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 df, giving df ≃ δf. OBSERVATIONS Consider the formula, df = ∂f ∂xδx + ∂f ∂yδy + . . .

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(a) If f(x, y, ...) ≡ x, then df = δx. Hence, dx = δx. (b) If f(x, y, ...) ≡ y, then df = δy. Hence, dy = 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

• nly approximately equal to the increment (or error) in

that variable. (d) The observations may be summarised by the formula, df = ∂f ∂xdx + ∂f ∂ydy + . . .

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14.4.2 TESTING FOR EXACT DIFFERENTIALS In general, an expression of the form, P(x, y, ...)dx + Q(x, y, ...)dy + . . ., 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)dx + Q(x, y)dy, is an exact differential if and only if ∂P ∂y ≡ ∂Q ∂x .

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Proof: (a) If the expression, P(x, y)dx + Q(x, y)dy, is an exact differential, df, then ∂f ∂x ≡ P(x, y) and ∂f ∂y ≡ Q(x, y). Hence, ∂P ∂y ≡ ∂Q ∂x

   ≡ ∂2f

∂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.

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But then, ∂Q ∂x ≡ ∂P ∂y ≡ ∂2u ∂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)dx + Q(x, y)dy = ∂u ∂xdx +

   ∂u

∂y + A(y)

    dy.

The right-hand side is the exact differential of u(x, y) +

A(y) dy.

(ii) By similar reasoning, it may be shown that P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz is an exact differential when ∂P ∂y ≡ ∂Q ∂x , ∂Q ∂z = ∂R ∂y , and ∂R ∂x = ∂P ∂z .

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ILLUSTRATIONS 1. xdx + ydy = d

  1

2

• x2 + y2
• 

  .

2. ydx + xdy = d[xy]. 3. ydx − xdy is not an exact differential, since ∂y ∂y = 1 and ∂(−x) ∂x = −1. 4. 2 ln ydx + (x + z)dy + z2dz is not an exact differential, since ∂(2 ln y) ∂y = 2 y, and ∂(x + z) ∂x = 1.

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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)dx + (1 + sin x)dy, 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)

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Integrating (1) partially with respect to x gives f(x, y) ≡ x2 2 + y sin x + A(y), where A(y) is an arbitrary function of y only. Substituting this result into (2) gives sin x + dA dy ≡ 1 + sin x. That is, dA dy ≡ 1, Hence, A(y) ≡ y + constant. We conclude that f(x, y) ≡ x2 2 + y sin x + y + constant.

• 2. Verify that the expression,

(yz + 2)dx + (xz + 6y)dy + (xy + 3z2)dz, is an exact differential and obtain the function of which it is the total differential.

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Solution Firstly, ∂ ∂y(yz + 2) ≡ ∂ ∂x(xz + 6y) ≡ z, ∂ ∂z(xz + 6y) ≡ ∂ ∂y(xy + 3z2) ≡ x, and ∂ ∂x(xy + 3z2) ≡ ∂ ∂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 + 6y, − − − − − − − (2) ∂F ∂z ≡ xy + 3z2. − − − − − − − (3) Integrating (1) partially with respect to x gives F(x, y, z) ≡ xyz + 2x + A(y, z), where A(y, z) is an arbitrary function of y and z only.

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Substituting into both (2) and (3) gives xz + ∂A ∂y ≡ xz + 6y, xy + ∂A ∂z ≡ xy + 3z2. That is, ∂A ∂y ≡ 6y, − − − − − −(4) ∂A ∂z ≡ 3z2. − − − − − −(5) Integrating (4) partially with respect to y gives A(y, z) ≡ 3y2 + B(z), where B(z) is an arbitrary function of z only. Substituting into (5) gives dB dz ≡ 3z2, which implies that B(z) ≡ z3 + constant and, hence, F(x, y, z) ≡ xyz + 2x + 3y2 + z3 + constant.

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