JUST THE MATHS SLIDES NUMBER 14.5 PARTIAL DIFFERENTIATION 5 - - PDF document

“JUST THE MATHS” SLIDES NUMBER 14.5 PARTIAL DIFFERENTIATION 5 (Partial derivatives of composite functions) by A.J.Hobson

14.5.1 Single independent variables 14.5.2 Several independent variables

UNIT 14.5 PARTIAL DIFFERENTIATION 5 PARTIAL DERIVATIVES OF COMPOSITE FUNCTIONS 14.5.1 SINGLE INDEPENDENT VARIABLES We shall be concerned with functions, f(x, y...), of two

• r more variables in which those variables are not inde-

pendent, but are themselves dependent on some other variable, t. The problem is to calculate the rate of increase (positive

• r negative) of such functions with respect to t.

Let t be subject to a small increment of δt, so that the variables, x, y..., are subject to small increments of δx, δy, . . . , respectively. The corresponding increment, δf, in f(x, y...), is given by δf ≃ ∂f ∂xδx + ∂f ∂yδy + . . .

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Note: It is not essential to use a specific formula, such as w = f(x, y..). Dividing throughout by δt gives δf δt ≃ ∂f ∂x.δx δt + ∂f ∂y.δy δt + . . . Allowing δt to tend to zero, we obtain the standard result for the “total derivative” of f(x, y..) with respect to t, df dt = ∂f ∂x.dx dt + ∂f ∂y.dy dt + . . . This rule may be referred to as the “chain rule”, but more advanced versions of it will appear later. EXAMPLES

• 1. A point, P, is moving along the curve of intersection
• f the surface whose cartesian equation is

x2 16 − y2 9 = z (a Paraboloid) and the surface whose cartesian equation is x2 + y2 = 5 (a Cylinder).

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If x is increasing at 0.2 cms/sec, how fast is z changing when x = 2 ? Solution We may use the formula dz dt = ∂z ∂x.dx dt + ∂z ∂y.dy dt, where dx dt = 0.2 and dy dt = dy dx.dx dt = 0.2dy dx. From the equation of the paraboloid, ∂z ∂x = x 8 and ∂z ∂y = −2y 9 . From the equation of the cylinder, dy dx = −x y. Substituting x = 2 gives y = ±1 on the curve of intersection, so that dz dt =

  2

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   (0.2) +   −2

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   (±1)(0.2)   −2

±1

   = 0.2   1

4 + 4 9

  

= 5 36 cms/sec.

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• 2. Determine the total derivative of u with respect to t

in the case when u = xy + yz + zx, x = et, y = e−t and z = x + y. Solution We use the formula du dt = ∂u ∂x.dx dt + ∂u ∂y.dy dt + ∂u ∂z.dz dt, where ∂u ∂x = y + z, ∂u ∂y = z + x, ∂u ∂z = x + y and dx dt = et = x, dy dt = −e−t = −y, dz dt = et−e−t = x−y. Hence, du dt = (y + z)x − (z + x)y + (x + y)(x − y) = −zy + zx + x2 − y2 = z(x − y) + (x − y)(x + y). That is, du dt = (x − y)(x + y + z).

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14.5.2 SEVERAL INDEPENDENT VARIABLES We may now extend the previous work to functions, f(x, y..),

• f two or more variables in which x, y.. are each depen-

dent on two or more variables, s, t.. Since the function f(x, y..) is dependent on s, t.., we may wish to determine its partial derivatives with respect to any one of these (independent) variables. The result previously established for a single indepen- dent variable may easily be adapted as follows: ∂f ∂s = ∂f ∂x.∂x ∂s + ∂f ∂y.∂y ∂s + . . . ∂f ∂t = ∂f ∂x.∂x ∂t + ∂f ∂y.∂y ∂t + . . . Again, this is referred to as the “chain rule”. EXAMPLES

• 1. Determine the first-order partial derivatives of z with

respect to r and θ in the case when z = x2 + y2, where x = r cos θ and y = r sin 2θ.

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Solution We may use the formulae ∂z ∂r = ∂z ∂x.∂x ∂r + ∂z ∂y.∂y ∂r, ∂z ∂θ = ∂z ∂x.∂x ∂θ + ∂z ∂y.∂y ∂θ. These give ∂z ∂r = 2x cos θ + 2y sin 2θ = 2r

• cos2θ + sin22θ
• .

∂z ∂θ = 2x(−r sin θ) + 2y(2r cos 2θ) = 2r2 (2 cos 2θ sin 2θ − cos θ sin θ) .

• 2. Determine the first-order partial derivatives of w with

respect to u, θ and φ in the case when w = x2 + 2y2 + 2z2, where x = u sin φ cos θ, y = u sin φ sin θ and z = u cos φ.

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Solution ∂w ∂u = ∂w ∂x.∂x ∂u + ∂w ∂y .∂y ∂u + ∂w ∂z .∂z ∂u, ∂w ∂θ = ∂w ∂x.∂x ∂θ + ∂w ∂y .∂y ∂θ + ∂w ∂z .∂z ∂θ ∂w ∂φ = ∂w ∂x.∂x ∂φ + ∂w ∂y .∂y ∂φ + ∂w ∂z .∂z ∂φ These give ∂w ∂u = 2x sin φ cos θ + 4y sin φ sin θ + 4z cos φ = 2usin2φcos2θ + 4usin2φsin2θ + 4ucos2φ, ∂w ∂θ = −2xu sin φ sin θ + 4yu sin φ cos θ = −2u2sin2φ sin θ cos θ + 4u2sin2φ sin θ cos θ = 2u2sin2φ sin θ cos θ ∂w ∂φ = 2xu cos φ cos θ + 4yu cos φ sin θ − 4zu sin φ = 2u2 sin φ cos φcos2θ + 4u2 sin φ cos φsin2θ −4u2 sin φ cos φ = 2u2 sin φ cos φ

• cos2θ + 2sin2θ − 2
• .

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