JUST THE MATHS SLIDES NUMBER 14.4 PARTIAL DIFFERENTIATION 4 - PDF document

“JUST THE MATHS” SLIDES NUMBER 14.4 PARTIAL DIFFERENTIATION 4 (Exact differentials) by A.J.Hobson 14.4.1 Total differentials 14.4.2 Testing for exact differentials 14.4.3 Integration of exact differentials

UNIT 14.4 PARTIAL DIFFERENTIATION 4 EXACT DIFFERENTIALS 14.4.1 TOTAL DIFFERENTIALS The expression, ∂f ∂xδx + ∂f ∂yδy + . . ., is an approximation for the increment (or error), δf in the function f ( x, y, ... ) when x , y etc. are subject to increments (or errors) of δx , δy etc., respectively The expression may be called the “total differential” of f ( x, y, ... ), and may be denoted by d f , giving d f ≃ δf. OBSERVATIONS Consider the formula, d f = ∂f ∂xδx + ∂f ∂yδy + . . . 1

(a) If f ( x, y, ... ) ≡ x , then d f = δx . Hence, d x = δx. (b) If f ( x, y, ... ) ≡ y , then d f = δy . Hence, d y = δy. (c) The total differential of each independent variable is the same as the small increment (or error) in that vari- able; but the total differential of the dependent variable is only approximately equal to the increment (or error) in that variable. (d) The observations may be summarised by the formula, d f = ∂f ∂x d x + ∂f ∂y d y + . . . 2

14.4.2 TESTING FOR EXACT DIFFERENTIALS In general, an expression of the form, P ( x, y, ... )d x + Q ( x, y, ... )d y + . . ., will not be the total differential of a function f ( x, y, ... ) unless P ( x, y, ... ), Q ( x, y, ... ) etc. can be identified with ∂f ∂x , ∂f ∂y etc., respectively. If this IS possible, then the expression is known as an “exact differential” . RESULTS (i) The expression, P ( x, y )d x + Q ( x, y )d y, is an exact differential if and only if ∂P ∂y ≡ ∂Q ∂x . 3

Proof: (a) If the expression, P ( x, y )d x + Q ( x, y )d y, is an exact differential, d f , then ∂f ∂x ≡ P ( x, y ) and ∂f ∂y ≡ Q ( x, y ) . Hence,  ≡ ∂ 2 f ∂P ∂y ≡ ∂Q    .     ∂x ∂x∂y (b) Conversely, suppose that ∂P ∂y ≡ ∂Q ∂x . We can certainly say that P ( x, y ) ≡ ∂u ∂x for some function u ( x, y ), since P ( x, y ) could be inte- grated partially with respect to x . 4

But then, ∂y ≡ ∂ 2 u ∂Q ∂x ≡ ∂P ∂y∂x. On integrating partially with respect to x , Q ( x, y ) = ∂u ∂y + A ( y ) , where A ( y ) is an arbitrary function of y . Thus, P ( x, y )d x + Q ( x, y )d y = ∂u   ∂u  ∂x d x + ∂y + A ( y )  d y.     The right-hand side is the exact differential of � A ( y ) d y. u ( x, y ) + (ii) By similar reasoning, it may be shown that P ( x, y, z )d x + Q ( x, y, z )d y + R ( x, y, z )d z is an exact differential when ∂P ∂y ≡ ∂Q ∂Q ∂z = ∂R ∂y , and ∂R ∂x = ∂P ∂x , ∂z . 5

ILLUSTRATIONS 1.  1   x 2 + y 2 � �  . x d x + y d y = d   2 2. y d x + x d y = d[ xy ] . 3. y d x − x d y is not an exact differential, since ∂y = 1 and ∂ ( − x ) ∂y = − 1 . ∂x 4. 2 ln y d x + ( x + z )d y + z 2 d z is not an exact differential, since ∂ (2 ln y ) = 2 y, and ∂ ( x + z ) = 1 . ∂y ∂x 6

14.4.3 INTEGRATION OF EXACT DIFFERENTIALS The method may be illustrated by the following examples: EXAMPLES 1. Verify that the expression, ( x + y cos x )d x + (1 + sin x )d y, is an exact differential and obtain the function of which it is the total differential Solution Firstly, ∂y ( x + y cos x ) ≡ ∂ ∂ ∂x (1 + sin x ) ≡ cos x and, hence, the expression is an exact differential. Secondly, suppose that the expression is the total dif- ferential of the function, f ( x, y ). Then, ∂f ∂x ≡ x + y cos x − − − − − − − − − (1) and ∂f ∂y ≡ 1 + sin x. − − − − − − − − − − (2) 7

Integrating (1) partially with respect to x gives f ( x, y ) ≡ x 2 2 + y sin x + A ( y ) , where A ( y ) is an arbitrary function of y only. Substituting this result into (2) gives sin x + d A d y ≡ 1 + sin x. That is, d A d y ≡ 1 , Hence, A ( y ) ≡ y + constant . We conclude that f ( x, y ) ≡ x 2 2 + y sin x + y + constant . 2. Verify that the expression, ( yz + 2)d x + ( xz + 6 y )d y + ( xy + 3 z 2 )d z, is an exact differential and obtain the function of which it is the total differential. 8

Solution Firstly, ∂y ( yz + 2) ≡ ∂ ∂ ∂x ( xz + 6 y ) ≡ z, ∂z ( xz + 6 y ) ≡ ∂ ∂ ∂y ( xy + 3 z 2 ) ≡ x, and ∂x ( xy + 3 z 2 ) ≡ ∂ ∂ ∂z ( yz + 2) ≡ y ; so that the given expression is an exact differential. Suppose it is the total differential of the function, F ( x, y, z ). Then, ∂F ∂x ≡ yz + 2 , − − − − − − − − (1) ∂F ∂y ≡ xz + 6 y, − − − − − − − (2) ∂F ∂z ≡ xy + 3 z 2 . − − − − − − − (3) Integrating (1) partially with respect to x gives F ( x, y, z ) ≡ xyz + 2 x + A ( y, z ) , where A ( y, z ) is an arbitrary function of y and z only. 9

Substituting into both (2) and (3) gives xz + ∂A ∂y ≡ xz + 6 y, xy + ∂A ∂z ≡ xy + 3 z 2 . That is, ∂A ∂y ≡ 6 y, − − − − − − (4) ∂A ∂z ≡ 3 z 2 . − − − − − − (5) Integrating (4) partially with respect to y gives A ( y, z ) ≡ 3 y 2 + B ( z ) , where B ( z ) is an arbitrary function of z only. Substituting into (5) gives d B d z ≡ 3 z 2 , which implies that B ( z ) ≡ z 3 + constant and, hence, F ( x, y, z ) ≡ xyz + 2 x + 3 y 2 + z 3 + constant . 10