝑓(𝑥, 𝑦) = 9 − cos(𝑥) + sin(𝑥2 + 𝑦2), (a function of 2 variables defined for all (𝑥, 𝑦) ∈ 𝑅2 )
- 3. 2. 𝑓 𝑥, 𝑦, 𝑧 = 1 𝑥2+𝑦2+𝑧2 Then f is a function of 3 variables, defined whenever 𝑥2 + 𝑦2 + 𝑧2 ≠ 0 This is all (𝑥, 𝑦, 𝑧) ∈ 𝑅3 except for (𝑥, 𝑦, 𝑧) = (0, 0, 0).
- 4. Partial Derivative If 𝑓(𝑥, 𝑦) is a function of two variables 𝑥 and 𝑦, the partial derivative of 𝑓 with respect to 𝑥, is given by 𝑓𝑥 𝑥, 𝑦 = lim ℎ→0 𝑓 𝑥+ℎ,𝑦 −𝑓 𝑥,𝑦 ℎ The partial derivative of 𝑓 with respect to 𝑦, is given by 𝑓𝑦 𝑥, 𝑦 = lim ℎ→0 𝑓 𝑥,𝑦+ℎ −𝑓 𝑥,𝑦 ℎ
- 5. Note • If 𝑓(𝑥, 𝑦) is a function of two variables 𝑥 and 𝑦, then Partial derivative of 𝑓(𝑥, 𝑦) with respect to 𝑥 is 𝜕𝑓 𝜕𝑥 , it is denoted by 𝑓𝑥 • Partial derivative of 𝑓(𝑥, 𝑦) with respect to 𝑦 is 𝜕𝑓 𝜕𝑦 , it is denoted by 𝑓𝑦 • Partial second derivative of 𝑓(𝑥, 𝑦) with respect to 𝑥 is 𝜕2 𝑓 𝜕𝑥2 , it is denoted by 𝑓𝑥𝑥 • Partial second derivative of 𝑓(𝑥, 𝑦) with respect to 𝑦 is 𝜕2 𝑓 𝜕𝑦2 , it is denoted by 𝑓𝑦𝑦 • Partial derivative of 𝜕𝑓 𝜕𝑥 with respect to 𝑦 is 𝜕2 𝑓 𝜕𝑦𝜕𝑥 , it is denoted by 𝑓𝑥𝑦 and so on
- 6. Problem 1: If 𝒇 𝒙, 𝒚 = 𝟑𝒙𝒚 𝟐 − 𝟐𝒙 𝟐 𝒚,then find 𝒇 𝒙, 𝒇 𝒚, 𝒇 𝒙𝒙, 𝒇 𝒚𝒚, 𝒇 𝒙𝒚 Solution: Given 𝑓 𝑥, 𝑦 = 3𝑥𝑦2 − 2𝑥2 𝑦 ---(1) Diff. (1) partially w.r.to x, 𝑓𝑥 = 3 1 𝑦2 − 2 2𝑥 𝑦 𝑓𝑥 = 3𝑦2 − 4𝑥𝑦 --------- (2) Diff. (1) partially w.r.to y, 𝑓𝑦 = 3𝑥 (2𝑦) − 2𝑥2 (1) 𝑓𝑦 = 6𝑥𝑦 − 2𝑥2 -----------(3) Diff. (2) p.w.r.to x, 𝑓𝑥𝑥 = 0 − 4 1 𝑦 𝑓𝑥𝑥 = −4𝑦 Diff. (3) p.w.r.to y, 𝑓𝑦𝑦 = 6𝑥 1 − 0 𝑓𝑦𝑦 = 6𝑥 Diff. (2) p.w.r.to y, 𝑓𝑥𝑦 = 3 2𝑦 − 4𝑥(1) 𝑓𝑥𝑦 = 6𝑦 − 4𝑥 Note : Diff.(3) p.w.r.to x, 𝑓𝑦𝑥 = 6 1 𝑦 − 2 2𝑥 = 6𝑦 − 4𝑥 𝑓𝑥𝑦 = 𝑓𝑦𝑥
- 7. Problem 2: Find 𝝏𝒖 𝝏𝒙 , 𝝏𝒖 𝝏𝒚 , if 𝒖 = 𝒙𝒆 𝒚 + 𝒚 𝒆 𝒙 Solution: Given 𝑢 = 𝑥𝑒 𝑦 + 𝑦 𝑒 𝑥 -------(1) Diff. (1) p.w.r.to x, 𝜕𝑢 𝜕𝑥 = 1 𝑒 𝑦 + 𝑦 𝑒 𝑥 = 𝑒 𝑦 + 𝑦 𝑒 𝑥 Diff.(1) p.w.r.to y, 𝜕𝑢 𝜕𝑦 = 𝑥 𝑒 𝑦 + 1 𝑒 𝑥 = 𝑥𝑒 𝑦 + 𝑒 𝑥
- 8. Problem 3: If 𝒖 = 𝒙 𝟐 + 𝒚 𝟐 + 𝒛 𝟐 − 𝟏 𝟐 then find the value of 𝝏 𝟐 𝒖 𝝏𝒙 𝟐 + 𝝏 𝟐 𝒖 𝝏𝒚 𝟐 + 𝝏 𝟐 𝒖 𝝏𝒛 𝟐 Solution: Given 𝑢 = 𝑥2 + 𝑦2 + 𝑧2 −1 2 ------(1) Diff. (1) p. w.r.to x, 𝜕𝑢 𝜕𝑥 = − 1 2 𝑥2 + 𝑦2 + 𝑧2 − 1 2 −1 2𝑥 + 0 + 0 = − 1 2 𝑥2 + 𝑦2 + 𝑧2 − 3 2(2𝑥) 𝜕𝑢 𝜕𝑥 = −𝑥 𝑥2 + 𝑦2 + 𝑧2 − 3 2 ------- (2) Diff. (2) p. w.r.to x, 𝜕2 𝑢 𝜕𝑥2 = −𝑥 − 3 2 𝑥2 + 𝑦2 + 𝑧2 − 3 2 −1 (2𝑥) + 𝑥2 + 𝑦2 + 𝑧2 − 3 2(−1) = 3𝑥2 𝑥2 + 𝑦2 + 𝑧2 − 3 2 −1 − 𝑥2 + 𝑦2 + 𝑧2 − 3 2 = 𝑥2 + 𝑦2 + 𝑧2 − 3 2 3𝑥2 𝑥2 + 𝑦2 + 𝑧2 −1 − 1
- 9. = 𝑥2 + 𝑦2 + 𝑧2 − 3 2 3𝑥2 𝑥2+𝑦2+𝑧2 1 − 1 = 𝑥2+𝑦2+𝑧2 − 3 2 3𝑥2− 𝑥2+𝑦2+𝑧2 𝑥2+𝑦2+𝑧2 = 𝑥2 + 𝑦2 + 𝑧2 − 3 2 𝑥2 + 𝑦2 + 𝑧2 −1 3𝑥2 − 𝑥2 − 𝑦2 − 𝑧2 = 𝑥2 + 𝑦2 + 𝑧2 − 3 2 −1 2𝑥2 − 𝑦2 − 𝑧2 𝜕2 𝑢 𝜕𝑥2 = 𝑥2 + 𝑦2 + 𝑧2 − 5 2 2𝑥2 − 𝑦2 − 𝑧2 ------ (3) Similarly, 𝜕2 𝑢 𝜕𝑦2 = 𝑥2 + 𝑦2 + 𝑧2 − 5 2 2𝑦2 − 𝑧2 − 𝑥2 ------- (4) 𝜕2 𝑢 𝜕𝑧2 = 𝑥2 + 𝑦2 + 𝑧2 − 5 2 2𝑧2 − 𝑥2 − 𝑦2 ------ (5) (3)+(4)+(5) ⇒ 𝜕2 𝑢 𝜕𝑥2+ 𝜕2 𝑢 𝜕𝑦2+ 𝜕2 𝑢 𝜕𝑧2 = 𝑥2 + 𝑦2 + 𝑧2 − 5 2(2𝑥2 − 𝑦2 − 𝑧2 +2𝑦2 − 𝑧2 − 𝑥2 +2𝑧2 − 𝑥2 − 𝑦2 ) = 𝑥2 + 𝑦2 + 𝑧2 − 5 2 2𝑥2 + 2𝑦2 + 2𝑧2 − 2𝑥2 − 2𝑦2 − 2𝑧2 = 0
- 10. Problem 4 If 𝒖 = 𝒍𝒐𝒈 𝒕𝒂𝒏𝒙 + 𝒕𝒂𝒏𝒚 + 𝒕𝒂𝒏𝒛 𝒇𝒊𝒏𝒅 𝒔𝒊𝒏𝟐𝒙. 𝝏𝒖 𝝏𝒙 . Solution: Given 𝑢 = log 𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛𝑦 + 𝑡𝑎𝑛𝑧 ------- (1) Diff. (1) p.w.r.to x, 𝜕𝑢 𝜕𝑥 = 1 𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 (sec2 𝑥) = 1 cos2 𝑥(𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧) , (∵ sec2 𝑥 = 1 cos2 𝑥 ) Now 𝑠𝑖𝑛2𝑥. 𝜕𝑢 𝜕𝑥 = 𝑠𝑖𝑛2𝑥 1 cos2 𝑥(𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧) = 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 cos2 𝑥(𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧) , (∵ 𝑠𝑖𝑛2𝑥 = 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥) = 2𝑠𝑖𝑛𝑥 cos 𝑥(𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧) = 2 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 ( 1 𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 )
- 11. ∴ 𝑠𝑖𝑛2𝑥. 𝜕𝑢 𝜕𝑥 = 2𝑡𝑎𝑛𝑥 ( 1 𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 )----(2) Similarly, 𝑠𝑖𝑛2𝑦. 𝜕𝑢 𝜕𝑦 = 2𝑡𝑎𝑛𝑦 ( 1 𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 )----- (3) 𝑠𝑖𝑛2𝑧. 𝜕𝑢 𝜕𝑧 = 2𝑡𝑎𝑛𝑧 ( 1 𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 )------ (4) (3)+(4)+(5)⟹ 𝑠𝑖𝑛2𝑥. 𝜕𝑢 𝜕𝑥 +𝑠𝑖𝑛2𝑦. 𝜕𝑢 𝜕𝑦 +𝑠𝑖𝑛2𝑧. 𝜕𝑢 𝜕𝑧 =2𝑡𝑎𝑛𝑥 1 𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 + 2𝑡𝑎𝑛𝑦 1 𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 +2𝑡𝑎𝑛𝑧 1 𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 =2 1 𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛𝑦 + 𝑡𝑎𝑛𝑧 = 2
- 12. Homogeneous Function: A function 𝑓(𝑥, 𝑦) is called a homogeneous function of the degree ′𝑛′ if the following relationship is valid for all 𝑡 > 0: 𝑓 𝑡𝑥, 𝑡𝑦 = 𝑡 𝑛 𝑓 𝑥, 𝑦 . Example: Consider 𝑓 𝑥, 𝑦 = 𝑥2+𝑦2 𝑥𝑦 𝑓 𝑡𝑥, 𝑡𝑦 = 𝑡𝑥 2+ 𝑡𝑦 2 𝑡𝑥 𝑡𝑦 = 𝑡2 𝑥2+𝑡2 𝑦2 𝑡2 𝑥𝑦 = 𝑡2 𝑥2+𝑦2 𝑡2 𝑥𝑦 = 𝑥2+𝑦2 𝑥𝑦 = 𝑡0 𝑓(𝑥, 𝑦) Hence f(x, y ) is homogeneous of order zero
- 13. Euler’s Theorem: If 𝑢 = 𝑓 𝑥, 𝑦 is a homogeneous function of degree ‘n’, then 𝑥 𝜕𝑢 𝜕𝑥 + 𝑦 𝜕𝑢 𝜕𝑦 = 𝑛𝑢 (OR) 𝑥 𝜕𝑓 𝜕𝑥 + 𝑦 𝜕𝑓 𝜕𝑦 = 𝑛𝑓 Problem 1: If 𝒖 = 𝒇 𝒙 𝒚 , prove that 𝒙 𝝏𝒖 𝝏𝒙 + 𝒚 𝝏𝒖 𝝏𝒚 = 𝟎 Solution: Given 𝑢 𝑥, 𝑦 = 𝑓 𝑥 𝑦 For any 𝑡 > 0, 𝑢 𝑡𝑥, 𝑡𝑦 = 𝑓 𝑡𝑥 𝑡𝑦 = 𝑓 𝑥 𝑦 = 𝑡0 𝑓 𝑥 𝑦 = 𝑡0 𝑢(𝑥, 𝑦) Hence 𝑢 = 𝑓 𝑥 𝑦 is a homogeneous function of order zero 𝑖. 𝑒 𝑛 = 0 By Euler’s Theorem , 𝑥 𝜕𝑢 𝜕𝑥 + 𝑦 𝜕𝑢 𝜕𝑦 = 𝑛𝑢 ⟹ 𝑥 𝜕𝑢 𝜕𝑥 + 𝑦 𝜕𝑢 𝜕𝑦 = 0 u = 0 Hence proved. .
- 14. Problem 2: If 𝒖 = 𝐬𝐢𝐧−𝟏 𝒙 𝒚 + 𝐭𝐚𝐧−𝟏 𝒚 𝒙 , prove that 𝒙 𝝏𝒖 𝝏𝒙 + 𝒚 𝝏𝒖 𝝏𝒚 = 𝟎. Solution: Given 𝑢(𝑥, 𝑦) = sin−1 𝑥 𝑦 + tan−1 𝑦 𝑥 For any 𝑡 > 0, 𝑢(𝑡𝑥, 𝑡𝑦) = sin−1 𝑡𝑥 𝑡𝑦 + tan−1 𝑡𝑦 𝑡𝑥 = sin−1 𝑥 𝑦 + tan−1 𝑦 𝑥 = 𝑡0 ( sin−1 𝑥 𝑦 + tan−1 𝑦 𝑥 ) Hence 𝑢(𝑥, 𝑦) is a homogeneous function of order zero 𝑖. 𝑒 𝑛 = 0 By Euler’s Theorem, 𝑥 𝜕𝑢 𝜕𝑥 + 𝑦 𝜕𝑢 𝜕𝑦 = 𝑛𝑢 = 0 𝑢 = 0
- 15. Total differential coefficient If 𝑢 = 𝑓 𝑥, 𝑦 , the total differential of 𝑢 is given by 𝑑𝑢 = 𝑢 𝑥 𝑑𝑥 + 𝑢 𝑦 𝑑𝑦 (OR) 𝑑𝑓 = 𝑓𝑥 𝑑𝑥 + 𝑓𝑦 𝑑𝑦 If 𝑢 = 𝑓 𝑥, 𝑦 , and x = g(t) , y = h(t) , the total differential of 𝑢 is given by 𝑑𝑢 𝑑𝑡 = 𝜕𝑢 𝜕𝑥 𝑑𝑥 𝑑𝑡 + 𝜕𝑢 𝜕𝑦 𝑑𝑦 𝑑𝑡 = 𝑢 𝑥 𝑑𝑥 𝑑𝑡 + 𝑢 𝑦 𝑑𝑦 𝑑𝑡 (OR) 𝑑𝑓 𝑑𝑡 = 𝜕𝑓 𝜕𝑥 𝑑𝑥 𝑑𝑡 + 𝜕𝑓 𝜕𝑦 𝑑𝑦 𝑑𝑡 = 𝑓𝑥 𝑑𝑥 𝑑𝑡 + 𝑓𝑦 𝑑𝑦 𝑑𝑡
- 16. Note : If 𝑢 = 𝑓 𝑝, 𝑞, 𝑟 and p = 𝑔 𝑥, 𝑦, 𝑧 , 𝑞 = ℎ(𝑥, 𝑦, 𝑧) 𝑟 = 𝑘(𝑥, 𝑦, 𝑧) Total differential 𝑑𝑢 𝑑𝑥 = 𝑢 𝑝 𝑑𝑝 𝑑𝑥 + 𝑢 𝑞 𝑑𝑞 𝑑𝑥 + 𝑢 𝑟 𝑑𝑟 𝑑𝑥 𝑑𝑢 𝑑𝑦 = 𝑢 𝑝 𝑑𝑝 𝑑𝑦 + 𝑢 𝑞 𝑑𝑞 𝑑𝑦 + 𝑢 𝑟 𝑑𝑟 𝑑𝑦 𝑑𝑢 𝑑𝑧 = 𝑢 𝑝 𝑑𝑝 𝑑𝑧 + 𝑢 𝑞 𝑑𝑞 𝑑𝑧 + 𝑢 𝑟 𝑑𝑟 𝑑𝑧
- 17. Partial derivative: If 𝑧 = 𝑓 𝑢, 𝑣 and 𝑢 = 𝑔 𝑥, 𝑦 , 𝑣 = ℎ(𝑥, 𝑦) then 𝜕𝑧 𝜕𝑥 = 𝜕𝑧 𝜕𝑢 𝜕𝑢 𝜕𝑥 + 𝜕𝑧 𝜕𝑣 𝜕𝑣 𝜕𝑥 = 𝑧 𝑢 𝑢 𝑥 + 𝑧 𝑣 𝑣 𝑥 𝜕𝑧 𝜕𝑦 = 𝜕𝑧 𝜕𝑢 𝜕𝑢 𝜕𝑦 + 𝜕𝑧 𝜕𝑣 𝜕𝑣 𝜕𝑦 = 𝑧 𝑢 𝑢 𝑦 + 𝑧 𝑣 𝑣 𝑦 Note : If 𝑢 = 𝑓 𝑝, 𝑞, 𝑟 and p = 𝑔 𝑥, 𝑦, 𝑧 , 𝑞 = ℎ(𝑥, 𝑦, 𝑧) 𝑟 = 𝑘(𝑥, 𝑦, 𝑧) Partial differential 𝜕𝑢 𝜕𝑥 = 𝑢 𝑝 𝑝 𝑥 + 𝑢 𝑞 𝑞 𝑥 + 𝑢 𝑟 𝑟𝑥 𝜕𝑢 𝜕𝑦 = 𝑢 𝑝 𝑝 𝑦 + 𝑢 𝑞 𝑞 𝑦 + 𝑢 𝑟 𝑟𝑦 𝜕𝑢 𝜕𝑧 = 𝑢 𝑝 𝑝 𝑧 + 𝑢 𝑞 𝑞 𝑧 + 𝑢 𝑟 𝑟𝑧
- 18. Problem 1: If 𝒘 = 𝒇(𝒚 − 𝒛, 𝒛 − 𝒙, 𝒙 − 𝒚) then show that 𝝏𝒘 𝝏𝒙 + 𝝏𝒘 𝝏𝒚 + 𝝏𝒘 𝝏𝒛 = 𝟎 Solution: Given 𝑤 = 𝑓(𝑦 − 𝑧, 𝑧 − 𝑥, 𝑥 − 𝑦) Take p = 𝑦 − 𝑧, 𝑞 = 𝑧 − 𝑥, 𝑟 = 𝑥 − 𝑦 ⇒ 𝑤 = 𝑓(𝑝, 𝑞, 𝑟) and 𝑝 𝑦, 𝑧 , 𝑞 𝑧, 𝑥 , 𝑟(𝑥, 𝑦) p = 𝑦 − 𝑧, 𝑞 = 𝑧 − 𝑥, 𝑟 = 𝑥 − 𝑦 𝑝 𝑥 = 0 𝑞 𝑥 = −1 𝑟𝑥 = 1 𝑝 𝑦 = 1 𝑞 𝑦 = 0 𝑟𝑦 = −1 𝑝 𝑧 = −1 𝑞 𝑧 = 1 𝑟𝑧 = 0 𝜕𝑤 𝜕𝑥 = 𝑤 𝑝 𝑝 𝑥 + 𝑤 𝑞 𝑞 𝑥 + 𝑤𝑟 𝑟𝑥 = 𝑤 𝑝 0 + 𝑤 𝑞 −1 + 𝑤𝑟 1 𝜕𝑤 𝜕𝑥 = −𝑤 𝑞 + 𝑤𝑟 ------ (1) 𝜕𝑤 𝜕𝑦 = 𝑤 𝑝 𝑝 𝑦 + 𝑤 𝑞 𝑞 𝑦 + 𝑤𝑟 𝑟𝑦 = 𝑤 𝑝 1 + 𝑤 𝑞 0 + 𝑤𝑟 −1 𝜕𝑤 𝜕𝑦 = 𝑤 𝑝 − 𝑤𝑟 ------ (2) 𝜕𝑤 𝜕𝑧 = 𝑤 𝑝 𝑝 𝑧 + 𝑤 𝑞 𝑞 𝑧 + 𝑤𝑟 𝑟𝑧 = 𝑤 𝑝 −1 + 𝑤 𝑞 1 + 𝑤𝑟 0 𝜕𝑤 𝜕𝑧 = −𝑤 𝑝 + 𝑤 𝑞 ------ (3) (1)+(2)+(3) ⇒ 𝜕𝑤 𝜕𝑥 + 𝜕𝑤 𝜕𝑦 + 𝜕𝑤 𝜕𝑧 = −𝑤 𝑞 + 𝑤𝑟+𝑤 𝑝 − 𝑤𝑟 −𝑤 𝑝 +𝑤 𝑞 = 0
- 19. Problem2: If 𝒖 = 𝒇(𝟐𝒙 − 𝟑𝒚, 𝟑𝒚 − 𝟒𝒛, 𝟒𝒛 − 𝟐𝒙) then find 𝟏 𝟐 𝝏𝒖 𝝏𝒙 + 𝟏 𝟑 𝝏𝒖 𝝏𝒚 + 𝟏 𝟒 𝝏𝒖 𝝏𝒛 Solution: Given 𝑢 = 𝑓(2𝑥 − 3𝑦, 3𝑦 − 4𝑧, 4𝑧 − 2𝑥) Take p = 2𝑥 − 3𝑦, 𝑞 = 3𝑦 − 4𝑧, 𝑟 = 4𝑧 − 2𝑥 ⇒ 𝑢 = 𝑓(𝑝, 𝑞, 𝑟) and 𝑝 𝑥, 𝑦 , 𝑞 𝑦, 𝑧 , 𝑟(𝑧, 𝑥) p = 2𝑥 − 3𝑦, 𝑞 = 3𝑦 − 4𝑧, 𝑟 = 4𝑧 − 2𝑥 𝑝 𝑥 = 2 𝑞 𝑥 = 0 𝑟𝑥 = −2 𝑝 𝑦 = −3 𝑞 𝑦 = 3 𝑟𝑦 = 0 𝑝 𝑧 = 0 𝑞 𝑧 = −4 𝑟𝑧 = 4 𝜕𝑢 𝜕𝑥 = 𝑢 𝑝 𝑝 𝑥 + 𝑢 𝑞 𝑞 𝑥 + 𝑢 𝑟 𝑟𝑥 = 𝑢 𝑝 2 + 𝑢 𝑞 0 + 𝑢 𝑟 −2 1 2 𝜕𝑢 𝜕𝑥 = 1 2 2𝑢 𝑝 − 2𝑢 𝑟 = 1 2 (2 𝑢 𝑝 − 𝑢 𝑟 ) = 𝑢 𝑝 − 𝑢 𝑟 --------- (1) 𝜕𝑢 𝜕𝑦 = 𝑢 𝑝 𝑝 𝑦 + 𝑢 𝑞 𝑞 𝑦 + 𝑢 𝑟 𝑟𝑦 = 𝑢 𝑝 −3 + 𝑢 𝑞 3 + 𝑢 𝑟 0 1 3 𝜕𝑢 𝜕𝑦 = 1 3 (−3𝑢 𝑝+3𝑢 𝑞) = 1 3 3 −up + uq = −𝑢 𝑝 + 𝑢 𝑞 ---------(2) 𝜕𝑢 𝜕𝑧 = 𝑢 𝑝 𝑝 𝑧 + 𝑢 𝑞 𝑞 𝑧 + 𝑢 𝑟 𝑟𝑧 = 𝑢 𝑝 0 + 𝑢 𝑞 −4 + 𝑢 𝑟 4 1 4 𝜕𝑢 𝜕𝑧 = 1 4 −4𝑢 𝑞 + 4𝑢 𝑟 = 1 4 4 −𝑢 𝑞 ∓ 𝑢 𝑟 = −𝑢 𝑞 + 𝑢 𝑟 ------- (3) (1)+(2)+(3) ⇒ 1 2 𝜕𝑢 𝜕𝑥 + 1 3 𝜕𝑢 𝜕𝑦 + 1 4 𝜕𝑢 𝜕𝑧 = 𝑢 𝑝 − 𝑢 𝑟 − 𝑢 𝑝 + 𝑢 𝑞 − 𝑢 𝑞 + 𝑢 𝑟 = 0
- 20. Problem3: If 𝒖 = 𝒇( 𝒚−𝒙 𝒙𝒚 , 𝒛−𝒙 𝒙𝒛 ) then find 𝒙 𝟐 𝝏𝒖 𝝏𝒙 + 𝒚 𝟐 𝝏𝒖 𝝏𝒚 + 𝒛 𝟐 𝝏𝒖 𝝏𝒛 Solution: Given𝒖 = 𝒇( 𝒚−𝒙 𝒙𝒚 , 𝒛−𝒙 𝒙𝒛 ) Take p = 𝒚−𝒙 𝒙𝒚 , 𝑞 = 𝒛−𝒙 𝒙𝒛 ⇒ 𝑢 = 𝑓(𝑝, 𝑞) and 𝑝 𝑥, 𝑦 , 𝑞 𝑥, 𝑧 𝐩 = 𝐲 − 𝐱 𝐱𝐲 𝐪 = 𝐳 − 𝐱 𝐱𝐳 𝐩 𝐱 = 𝐱𝐲 −𝟏 − 𝐲−𝐱 𝐲 𝐱𝐲 𝟐 = −𝐱𝐲−𝐲 𝟐+𝐱𝐲 𝐱 𝟐 𝐲 𝟐 = − 𝐲 𝟐 𝐱 𝟐 𝐲 𝟐 = − 𝟏 𝐱 𝟐 𝐪 𝐱 = 𝐱𝐳 −𝟏 − 𝐳−𝐱 𝐳 𝐱𝐳 𝟐 = −𝐱𝐳−𝐳 𝟐+𝐱𝐳 𝐱 𝟐 𝐳 𝟐 = − 𝐳 𝟐 𝐱 𝟐 𝐳 𝟐 = − 𝟏 𝐱 𝟐 𝐩 𝐲 = 𝐱𝐲 𝟏 − 𝐲−𝐱 𝐱 𝐱𝐲 𝟐 = 𝐱𝐲−𝐱𝐲+𝐱 𝟐 𝐱 𝟐 𝐲 𝟐 = 𝐱 𝟐 𝐱 𝟐 𝐲 𝟐 = 𝟏 𝐲 𝟐 𝐪 𝐲 = 𝟎 𝐩 𝐳 = 𝟎 𝐪 𝐳 = 𝐱𝐳 𝟏 − 𝐳−𝐱 𝐱 𝐱𝐳 𝟐 = 𝐱𝐳−𝐳𝐱+𝐱 𝟐 𝐱 𝟐 𝐳 𝟐 = 𝐱 𝟐 𝐱 𝟐 𝐳 𝟐 = 𝟏 𝐳 𝟐
- 21. 𝜕𝑢 𝜕𝑥 = 𝑢 𝑝 𝑝 𝑥 + 𝑢 𝑞 𝑞 𝑥 = 𝑢 𝑝 − 1 𝑥2 + 𝑢 𝑞 − 1 𝑥2 = − 1 𝑥2 (𝑢 𝑝 + 𝑢 𝑞) 𝑥2 𝜕𝑢 𝜕𝑥 = 𝑥2 − 1 𝑥2 𝑢 𝑝 + 𝑢 𝑞 = −𝑢 𝑝 − 𝑢 𝑞 --------- (1) 𝜕𝑢 𝜕𝑦 = 𝑢 𝑝 𝑝 𝑦 + 𝑢 𝑞 𝑞 𝑦 = 𝑢 𝑝 1 𝑦2 + 𝑢 𝑞 0 𝑦2 𝜕𝑢 𝜕𝑦 = 𝑦2 ( 1 𝑦2 𝑢 𝑝) = 𝑢 𝑝 ---------(2) 𝜕𝑢 𝜕𝑧 = 𝑢 𝑝 𝑝 𝑧 + 𝑢 𝑞 𝑞 𝑧 = 𝑢 𝑝 0 + 𝑢 𝑞 1 𝑧2 𝑧2 𝜕𝑢 𝜕𝑧 = 𝑧2 1 𝑧2 (𝑢 𝑞) = 𝑢 𝑞 ------- (3) (1)+(2)+(3) ⇒ 𝑥2 𝜕𝑢 𝜕𝑥 + 𝑦2 𝜕𝑢 𝜕𝑦 + 𝑧2 𝜕𝑢 𝜕𝑧 = −𝑢 𝑝 − 𝑢 𝑞 + 𝑢 𝑝 + 𝑢 𝑞 = 0
- 22. • If 𝑓(𝑥, 𝑦) is a function of two variables x and y, then Taylor’s expansion of 𝑓(𝑥, 𝑦) about the point (𝑎, 𝑏) is given by 𝑓 𝑥, 𝑦 = 𝑓 𝑎, 𝑏 + 𝑥 − 𝑎 𝑓𝑥 𝑎, 𝑏 + 𝑦 − 𝑏 𝑓𝑦 𝑎, 𝑏 + 1 2! 𝑥 − 𝑎 2 𝑓𝑥𝑥 𝑎, 𝑏 + 2 𝑥 − 𝑎 𝑦 − 𝑏 𝑓𝑥𝑦 𝑎, 𝑏 + 𝑦 − 𝑏 2 𝑓𝑦𝑦 𝑎, 𝑏 + 1 3! 𝑥 − 𝑎 3 𝑓𝑥𝑥𝑥 𝑎, 𝑏 + 3 𝑥 − 𝑎 2 𝑦 − 𝑏 𝑓𝑥𝑥𝑦 + 3 𝑥 − 𝑎 𝑦 − 𝑏 2 𝑓𝑥𝑦𝑦 + 𝑦 − 𝑏 3 𝑓𝑦𝑦𝑦 𝑎, 𝑏 + ⋯ • Taylor’s expansion of 𝑓(𝑥, 𝑦) about the origin or (0,0) is given by 𝑓 𝑥, 𝑦 = 𝑓 0,0 + 𝑥𝑓𝑥 0,0 + 𝑦𝑓𝑦 0,0 + 1 2! 𝑥2 𝑓𝑥𝑥 0,0 + 2𝑥𝑦𝑓𝑥𝑦 0,0 + 𝑦2 𝑓𝑦𝑦 0,0 + ⋯ Taylor’s Theorem for functions of two variables
- 23. Problem 1: Obtain the Taylor series of 𝒙 𝟑 + 𝒚 𝟑 + 𝒙𝒚 𝟐 in powers of 𝒙 − 𝟏 𝒂𝒏𝒅 𝒚 − 𝟐. Solution: W.kt, Taylor’s expansion of 𝑓(𝑥, 𝑦) about the point (𝑎, 𝑏) is given by 𝑓 𝑥, 𝑦 = 𝑓 𝑎, 𝑏 + 𝑥 − 𝑎 𝑓𝑥 𝑎, 𝑏 + 𝑦 − 𝑏 𝑓𝑦 𝑎, 𝑏 + 1 2! 𝑥 − 𝑎 2 𝑓𝑥𝑥 𝑎, 𝑏 + 2 𝑥 − 𝑎 𝑦 − 𝑏 𝑓𝑥𝑦 𝑎, 𝑏 + 𝑦 − 𝑏 2 𝑓𝑦𝑦 𝑎, 𝑏 + 1 3! 𝑥 − 𝑎 3 𝑓𝑥𝑥𝑥 𝑎, 𝑏 + 3 𝑥 − 𝑎 2 𝑦 − 𝑏 𝑓𝑥𝑥𝑦(𝑎, 𝑏) + 3 𝑥 − 𝑎 𝑦 − 𝑏 2 𝑓𝑥𝑦𝑦(𝑎, 𝑏) + 𝑦 − 𝑏 3 𝑓𝑦𝑦𝑦 𝑎, 𝑏 + ⋯ Here 𝑓 𝑥, 𝑦 = 𝑥3 + 𝑦3 + 𝑥𝑦2 , the points are 𝑎 = 1, 𝑏 = 2 , (∵ 𝑥 − 1 = 0 & 𝑦 − 2 = 0 ⇒ 𝑥 = 1, 𝑦 = 2 )
- 24. 𝑓 𝑥, 𝑦 = 𝑥3 + 𝑦3 + 𝑥𝑦2 𝑓 1,2 = 13 + 23 + 1 22 = 1 + 8 + 4 = 13 𝑓𝑥 = 3𝑥2 + 𝑦2 𝑓𝑥 1,2 = 3 1 2 + 22 = 3 + 4 = 7 𝑓𝑦 = 3𝑦2 + 𝑥 2𝑦 = 3𝑦2 + 2𝑥𝑦 𝑓𝑦 1,2 = 3 2 2 + 2 1 2 = 12 + 4 = 16 𝑓𝑥𝑥 = 6𝑥 + 0 = 6𝑥 𝑓𝑥𝑥 1,2 = 6 1 = 6 𝑓𝑥𝑦 = 0 + 2𝑦 = 2𝑦 𝑓𝑥𝑦 1,2 = 2 2 = 4 𝑓𝑦𝑦 = 3 2𝑦 + 2𝑥 1 = 6𝑦 + 2𝑥 𝑓𝑦𝑦 1,2 = 6 2 + 2 1 = 12 + 2 = 14 𝑓𝑥𝑥𝑥 = 6 1 = 6 𝑓𝑥𝑥𝑥 1,2 = 6 𝑓𝑥𝑥𝑦 =0 𝑓𝑥𝑥𝑦(1,2) = 0 𝑓𝑥𝑦𝑦 = 2 1 = 2 𝑓𝑥𝑦𝑦(1,2) = 2 𝑓𝑦𝑦𝑦 = 6 1 + 0 = 6 𝑓𝑦𝑦𝑦(1,2) = 6
- 25. Taylor’s expansion of 𝑓(𝑥, 𝑦) about the point (1,2) is given by 𝑓 𝑥, 𝑦 = 𝑓 1,2 + 𝑥 − 1 𝑓𝑥 1,2 + 𝑦 − 2 𝑓𝑦 1,2 + 1 2! 𝑥 − 1 2 𝑓𝑥𝑥 1,2 + 2 𝑥 − 1 𝑦 − 2 𝑓𝑥𝑦 1,2 + 𝑦 − 2 2 𝑓𝑦𝑦 1,2 + 1 3! 𝑥 − 1 3 𝑓𝑥𝑥𝑥 1,2 + 3 𝑥 − 1 2 𝑦 − 2 𝑓𝑥𝑥𝑦(1,2) + 3 𝑥 − 1 𝑦 − 2 2 𝑓𝑥𝑦𝑦 (1,2) + 𝑦 − 2 3 𝑓𝑦𝑦𝑦 1,2 + ⋯ 𝑓 𝑥, 𝑦 = 13 + 7 𝑥 − 1 + 16 𝑦 − 2 + 1 2! [6 𝑥 − 1 2 + 2 4 𝑥 − 1 𝑦 − 2 + 14 𝑦 − 2 2 ] + 1 3! 6 𝑥 − 1 3 + 3 0 𝑥 − 1 2 𝑦 − 2 + 3 2 𝑥 − 1 𝑦 − 2 2 + 6 𝑦 − 2 3 + ⋯ = 13 + 7 𝑥 − 1 + 16 𝑦 − 2 + 1 2! [6 𝑥 − 1 2 + 8 𝑥 − 1 𝑦 − 2 + 14 𝑦 − 2 2 ] + 1 3! 6 𝑥 − 1 3 + 0 + 6 𝑥 − 1 𝑦 − 2 2 + 6 𝑦 − 2 3 + ⋯ = 13 + 7 𝑥 − 1 + 16 𝑦 − 2 + 6 2 𝑥 − 1 2 + 8 2 𝑥 − 1 𝑦 − 2 + 14 2 𝑦 − 2 2 + 6 6 𝑥 − 1 3 + 6 6 𝑥 − 1 𝑦 − 2 2 + 6 6 𝑦 − 2 3 +… , (2! = 1.2 = 2, 3! = 1.2.3 = 6)
- 26. = 13 + 7 𝑥 − 1 + 16 𝑦 − 2 + 6 2 𝑥 − 1 2 + 8 2 𝑥 − 1 𝑦 − 2 + 14 2 𝑦 − 2 2 + 𝑥 − 1 3 + 𝑥 − 1 𝑦 − 2 2 + 𝑦 − 2 3 + ⋯ Problem 2 Expand 𝒆 𝒙 𝒄𝒐𝒔𝒚 at 𝟎, 𝝅 𝟐 upto the third term by using Taylor’s series. Solution W.kt, Taylor’s expansion of 𝑓(𝑥, 𝑦) about the point (𝑎, 𝑏) is given by 𝑓 𝑥, 𝑦 = 𝑓 𝑎, 𝑏 + 𝑥 − 𝑎 𝑓𝑥 𝑎, 𝑏 + 𝑦 − 𝑏 𝑓𝑦 𝑎, 𝑏 + 1 2! 𝑥 − 𝑎 2 𝑓𝑥𝑥 𝑎, 𝑏 + 2 𝑥 − 𝑎 𝑦 − 𝑏 𝑓𝑥𝑦 𝑎, 𝑏 + 𝑦 − 𝑏 2 𝑓𝑦𝑦 𝑎, 𝑏 + 1 3! 𝑥 − 𝑎 3 𝑓𝑥𝑥𝑥 𝑎, 𝑏 + 3 𝑥 − 𝑎 2 𝑦 − 𝑏 𝑓𝑥𝑥𝑦(𝑎, 𝑏) + 3 𝑥 − 𝑎 𝑦 − 𝑏 2 𝑓𝑥𝑦𝑦(𝑎, 𝑏) + 𝑦 − 𝑏 3 𝑓𝑦𝑦𝑦 𝑎, 𝑏 + ⋯ Here 𝑓 𝑥, 𝑦 = 𝑒 𝑥 𝑐𝑜𝑠𝑦 the points are 𝑎 = 0, 𝑏 = 𝜋 2 ,
- 27. 𝑓 𝑥, 𝑦 = 𝑒 𝑥 𝑐𝑜𝑠𝑦 𝑓 0, 𝜋 2 = 𝑒0 cos 𝜋 2 = 1.0 = 0, (∵ 𝑒0 = 1, cos 𝜋 2 = 0 ) 𝑓𝑥 = 𝑒 𝑥 𝑐𝑜𝑠𝑦 𝑓𝑥 0, 𝜋 2 = 𝑒0 cos 𝜋 2 = 0 𝑓𝑦 = 𝑒 𝑥 −𝑠𝑖𝑛𝑦 = −𝑒 𝑥 𝑠𝑖𝑛𝑦 𝑓𝑦 0, 𝜋 2 = −𝑒0 sin 𝜋 2 = −1.1 = −1 (∵ 𝑒0 = 1, sin 𝜋 2 = 1) 𝑓𝑥𝑥 = 𝑒 𝑥 𝑐𝑜𝑠𝑦 𝑓𝑥𝑥 0, 𝜋 2 = 𝑒0 cos 𝜋 2 = 0 𝑓𝑥𝑦 = −𝑒 𝑥 𝑠𝑖𝑛𝑦 𝑓𝑥𝑦 0, 𝜋 2 = −1 𝑓𝑦𝑦 = −𝑒 𝑥 (𝑐𝑜𝑠𝑦) 𝑓𝑦𝑦 0, 𝜋 2 = −𝑒0 cos 𝜋 2 = 0 𝑓𝑥𝑥𝑥 = 𝑒 𝑥 𝑐𝑜𝑠𝑦 𝑓𝑥𝑥𝑥 0, 𝜋 2 = 0 𝑓𝑥𝑥𝑦 = −𝑒 𝑥 𝑠𝑖𝑛𝑦 𝑓𝑥𝑥𝑦 0, 𝜋 2 = −1 𝑓𝑥𝑦𝑦 = −𝑒 𝑥 𝑐𝑜𝑠𝑦 𝑓𝑥𝑦𝑦(0, 𝜋 2 ) = 0 𝑓𝑦𝑦𝑦 = −𝑒 𝑥 −𝑠𝑖𝑛𝑦 = 𝑒 𝑥 𝑠𝑖𝑛𝑦 𝑓𝑦𝑦𝑦 0, 𝜋 2 = 1.1 = 1
- 28. Taylor’s expansion of 𝑓(𝑥, 𝑦) about the point (0, 𝜋 2 ) is given by 𝑓 𝑥, 𝑦 = 𝑓 0, 𝜋 2 + 𝑥 − 0 𝑓𝑥 0, 𝜋 2 + 𝑦 − 𝜋 2 𝑓𝑦 0, 𝜋 2 + 1 2! 𝑥 − 0 2 𝑓𝑥𝑥 0, 𝜋 2 + 2 𝑥 − 0 𝑦 − 𝜋 2 𝑓𝑥𝑦 0, 𝜋 2 + 𝑦 − 𝜋 2 2 𝑓𝑦𝑦 0, 𝜋 2 + 1 3! 𝑥 − 0 3 𝑓𝑥𝑥𝑥 0, 𝜋 2 + 3 𝑥 − 0 2 𝑦 − 𝜋 2 𝑓𝑥𝑥𝑦(0, 𝜋 2 ) + 3 𝑥 − 0 𝑦 − 𝜋 2 2 𝑓𝑥𝑦𝑦(0, 𝜋 2 ) + 𝑦 − 𝜋 2 3 𝑓𝑦𝑦𝑦 0, 𝜋 2 + ⋯ = 0 + 𝑥 0 + 𝑦 − 𝜋 2 −1 + 1 2! 𝑥 2 0 + 2 𝑥 𝑦 − 𝜋 2 −1 + 𝑦 − 𝜋 2 2 0 + 1 3! 𝑥 3 0 + 3 𝑥 2 𝑦 − 𝜋 2 −1 + 3 𝑥 𝑦 − 𝜋 2 2 (0) + 𝑦 − 𝜋 2 3 (1) + ⋯
- 29. = − 𝑦 − 𝜋 2 + − 2 2 𝑥 𝑦 − 𝜋 2 + − 3 6 𝑥 2 𝑦 − 𝜋 2 + 1 6 𝑦 − 𝜋 2 3 + ⋯ (∵ 2! = 1.2 = 2, 3! = 1.2.3 = 6) = − 𝑦 − 𝜋 2 − 𝑥 𝑦 − 𝜋 2 − 1 2 𝑥 2 𝑦 − 𝜋 2 + 1 6 𝑦 − 𝜋 2 3 + ⋯ Problem 3 Find the Taylor’s series expansion of 𝟏 + 𝒙 + 𝒚 𝟐 in powers of 𝒙 − 𝟏 and 𝒚 upto second degree term. Solution: W.kt, Taylor’s expansion (upto second degree) of 𝑓(𝑥, 𝑦) about the point (𝑎, 𝑏) is given by 𝑓 𝑥, 𝑦 = 𝑓 𝑎, 𝑏 + 𝑥 − 𝑎 𝑓𝑥 𝑎, 𝑏 + 𝑦 − 𝑏 𝑓𝑦 𝑎, 𝑏 + 1 2! 𝑥 − 𝑎 2 𝑓𝑥𝑥 𝑎, 𝑏 + 2 𝑥 − 𝑎 𝑦 − 𝑏 𝑓𝑥𝑦 𝑎, 𝑏 + 𝑦 − 𝑏 2 𝑓𝑦𝑦 𝑎, 𝑏 + ⋯ Here 𝑓 𝑥, 𝑦 = 1 + 𝑥 + 𝑦2 the points are 𝑎 = 1, 𝑏 = 0,
- 30. 𝑓 𝑥, 𝑦 = 1 + 𝑥 + 𝑦2 = 1 + 𝑥 + 𝑦2 1 2 𝑓 1,0 = 1 + 1 + 0 = 2 1 2 𝑓𝑥 = 1 2 1 + 𝑥 + 𝑦2 1 2 −1 1 = 1 2 (1 + 𝑥 + 𝑓𝑥 1,0 = 1 2 1 + 1 + 0 − 1 2 = 1 2 2 − 1 2 = 2− 1 2 −1 = 2− 3 2 𝑓𝑦 = 1 2 1 + 𝑥 + 𝑦2 1 2 −1 2𝑦 = 𝑦 1 + 𝑥 + 𝑦2 − 1 2 𝑓𝑦 1,0 = 0 𝑓𝑥𝑥 = 1 2 − 1 2 1 + 𝑥 + 𝑦2 − 1 2 −1 1 = −1 4 1 + 𝑥 + 𝑦2 − 3 2 𝑓𝑥𝑥 1,0 = − 1 4 1 + 1 + 0 − 3 2 = 1 4 2 − 3 2 = − 1 22 2 − 3 2 = −2− 3 2 −2 = −2− 7 2 𝑓𝑥𝑦 = 1 2 − 1 2 1 + 𝑥 + 𝑦2 − 1 2 −1 2𝑦 = −𝑦 2 1 + 𝑥 + 𝑦2 − 3 2 𝑓𝑥𝑦 1,0 = 0 𝑓𝑦𝑦 = 𝑦 − 1 2 1 + 𝑥 + 𝑦2 − 1 2 −1 2𝑦 + 1 + 𝑥 + 𝑦2 − 1 2 1 = −𝑦 1 + 𝑥 + 𝑦2 − 3 2 +(1 + 𝑥 + 𝑓𝑦𝑦 1,0 = 0 + 1 + 1 + 0 − 1 2 = 2− 1 2
- 31. Taylor’s expansion of 𝑓(𝑥, 𝑦) about the point (1,0) is given by 𝑓 𝑥, 𝑦 = 𝑓 1,0 + 𝑥 − 1 𝑓𝑥 1,0 + 𝑦 𝑓𝑦 1,0 + 1 2! 𝑥 − 1 2 𝑓𝑥𝑥 1,0 + 2 𝑥 − 1 𝑦 𝑓𝑥𝑦 1,0 + 𝑦 2 𝑓𝑦𝑦 1,0 + ⋯ = 2 1 2 + 𝑥 − 1 2− 3 2 + 𝑦 0 + 1 2 𝑥 − 1 2 −2 − 7 2 + 2 2 𝑥 − 1 𝑦 0 + 𝑦22− 1 2 = 2 1 2 + 2− 3 2 𝑥 − 1 − −2 − 7 2 −1 𝑥 − 1 2 + 2− 1 2 𝑦2 = 2 1 2 + 2− 3 2 𝑥 − 1 − −2 − 9 2 𝑥 − 1 2 + 2− 1 2 𝑦2
- 32. Maximum and Minimum (Extreme Values) of two variable function
- 33. Necessary Conditions The necessary conditions for 𝑓(𝑥, 𝑦) to have a maximum or minimum at a point (𝑎, 𝑏) are that 𝑓𝑥 𝑎, 𝑏 = 0 𝑎𝑛𝑑 𝑓𝑦(𝑎, 𝑏) = 0 𝑎𝑡 (𝑎, 𝑏) Sufficient Conditions The sufficient conditions for 𝑓(𝑥, 𝑦) to have a maximum or minimum at a point (𝑎, 𝑏) are as follows Let 𝑟 = 𝑓𝑥𝑥 𝑎, 𝑏 , 𝑠 = 𝑓𝑥𝑦 𝑎, 𝑏 𝑎𝑛𝑑 𝑡 = 𝑓𝑦𝑦 𝑎, 𝑏 The function 𝑓(𝑥, 𝑦) has maximum or minimum (extreme values) at ( a, b) if 1. 𝑓𝑥 𝑎, 𝑏 = 0 𝑎𝑛𝑑 𝑓𝑦(𝑎, 𝑏) = 0 2. 𝑟𝑡 − 𝑠2 > 0 3. 𝑓(𝑥, 𝑦) has maximum or minimum at (𝑎, 𝑏) according as 𝑟 < 0 or 𝑟 > 0
- 34. Procedure to find maximum or minimum (extreme values): Step 1. Find 𝑓𝑥 𝑎𝑛𝑑 𝑓𝑦 Step 2. Put 𝑓𝑥 = 0 𝑎𝑛𝑑 𝑓𝑦 = 0 Step 3. Solve the simultaneous equations in a and y, find the values of x and y Say (𝑎, 𝑏), (𝑐, 𝑑) … . ( these points are called stationary / critical points) Step 4.Find 𝑟 = 𝑓𝑥𝑥 , 𝑡 = 𝑓𝑦𝑦 𝑎𝑛𝑑 𝑠 = 𝑓𝑥𝑦 (at each pair (𝑎, 𝑏) , (𝑐, 𝑑) … . ) Step 5. Find 𝑟𝑡 − 𝑠2 (at each pair (𝑎, 𝑏) , (𝑐, 𝑑) … . ) Step 6. (i) If r𝑡 − 𝑠2 > 0 𝑎𝑛𝑑 𝑟 < 0 𝑎𝑡 (𝑎, 𝑏) , then 𝑓(𝑥, 𝑦) has maximum at (𝑎, 𝑏) And the maximum value is 𝑓(𝑎, 𝑏) (ii) If r𝑡 − 𝑠2 > 0 𝑎𝑛𝑑 𝑟 > 0 𝑎𝑡 (𝑎, 𝑏) , then 𝑓(𝑥, 𝑦) has minimum at (𝑎, 𝑏) And the minimum value is 𝑓(𝑎, 𝑏) (iii) If r𝑡 − 𝑠2 < 0 𝑎𝑡 (𝑎, 𝑏) , then 𝑓(𝑥, 𝑦) has neither maximum nor minimum at (𝑎, 𝑏) (no extreme values at (𝑎, 𝑏)) and the point (𝑎, 𝑏) is called saddle point. (iv) If 𝑟𝑡 − 𝑠2 = 0 𝑎𝑡 (𝑎, 𝑏), the case is doubtful and need further investigation. Step 7. Step 6 to be repeated for other pair of values (c, d) … to examine extreme values
- 35. Problem 1: Discuss the maxima and minima of the function 𝒇 𝒙, 𝒚 = 𝒙 𝟒 + 𝒚 𝟒 − 𝟐𝒙 𝟐 + 𝟒𝒙𝒚 − 𝟐𝒚 𝟐 Solution: Given 𝑓 𝑥, 𝑦 = 𝑥4 + 𝑦4 − 2𝑥2 + 4𝑥𝑦 − 2𝑦2 --------- (1) To find the stationary point: Put 𝝏𝒇 𝝏𝒙 = 𝑓𝑥 = 0 & 𝝏𝒇 𝝏𝒚 = 𝒇 𝒚 = 0 𝝏𝒇 𝝏𝒙 = 0 ⟹ 4𝑥3 − 4𝑥 + 4𝑦 = 0 ⟹ 4(𝑥3 − 𝑥 + 𝑦) = 0 ⟹ 𝑥3 − 𝑥 + 𝑦 = 0 ---------(2) 𝝏𝒇 𝝏𝒚 = 0 ⟹ 4𝑦3 + 4𝑥 − 4𝑦 = 0 ⟹ 4 𝑦3 + 𝑥 − 𝑦 = 0 ⟹ 𝑦3 + 𝑥 − 𝑦 = 0 ------(3) 𝐩 = 𝛛𝐟 𝛛𝐱 = 𝒇 𝒙 = 𝟒𝐱 𝟑 − 𝟒𝐱 + 𝟒𝐲 𝐫 = 𝛛 𝟐 𝐟 𝛛𝐱 𝟐 = 𝐟 𝐱𝐱 = = 𝟏𝟐𝐱 𝟐 − 𝟒 𝐬 = 𝛛 𝟐 𝐟 𝛛𝐱𝛛𝐲 = 𝒇 𝒙𝒚 = 𝟒 𝐪 = 𝛛𝐟 𝛛𝐲 = 𝐟 𝐲 = 𝟒𝐲 𝟑 + 𝟒𝐱 − 𝟒𝐲 𝐭 = 𝛛 𝟐𝐟 𝛛𝐲 𝟐 = 𝐟 𝐲𝐲 = 𝟏𝟐𝐲 𝟐 − 𝟒
- 36. 2 ⟹ 𝑥3 − 𝑥 + 𝑦 = 0 3 ⟹ 𝑦3 + 𝑥 − 𝑦 = 0 Add ------------------------ 𝑥3 + 𝑦3 = 0 ⟹ 𝑥 + 𝑦 𝑥2 − 𝑥𝑦 + 𝑦2 = 0 , (∵ 𝑎3 + 𝑏3 = (𝑎 + 𝑏)(𝑎2 − 𝑎𝑏 + 𝑏2 ) ⟹ 𝑥 + 𝑦 = 0 𝑜𝑟 (𝑥2 − 𝑥𝑦 + 𝑦2) = 0 ⟹ 𝑥 = −𝑦 𝑜𝑟 𝑥2 − 𝑥𝑦 + 𝑦2 = 0 Put 𝑥 = −𝑦 in (2) ⟹ −𝑦 3 − −𝑦 + 𝑦 = 0 ⟹ −𝑦3 + 𝑦 + 𝑦 = 0 ⟹ −𝑦3 + 2𝑦 = 0 ⟹ 𝑦(−𝑦2 + 2) = 0 ⟹ 𝑦 = 0 𝑜𝑟 − 𝑦2 + 2 = 0 ⟹ 𝑦 = 0 𝑜𝑟 𝑦2 = 2 ⟹ 𝑦 = 0 𝑜𝑟 𝑦 = ± 2 When 𝑦 = 0 , from (3) 0 + 𝑥 − 0 = 0 ⟹ 𝑥 = 0 The point is ( 𝑥, 𝑦) = (0,0) When 𝑦 = 2 , from (3) 2 3 + 𝑥 − 2 = 0 ⟹ 𝑥 = − 2 3 + 2 ⟹ 𝑥 = 2 (− 2 2 + 1) ⟹ 𝑥 = 2 −2 + 1 = 2 −1 = − 2 The point is ( 𝑥, 𝑦) = (− 2, 2)
- 37. When 𝑦 = − 2 , from (3), − 2 3 + 𝑥 − (− 2) = 0 ⟹ − 2 3 + 𝑥 + 2 = 0 ⟹ 𝑥 = 2 3 − 2 = 2 2 2 − 1 = 2 2 − 1 = 2 1 = 2 The point is ( 𝑥, 𝑦) = ( 2, − 2) The stationary / critical points are (0,0), − 2, 2 , ( 2, − 2) To find the point at which 𝒇(𝒙, 𝒚) has maximum/minimum: 𝑟𝑡 − 𝑠2 = 12𝑥2 − 4 12𝑦2 − 4 − 4 2 = 12𝑥2 − 4 12𝑦2 − 4 − 16 At the Point (0,0): 𝑟𝑡 − 𝑠2 𝑎𝑡 0,0 = 12 0 2 − 4 12 0 2 − 4 − 16 = −4 −4 − 16 = 16 − 16 = 𝟎 Hence there is no maximum / minimum for 𝑓(𝑥, 𝑦) at (0,0) (we can’t judge) At the Point − 𝟐, 𝟐 : 𝑟𝑡 − 𝑠2 𝑎𝑡 − 2, 2 = 12 − 2 2 − 4 12 2 2 − 4 − 16 = 12 2 − 4 12 2 − 4 − 16 = 24 − 4 24 − 4 − 16 = 20 20 − 16 = 400 − 16 = 394 > 𝟎 𝑟 𝑎𝑡 − 2, 2 = 12 − 2 2 − 4 = 12 2 − 4 = 24 − 4 = 20 > 𝟎 Hence 𝑓(𝑥, 𝑦) has minimum at − 2, 2
- 38. To find the minimum value : Put x, y = − 2, 2 in (1) The minimum value is 𝑓(− 2, 2) = − 2 4 + 2 4 − 2 − 2 2 + 4 − 2 2 − 2 2 2 = 4 + 4 − 4 − 8 − 4 = −8 Hence the minimum value is −8 At the Point 𝟐, − 𝟐 : 𝑟𝑡 − 𝑠2 𝑎𝑡 2,− 2 = 12 2 2 − 4 12 − 2 2 − 4 − 16 = 12 2 − 4 12 2 − 4 − 16 = 24 − 4 24 − 4 − 16 = 20 20 − 16 = 400 − 16 = 394 > 𝟎 𝑟 𝑎𝑡 − 2, 2 = 12 − 2 2 − 4 = 12 2 − 4 = 24 − 4 = 20 > 𝟎 Again 𝑓(𝑥, 𝑦) has minimum at 2, − 2
- 39. Problem 2: Find the maximum or minimum values of 𝒇 𝒙, 𝒚 = 𝟑𝒙 𝟐 − 𝒚 𝟐 + 𝒙 𝟑 Solution: Given 𝑓 𝑥, 𝑦 = 3𝑥2 − 𝑦2 + 𝑥3 ------- (1) To find the stationary point: Put 𝜕𝑓 𝜕𝑥 = 𝑓𝑥 = 0 & 𝜕𝑓 𝜕𝑦 = 𝑓𝑦 = 0 𝜕𝑓 𝜕𝑥 = 0 ⇒ 6𝑥 + 3𝑥2 = 0 ⟹ 3𝑥 2 + 𝑥 = 0 ⟹ 𝑥 = 0, 𝑥 + 2 = 0 ⟹ 𝑥 = 0, 𝑥 = −2 𝜕𝑓 𝜕𝑦 = 0 ⟹ −2𝑦 = 0 ⟹ 𝑦 = 0 The stationary points are 0,0 , (−2,0) 𝑝 = 𝐟 𝐱 = 𝜕𝑓 𝜕𝑥 = 6𝑥 + 3𝑥2 𝑟 = 𝐟 𝐱𝐱 = 𝜕2 𝑓 𝜕𝑥2 = 6 + 6𝑥 𝑠 = 𝐟 𝐱𝐲 = 𝜕2 𝑓 𝜕𝑥𝜕𝑦 = 0 𝑞 = 𝐟 𝐲 = 𝜕𝑓 𝜕𝑦 = −2𝑦 𝑡 = 𝑓𝑦𝑦 = 𝜕2𝑓 𝜕𝑦2 = −2
- 40. To find the point at which 𝒇(𝒙, 𝒚) has maximum/minimum: 𝑟𝑡 − 𝑠2 = 6 + 6𝑥 −2 − 0 = −12 − 12𝑥 At the Point (0,0): 𝑟𝑡 − 𝑠2 𝑎𝑡 0,0 = −12 − 12 0 = −12 < 𝟎 Hence the point (0,0) is the saddle point of 𝑓(𝑥, 𝑦) At the Point (-2,0): 𝑟𝑡 − 𝑠2 𝑎𝑡 −2,0 = −12 − 12 −2 = −12 + 24 = 12 > 𝟎 𝑟at −2,0 = 6 + 6 −2 = 6 − 12 = −6 < 𝟎 Hence at the point −2,0 the function 𝑓 𝑥, 𝑦 has maximum. To find the maximum value: Put 𝑥, 𝑦 = (−2,0) in (1), 𝑓 −2,0 = 3(−2)2 − 0 2 + −2 3 = 12 − 8 = 4
- 41. Constrained Maxima and Minima – Lagrangian Multiplier If 𝑓(𝑥, 𝑦, 𝑧) is a function of three variables 𝑥, 𝑦, 𝑧 , we will find the extreme values (maximum or minimum) of 𝑓(𝑥, 𝑦, 𝑧) with respect to a constraint ∅ 𝑥, 𝑦, 𝑧 = 0 Procedure Step 1. Identify the constraint equation ∅ 𝑥, 𝑦, 𝑧 = 0 Step 2. Identify the main function for which we have to find the extreme value, let it be 𝑓(𝑥, 𝑦, 𝑧) Step 3. Form the equation 𝐹 = 𝑓 + 𝜆∅ Step 4. Find 𝐹𝑥 , 𝐹𝑦, 𝐹𝑧 Step 5. Put 𝐹𝑥 = 0 , 𝐹𝑦 = 0, 𝐹𝑧 = 0 and solve all the equations including ∅ 𝑥, 𝑦, 𝑧 = 0 Step 6. Find the values of 𝑥, 𝑦, 𝑧 and 𝜆 Step 7. The values of 𝑥, 𝑦, 𝑧 gives the extreme values of 𝑓(𝑥, 𝑦, 𝑧)
- 42. Note : Distance of a point 𝑥1, 𝑦1, 𝑧1 𝑓𝑟𝑜𝑚 𝑥, 𝑦, 𝑧 is given by 𝑑 = 𝑥 − 𝑥1 2 + 𝑦 − 𝑦1 2 + 𝑧 − 𝑧1 2 Square of the distance is 𝑑2 = 𝑥 − 𝑥1 2 + 𝑦 − 𝑦1 2 + 𝑧 − 𝑧1 2
- 43. Problem 1: Find the length of the shortest line form the point (𝟎, 𝟎, 𝟐𝟓 𝟗 ) to the surface 𝒛 = 𝒙𝒚 Solution: The square of the distance from the point (0,0, 25 9 ) to (𝑥, 𝑦, 𝑧) is (𝑑2) 𝑓 𝑥, 𝑦, 𝑧 = 𝑥 − 0 2 + 𝑦 − 0 2 + 𝑧 − 25 9 2 𝑓 𝑥, 𝑦, 𝑧 = 𝑥2 + 𝑦2 + 𝑧 − 25 9 2 ------- (1) Subject to ( to the surface) ∅ 𝑥, 𝑦𝑧 = 𝑧 − 𝑥𝑦 = 0 ------------(2) (∵ 𝑧 = 𝑥𝑦 ⟹ 𝑧 − 𝑥𝑦 = 0) Consider the Lagrangian function F = 𝑓 𝑥, 𝑦, 𝑧 + 𝜆𝜙 𝑥, 𝑦, 𝑧 𝐹 = 𝑥2 + 𝑦2 + 𝑧 − 25 9 2 + 𝜆(𝑧 − 𝑥𝑦) 𝜕F 𝜕𝐱 = 2x − λy 𝜕F 𝜕y = 2y − λx 𝜕F 𝜕z = 2 z − 25 9 + λ
- 44. To find the stationary points: Put 𝜕F 𝜕𝑥 = 0, 𝜕F 𝜕𝑦 = 0 & 𝜕F 𝜕𝑧 = 0 𝜕𝐹 𝜕𝑥 = 0 ⟹ 2𝑥 − 𝜆𝑦 = 0 ⟹ 𝜆𝑦 = 2𝑥 ⟹ 𝜆 = 2𝑥 𝑦 ------- (3) 𝜕F 𝜕𝑦 = 0 ⟹ 2𝑦 − 𝜆x = 0 ⟹ 𝜆𝑥 = 2𝑦 ⟹ 𝜆 = 2𝑦 𝑥 ------ (4) 𝜕F 𝜕𝑧 = 0 ⟹ 2 z − 25 9 + λ = 0 ⟹ λ = −2 z − 25 9 ----- (5) From (3), (4), (5), 2𝑥 𝑦 = 2𝑦 𝑥 = −2 z − 25 9 ⟹ 𝑥 𝑦 = 𝑦 𝑥 = −z + 25 9 Consider 𝑥 𝑦 = 𝑦 𝑥 ⟹ 𝑥2 = 𝑦2 ⟹ 𝑥 = ±𝑦 Consider 𝑦 𝑥 = −z + 25 9 , If 𝑥 = 𝑦 −𝑧 + 25 9 = 𝑦 𝑦 = 1 ⟹ 𝑧 = 25 9 − 1 = 25−9 9 = 16 9 Given 𝑧 = 𝑥𝑦 = 𝑦. 𝑦 = 𝑦2 (∵ 𝑥 = 𝑦) ⟹ 𝑦2 = 16 9 ∵ 𝑧 = 16 9 ⟹ 𝑦 = ± 16 9 = ± 4 3 If 𝑥 = −𝑦 −𝑧 + 25 9 = 𝑦 −𝑦 = −1 ⟹ 𝑧 = 25 9 + 1 = 25+9 9 = 34 9 Given 𝑧 = 𝑥𝑦 = −𝑦. 𝑦 = −𝑦2 (∵ 𝑥 = −𝑦) ⟹ −𝑦2 = 34 9 ⟹ 𝑦2 = − 34 9 ∵ 𝑧 = 34 9 ⟹ 𝑦 = ± −34 9 (Imaginary , this is not possible) Hence 𝑥 = −𝑦 is ruled out.
- 45. Hence 𝑥 = 𝑦 = ± 4 3 & 𝑧 = 𝑥𝑦 = 𝑦2 = 4 3 2 = 16 9 From (1) square distance is 𝑑2 = 𝑥2 + 𝑦2 + 𝑧 − 25 9 2 = 4 3 2 + 4 3 2 + 16 9 − 25 9 2 = 16 9 + 16 9 + − 9 9 2 = 2 16 9 + −1 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 𝑥𝑦𝑧 = 32 , which is the given constrain (condition). Let ∅ 𝑥, 𝑦, 𝑧 = 𝑥𝑦𝑧 − 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 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 Let 𝑓 𝑥, 𝑦, 𝑧 = 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 ------ (2)
- 47. Note:
- 49. Hence the total surface area of a open rectangular box is 𝒙𝒚 + 𝟐𝒙𝒛 + 𝟐𝒚𝒛
- 50. Consider the Lagrangian function 𝐹 = 𝑓 + 𝜆𝜙 𝐹 = 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 + 𝜆 𝑥𝑦𝑧 − 32 ( by using (1) & (2)) To find the stationary points: Put 𝜕F 𝜕𝑥 = 0, 𝜕F 𝜕𝑦 = 0 & 𝜕F 𝜕𝑧 = 0 𝜕𝐹 𝜕𝑥 = 0 ⟹ y + 2z + 𝜆𝑦𝑧 = 0 ⟹ 𝜆𝑦𝑧 = − 𝑦 + 2𝑧 ⟹ 𝜆 = −(𝑦+2𝑧) 𝑦𝑧 ------ (3) 𝜕F 𝜕𝑦 = 0 ⟹ 𝑥 + 2𝑧 + 𝜆xz = 0 ⟹ 𝜆𝑥𝑧 = −(𝑥 + 2𝑧) ⟹ 𝜆 = −(𝑥+2𝑧) 𝑥𝑧 ------ (4) 𝜕F 𝜕𝑧 = 0 ⟹ 2x + 2y + λxy = 0 ⟹ λxy = −2x − 2y = −2 x + y 𝜆 = −2 𝑥+𝑦 𝑥𝑦 ----- (5) 𝜕𝐹 𝜕𝑥 = 𝑦 + 2𝑧 + 𝜆𝑦𝑧 𝜕𝐹 𝜕𝑦 = 𝑥 + 2𝑧 + 𝜆𝑥𝑧 𝜕𝐹 𝜕𝑧 = 2𝑥 + 2𝑦 + 𝜆𝑥𝑦
- 51. From (3), (4), (5), −(𝑦+2𝑧) 𝑦𝑧 = −(𝑥+2𝑧) 𝑥𝑧 = −2 𝑥+𝑦 𝑥𝑦 Consider −(𝑦+2𝑧) 𝑦𝑧 = −(𝑥+2𝑧) 𝑥𝑧 ⇒ −𝑥(𝑦+2𝑧) 𝑥𝑦𝑧 = −𝑦(𝑥+2𝑧) 𝑦𝑥𝑧 ⇒ 𝑥𝑦 + 2𝑧𝑥 = (𝑦𝑥 + 2𝑧𝑦) ⇒ 2𝑧𝑥 = 2𝑧𝑦 ⇒ 𝑥 = 𝑦 ------ (6) Consider −(𝑥+2𝑧) 𝑥𝑧 = −2 𝑥+𝑦 𝑥𝑦 ⇒ 𝑥+2𝑧 𝑥𝑧 = 2 𝑥+𝑦 𝑥𝑦 Using (6), 𝑥+2𝑧 𝑥𝑧 = 2 𝑥+𝑥 𝑥𝑥 ⇒ 𝑥+2𝑧 𝑥𝑧 = 4𝑥 𝑥2 ⇒ 𝑥+2𝑧 𝑥𝑧 = 4 𝑥 ⇒ 𝑥+2𝑧 𝑧 = 4
- 52. ⇒ 𝑥 + 2𝑧 = 4𝑧 ⇒ 4𝑧 − 2𝑧 = 𝑥 ⇒ 2𝑧 = 𝑥 ⇒ 𝑧 = 𝑥 2 -------- (7) Given volume 𝑥𝑦𝑧 = 32 From (6) & (7) 𝑥 . 𝑥 . 𝑥 2 = 32 , (∵ 𝑦 = 𝑥 & 𝑧 = 𝑥 2 ) 𝑥3 = 2 32 = 64 𝑥 = 64 1 3 𝑥 = 43 1 3 = 4 3 3 = 4 Hence 𝑥 = 4 , 𝑦 = 𝑥 = 4 and 𝑧 = 𝑥 2 = 4 2 = 2 The required dimension is 𝑥 = 4, 𝑦 = 4 𝑎𝑛𝑑 𝑧 = 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 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 = 432, which is the given constrain(condition) Let ∅ 𝑥, 𝑦, 𝑧 = 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 − 432 ------ (1) Requirement of a open rectangular box with maximum capacity (volume) Total volume of the open rectangular box is 𝑥𝑦𝑧 Let 𝑓 𝑥, 𝑦, 𝑧 = 𝑥𝑦𝑧 ------ (2)
- 54. Consider the Lagrangian function 𝐹 = 𝑓 + 𝜆𝜙 𝐹 = 𝑥𝑦𝑧 + 𝜆 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 − 432 ( by using (1) & (2)) To find the stationary points: Put 𝜕F 𝜕𝑥 = 0, 𝜕F 𝜕𝑦 = 0 & 𝜕F 𝜕𝑧 = 0 𝜕𝐹 𝜕𝑥 = 0 ⟹ yz + 𝜆 𝑦 + 2z = 0 ⟹ 𝜆 𝑦 + 2𝑧 = −𝑦𝑧 ⟹ 𝜆 = −𝑦𝑧 𝑦+2𝑧 ------ (3) 𝜕F 𝜕𝑦 = 0 ⟹ 𝑥𝑧 + 𝜆(x + 2z) = 0 ⟹ 𝜆(𝑥 + 2𝑧) = −𝑥𝑧 ⟹ 𝜆 = −𝑥𝑧 𝑥+2𝑧 ------ (4) 𝜕F 𝜕𝑧 = 0 ⟹ xy + λ 2x + 2y = 0 ⟹ λ 2x + 2y = −xy ⇒ 𝜆 = −𝑥𝑦 2𝑥+2𝑦 ----- (5) 𝜕𝐹 𝜕𝑥 = 𝑦𝑧 + 𝜆(𝑦 + 2𝑧) 𝜕𝐹 𝜕𝑦 = 𝑥𝑧 + 𝜆(𝑥 + 2𝑧) 𝜕𝐹 𝜕𝑧 = 𝑥𝑦 + 𝜆(2𝑥 + 2𝑦)
- 55. From (3), (4), (5), −𝑦𝑧 𝑦+2𝑧 = −𝑥𝑧 𝑥+2𝑧 = −𝑥𝑦 2𝑥+2𝑦 Consider −𝑦𝑧 𝑦+2𝑧 = −𝑥𝑧 𝑥+2𝑧 ⇒ 𝑦𝑧 𝑦+2𝑧 = 𝑥𝑧 𝑥+2𝑧 ⇒ 𝑦 𝑦+2𝑧 = 𝑥 𝑥+2𝑧 ⇒ 𝑦 𝑥 + 2𝑧 = 𝑥(𝑦 + 2𝑧) ⇒ 𝑥𝑦 + 2𝑧𝑦 = 𝑥𝑦 + 2𝑧𝑥 ⇒ 2𝑧𝑦 = 2𝑧𝑥 ⇒ 𝑦 = 𝑥 -------- (6) Consider −𝑥𝑧 𝑥+2𝑧 = −𝑥𝑦 2𝑥+2𝑦 ⇒ 𝑧 𝑥+2𝑧 = 𝑦 2 𝑥+𝑦 ⇒ 𝑧 𝑥+2𝑧 = 𝑥 2 𝑥+𝑥 (by using (6)) ⇒ 𝑧 𝑥+2𝑧 = 𝑥 2 2𝑥 ⇒ 𝑧 𝑥+2𝑧 = 1 4
- 56. ⇒ 𝑧 𝑥+2𝑧 = 1 4 ⇒ 4𝑧 = 𝑥 + 2𝑧 ⇒ 4𝑧 − 2𝑧 = 𝑥 ⇒ 2𝑧 = 𝑥 ⇒ 𝑧 = 𝑥 2 ------- (7) Given surface area 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 = 432 Using (6) & (7), 𝑥. 𝑥 + 2𝑥. 𝑥 2 + 2𝑥. 𝑥 2 = 432 ⇒ 𝑥2 + 𝑥2 + 𝑥2 = 432 ⇒ 3𝑥2 = 432 ⇒ 𝑥2 = 432 3 ⇒ 𝑥2 = 144 ⇒ 𝑥 = ± 144 ⇒ 𝑥 = ±12 (since 𝑥 is the length , 𝑥 = −12 is not possible ) Hence 𝑥 = 12 , 𝑦 = 𝑥 = 12 , & 𝑧 = 𝑥 2 = 12 2 = 6 The required dimension for maximum capacity is 𝑥 = 12 , 𝑦 = 12 & 𝑧 = 6
- 57. Problem 4: Find the minimum distance from the point (𝟏, 𝟐, 𝟎) to the cone 𝒛 𝟐 = 𝒙 𝟐 + 𝒚 𝟐 Solution: Wkt, the square of the distance (𝑑2 ) from the point (1, 2, 0) to (𝑥, 𝑦, 𝑧) is 𝑥 − 1 2 + 𝑦 − 2 2 + 𝑧 − 0 2 = 𝑥 − 1 2 + 𝑦 − 2 2 + 𝑧2 Let 𝑓 𝑥, 𝑦, 𝑧 = 𝑥 − 1 2 + 𝑦 − 2 2 + 𝑧2 --------- (1) Subject to 𝑧2 = 𝑥2 + 𝑦2 ⇒ 𝑧2 − 𝑥2 − 𝑦2 = 0 Let ∅ 𝑥, 𝑦, 𝑧 = 𝑧2 − 𝑥2 − 𝑦2 ---------- (2) Consider the Lagrangian function 𝐹 = 𝑓 + 𝜆𝜙 𝐹 = 𝑥 − 1 2 + 𝑦 − 2 2 + 𝑧2 + 𝜆(𝑧2 − 𝑥2 − 𝑦2) 𝜕F 𝜕x = 2 x − 1 − λ(2x) 𝜕F 𝜕y = 2 y − 2 − λ(2y) 𝜕F 𝜕z = 2z + λ 2z = 2z(1 + λ)
- 58. To find the stationary points: Put 𝜕F 𝜕𝑥 = 0, 𝜕F 𝜕𝑦 = 0 & 𝜕F 𝜕𝑧 = 0 𝜕𝐹 𝜕𝑥 = 0 ⟹ 2 𝑥 − 1 − 𝜆 2𝑥 = 0 ⟹ 2𝜆𝑥 = 2 𝑥 − 1 ⟹ 𝜆 = (𝑥−1) 𝑥 ------ (3) 𝜕F 𝜕𝑦 = 0 ⟹ 2 𝑦 − 2 − 𝜆 2𝑦 = 0 ⟹ 2𝜆𝑦 = 2 𝑦 − 2 ⟹ 𝜆 = (𝑦−2) 𝑦 ------ (4) 𝜕F 𝜕𝑧 = 0 ⟹ 2𝑧(1 + 𝜆) = 0 ----- (5) 5 ⟹ 𝑧 = 0 𝑜𝑟 1 + 𝜆 = 0 ⟹ 𝑧 = 0 𝑜𝑟 𝜆 = −1 Suppose 𝑧 = 0, Since 𝑥2 + 𝑦2 = 𝑧2 ⇒ 𝑥2 + 𝑦2 = 0 ⇒ 𝑥 = 0 & 𝑦 = 0 , this is not possible. Hence 𝜆 = −1
- 59. From (3) , -1= (𝑥−1) 𝑥 ⇒ −𝑥 = 𝑥 − 1 ⇒ −𝑥 − 𝑥 = −1 ⇒ −2𝑥 = −1 ⇒ 2𝑥 = 1 ⇒ 𝑥 = 1 2 From(4), −1 = 𝑦−2 𝑦 ⇒ −𝑦 = 𝑦 − 2 ⇒ −𝑦 − 𝑦 = −2 ⇒ −2𝑦 = −2 ⇒ 𝑦 = 1 Since 𝑥2 + 𝑦2 = 𝑧2 Substitute 𝑥 = 1 2 , 𝑦 = 1 𝑧2 = 1 2 2 + 1 2 = 1 4 + 1 = 1+4 4 = 5 4 ⇒ 𝑧 = ± 5 4 = ± 5 2 Hence , 𝑥 = 1 2 , 𝑦 = 1 , 𝑧 = ± 5 2 The square distance (𝑑2) is 𝑥 − 1 2 + 𝑦 − 2 2 + 𝑧2 = 1 2 − 1 2 + 1 − 2 2 + 5 2 2 = − 1 2 2 + −1 2 + 5 4 = 1 4 + 1 + 5 4 = 1+4+5 4 = 10 4 = 5 2 The minimum distance (𝑑) = 5 2
- 60. Jacobian • If 𝑢 = 𝑓(𝑥, 𝑦) and 𝑣 = 𝑔(𝑥, 𝑦) are two continuous functions of two independent variables x and y then the functional determinant 𝐽 = 𝜕𝑢 𝜕𝑥 𝜕𝑢 𝜕𝑦 𝜕𝑣 𝜕𝑥 𝜕𝑣 𝜕𝑦 = 𝑢 𝑥 𝑢 𝑦 𝑣 𝑥 𝑣 𝑦 is called Jacobian of 𝑢 , 𝑣 with respect to 𝑥, 𝑦 and it is denoted by 𝜕 𝑢,𝑣 𝜕 𝑥,𝑦 • 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 𝜕 𝑢,𝑣,𝑤 𝜕 𝑥,𝑦,𝑧 = 𝜕𝑢 𝜕𝑥 𝜕𝑢 𝜕𝑦 𝜕𝑢 𝜕𝑧 𝜕𝑣 𝜕𝑥 𝜕𝑣 𝜕𝑦 𝜕𝑣 𝜕𝑧 𝜕𝑤 𝜕𝑥 𝜕𝑤 𝜕𝑦 𝜕𝑤 𝜕𝑧 = 𝑢 𝑥 𝑢 𝑦 𝑢 𝑧 𝑣 𝑥 𝑣 𝑦 𝑣𝑧 𝑤 𝑥 𝑤 𝑦 𝑤𝑧
- 61. Two important Properties of Jacobian 1. If 𝑢, 𝑣 are functions of 𝑥, 𝑦 and 𝑥, 𝑦 are the function of 𝑟, 𝑠 then 𝜕 𝑢,𝑣 𝜕 𝑟,𝑠 = 𝜕 𝑢,𝑣 𝜕 𝑥,𝑦 𝜕 𝑥,𝑦 𝜕 𝑟,𝑠 2. If 𝑢, 𝑣 are the functions of 𝑥, 𝑦 then 𝜕 𝑥,𝑦 𝜕 𝑢,𝑣 = 1 𝜕 𝑢,𝑣 𝜕 𝑥,𝑦 Note : Two functions 𝑢(𝑥, 𝑦) & 𝑣(𝑥, 𝑦) are functionally depended if 𝜕 𝑢,𝑣 𝜕 𝑥,𝑦 =0
- 62. Problem 1: If 𝒖 = 𝒚𝒛 𝒙 , 𝒗 = 𝒛𝒙 𝒚 , 𝒘 = 𝒙𝒚 𝒛 find 𝝏 𝒖,𝒗,𝒘 𝝏 𝒙,𝒚,𝒛 Solution: Wkt, the jacobian of 𝑢, 𝑣, 𝑤 with respect to 𝑥, 𝑦, 𝑧 is given by 𝜕 𝑢,𝑣,𝑤 𝜕 𝑥,𝑦,𝑧 = 𝑢 𝑥 𝑢 𝑦 𝑢 𝑧 𝑣 𝑥 𝑣 𝑦 𝑣𝑧 𝑤 𝑥 𝑤 𝑦 𝑤𝑧 Given 𝑢 = 𝑦𝑧 𝑥 , 𝑣 = 𝑧𝑥 𝑦 , 𝑤 = 𝑥𝑦 𝑧
- 64. 𝜕 𝑢,𝑣,𝑤 𝜕 𝑥,𝑦,𝑧 = 𝑢 𝑥 𝑢 𝑦 𝑢 𝑧 𝑣 𝑥 𝑣 𝑦 𝑣𝑧 𝑤 𝑥 𝑤 𝑦 𝑤𝑧 = − 𝑦𝑧 𝑥2 𝑧 𝑥 𝑦 𝑥 𝑧 𝑦 − 𝑧𝑥 𝑦2 𝑥 𝑦 𝑦 𝑧 𝑥 𝑧 − 𝑥𝑦 𝑧2 = 1 𝑥 1 𝑦 1 𝑧 − 𝑦𝑧 𝑥 𝑧 𝑦 𝑧 − 𝑧𝑥 𝑦 𝑥 𝑦 𝑥 − 𝑥𝑦 𝑧 = 1 𝑥𝑦𝑧 − 𝑦𝑧 𝑥 − 𝑧𝑥 𝑦 − 𝑥𝑦 𝑧 − 𝑥2 − 𝑧 𝑧 − 𝑥𝑦 𝑧 − 𝑦𝑥 + 𝑦 𝑧𝑥 − − 𝑧𝑥 𝑦 𝑦 = 1 𝑥𝑦𝑧 [ − 𝑦𝑧 𝑥 𝑥2 − 𝑥2 − 𝑧 −𝑥𝑦 − 𝑥𝑦 + 𝑦 𝑧𝑥 + 𝑧𝑥 ] = 1 𝑥𝑦𝑧 − 𝑦𝑧 𝑥 0 − 𝑧 −2𝑥𝑦 + 𝑦 2𝑧𝑥 = 1 𝑥𝑦𝑧 2𝑥𝑦𝑧 + 2𝑥𝑦𝑧 = 4𝑥𝑦𝑧 𝑥𝑦𝑧 = 4
- 65. Problem 2: If 𝒙 = 𝒖 𝟐 – 𝒗 𝟐 , 𝒚 = 𝟐𝒖𝒗 find the Jacobian of 𝒙, 𝒚 with respect to 𝒖 𝒂𝒏𝒅 𝒗 Solution: Given 𝑥 = 𝑢2 – 𝑣2 , 𝑦 = 2𝑢𝑣 To find the jacobian of 𝑥, 𝑦 w . r. to 𝑢 & 𝑣 To find 𝜕 𝑥,𝑦 𝜕 𝑢,𝑣 = 𝑥 𝑢 𝑥 𝑣 𝑦 𝑢 𝑦𝑣 𝑥 = 𝑢2 – 𝑣2 𝑦 = 2𝑢𝑣 𝑥 𝑢 = 2𝑢 𝑦𝑢 = 2𝑣 𝑥 𝑣 = −2𝑣 𝑦𝑣 = 2𝑢
- 66. 𝜕 𝑥,𝑦 𝜕 𝑢,𝑣 = 𝑥 𝑢 𝑥 𝑣 𝑦 𝑢 𝑦𝑣 = 2𝑢 −2𝑣 −2𝑣 2𝑢 = 4𝑢2 − 4𝑣2 = 4 𝑢2 − 𝑣2 Problem 3: If 𝒙 = 𝒖𝒗 𝒂𝒏𝒅 𝒚 = 𝒖 𝒗 , then find 𝝏 𝒙,𝒚 𝝏 𝒖,𝒗 Solution: Given 𝑥 = 𝑢𝑣 𝑎𝑛𝑑 𝑦 = 𝑢 𝑣 𝑥 = 𝑢𝑣 𝑦 = 𝑢 𝑣 𝑥 𝑢 = 𝑣 𝑦𝑢 = 1 𝑣 𝑥 𝑣 = 𝑢 𝑦𝑣 = − 𝑢 𝑣2
- 67. 𝜕 𝑥,𝑦 𝜕 𝑢,𝑣 = 𝑥 𝑢 𝑥 𝑣 𝑦 𝑢 𝑦𝑣 = 𝑣 𝑢 1 𝑣 − 𝑢 𝑣2 = 𝑣 𝑢 𝑣2 − 1 𝑣 𝑢 = 𝑢 𝑣 − 𝑢 𝑣 = 0 Problem 4: If 𝒙 = 𝒓 𝐜𝐨𝐬 𝜽 and 𝒚 = 𝒓 𝐬𝐢𝐧 𝜽 then find 𝝏𝒓 𝝏𝒙 Solution: Given 𝑥 = 𝑟 cos 𝜃 and 𝑦 = 𝑟 sin 𝜃 𝑥2 + 𝑦2 = 𝑟2 cos2 𝜃 + 𝑟2 sin2 𝜃 = 𝑟2(cos2 𝜃 + sin2 𝜃) = 𝑟2 ∴ 𝑟2 = 𝑥2 + 𝑦2 2𝑟 𝜕𝑟 𝜕𝑥 = 2𝑥 ⟹ 𝜕𝑟 𝜕𝑥 = 2𝑥 2𝑟 = 𝑥 𝑟 = 𝑥 𝑥2+𝑦2 , ∵ 𝑟 = 𝑥2 + 𝑦2
- 68. Problem 5: If 𝒙 = 𝒖 (𝟏 – 𝒗) and 𝒚 = 𝒖𝒗, find 𝝏 𝒖,𝒗 𝝏 𝒙,𝒚 Solution: Given 𝑥 = 𝑢 (1 – 𝑣) and 𝑦 = 𝑢𝑣 𝑖. 𝑒 𝑥 = 𝑢 − 𝑢𝑣 and 𝑦 = 𝑢𝑣 Wkt 𝜕 𝑢,𝑣 𝜕 𝑥,𝑦 = 1 𝜕 𝑥,𝑦 𝜕 𝑢,𝑣 Now 𝜕 𝑥,𝑦 𝜕 𝑢,𝑣 = 𝑥 𝑢 𝑥 𝑣 𝑦 𝑢 𝑦𝑣 𝑥 = 𝑢 − 𝑢𝑣 𝑦 = 𝑢𝑣 𝑥 𝑢 = 1 − 𝑣 𝑦𝑢 = 𝑣 𝑥 𝑣 = −𝑢 𝑦𝑣 = 𝑢
- 69. 𝜕 𝑥,𝑦 𝜕 𝑢,𝑣 = 𝑥 𝑢 𝑥 𝑣 𝑦 𝑢 𝑦𝑣 = 1 − 𝑣 −𝑢 𝑣 𝑢 = 1 − 𝑣 𝑢 + 𝑢𝑣 = 𝑢 − 𝑢𝑣 + 𝑢𝑣 = 𝑢 Hence 𝜕 𝑢,𝑣 𝜕 𝑥,𝑦 = 1 𝜕 𝑥,𝑦 𝜕 𝑢,𝑣 = 1 𝑢 Problem 6: If 𝒙 = 𝒓 𝒄𝒐𝒔𝜽 𝒚 = 𝒓 𝐬𝐢𝐧 𝜽 then find 𝝏 𝒓,𝜽 𝝏 𝒙,𝒚 Solution: Given 𝑥 = 𝑟 𝑐𝑜𝑠𝜃 𝑦 = 𝑟 sin 𝜃 Wkt, 𝜕 𝑥,𝑦 𝜕 𝑟,𝜃 = 𝑥 𝑟 𝑥 𝜃 𝑦𝑟 𝑦 𝜃
- 70. 𝑥 = 𝑟 𝑐𝑜𝑠𝜃 𝑦 = 𝑟 𝑠𝑖𝑛𝜃 𝑥 𝑟 = 𝑐𝑜𝑠𝜃 𝑦𝑟 = 𝑠𝑖𝑛𝜃 𝑥 𝜃 = −𝑟𝑠𝑖𝑛𝜃 𝑦 𝜃 = 𝑟 𝑐𝑜𝑠𝜃 𝜕 𝑥,𝑦 𝜕 𝑟,𝜃 = 𝑥 𝑟 𝑥 𝜃 𝑦𝑟 𝑦 𝜃 = 𝑐𝑜𝑠𝜃 −𝑟𝑠𝑖𝑛𝜃 𝑠𝑖𝑛𝜃 𝑟 𝑐𝑜𝑠𝜃 = 𝑟 cos2 𝜃 + 𝑟𝑠𝑖𝑛2 𝜃 = 𝑟 (cos2 𝜃 + sin2 𝜃) = 𝑟 𝜕 𝑟,𝜃 𝜕 𝑥,𝑦 = 1 𝜕 𝑥,𝑦 𝜕 𝑟,𝜃 = 1 𝑟
- 71. Problem 7 If 𝒖 = 𝒙 + 𝒚 & 𝒚 = 𝒖𝒗, then find the jacobian 𝝏 𝒙,𝒚 𝝏 𝒖,𝒗 Solution: Given 𝑢 = 𝑥 + 𝑦 & 𝑦 = 𝑢𝑣 ⟹ 𝑢 = 𝑥 + 𝑢𝑣 ∵ 𝑦 = 𝑢𝑣 ⟹ 𝑥 = 𝑢 − 𝑢𝑣 & 𝑦 = 𝑢𝑣 Wkt, 𝜕 𝑥,𝑦 𝜕 𝑢,𝑣 = 𝑥 𝑢 𝑥 𝑣 𝑦 𝑢 𝑦𝑣 𝑥 = 𝑢 − 𝑢𝑣 𝑦 = 𝑢𝑣 𝑥 𝑢 = 1 − 𝑣 𝑦𝑢 = 𝑣 𝑥 𝑣 = −𝑢 𝑦𝑣 = 𝑢
- 72. 𝜕 𝑥,𝑦 𝜕 𝑢,𝑣 = 𝑥 𝑢 𝑥 𝑣 𝑦𝑢 𝑦𝑣 = 1 − 𝑣 −𝑢 𝑣 𝑢 = 1 − 𝑣 𝑢 + 𝑢𝑣 = 𝑢 − 𝑢𝑣 + 𝑢𝑣 = 𝑢 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 𝑦 = 𝑥2 + 3𝑥 – 8 here y is the dependent variable and is given in terms of the independent variable x. More Examples : 𝑦 = 𝑥 + 3 , 𝑦 = 𝑥2 − 𝑟2 𝑒𝑡𝑐., 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 𝑦 + 𝑥2 − 3𝑥 + 8 = 0 More Examples: 𝑥2 + 𝑦2 = 𝑎2 , 𝑥3 + 𝑥𝑦2 + 4𝑥 = 5, 𝑒𝑡𝑐. ,
- 73. Differentiation of implicit functions If 𝑓(𝑥, 𝑦 ) is the given implicit function , then 𝑑𝑦 𝑑𝑥 = − 𝜕𝑓 𝜕𝑥 𝜕𝑓 𝜕𝑦 Producer to find the differentiation: (i) Take 𝑓(𝑥, 𝑦) (ii) Find 𝜕𝑓 𝜕𝑥 & 𝜕𝑓 𝜕𝑦 (iii) Find 𝑑𝑦 𝑑𝑥 = − 𝜕𝑓 𝜕𝑥 𝜕𝑓 𝜕𝑦 Problem 1: If 𝒙 𝒚 + 𝒚 𝒙 = 𝒄, then find 𝒅𝒚 𝒅𝒙 Solution: Given 𝑥 𝑦 + 𝑦 𝑥 = 𝑐 ⟹ 𝑥 𝑦 + 𝑦 𝑥 − 𝑐 = 0 Take 𝑓 𝑥, 𝑦 = 𝑥 𝑦 + 𝑦 𝑥 − 𝑐
- 74. Now 𝜕𝑓 𝜕𝑥 = 𝜕 𝜕𝑥 𝑥 𝑦 + 𝑦 𝑥 − 𝑐 = 𝜕 𝜕𝑥 𝑥 𝑦 + 𝜕 𝜕𝑥 𝑦 𝑥 − 𝜕 𝜕𝑥 (𝑐) = 𝑦𝑥 𝑦−1 + 𝑦 𝑥 log 𝑦 − 0 , (∵ 𝑑 𝑑𝑥 𝑥 𝑛 = 𝑛𝑥 𝑛−1 & 𝑑 𝑑𝑥 𝑎 𝑥 = 𝑎 𝑥 log 𝑎 ) 𝜕𝑓 𝜕𝑥 = 𝑦𝑥 𝑦−1 + 𝑦 𝑥 log 𝑦 Now 𝜕𝑓 𝜕𝑦 = 𝜕 𝜕𝑦 𝑥 𝑦 + 𝑦 𝑥 − 𝑐 = 𝜕 𝜕𝑦 𝑥 𝑦 + 𝜕 𝜕𝑦 𝑦 𝑥 − 𝜕 𝜕𝑦 (𝑐) = 𝑥 𝑦 log 𝑥 + 𝑥 𝑦 𝑥−1 − 0 𝜕𝑓 𝜕𝑦 = 𝑥 𝑦 log 𝑥 + 𝑥 𝑦 𝑥−1 Wkt 𝑑𝑦 𝑑𝑥 = − 𝜕𝑓 𝜕𝑥 𝜕𝑓 𝜕𝑦 = − 𝑦𝑥 𝑦−1 +𝑦 𝑥 log 𝑦 𝑥 𝑦 log 𝑥+𝑥 𝑦 𝑥−1