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 or 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 or 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 + . . . 1

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 , d f d t = ∂f ∂x. d x d t + ∂f ∂y. d y d t + . . . 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 of the surface whose cartesian equation is x 2 16 − y 2 9 = z (a Paraboloid) and the surface whose cartesian equation is x 2 + y 2 = 5 (a Cylinder) . 2

If x is increasing at 0.2 cms/sec, how fast is z changing when x = 2 ? Solution We may use the formula d z d t = ∂z ∂x. d x d t + ∂z ∂y. d y d t, where d x d t = 0 . 2 and d y d t = d y d x. d x d t = 0 . 2d y d x. From the equation of the paraboloid, ∂z ∂x = x 8 and ∂z ∂y = − 2 y 9 . From the equation of the cylinder, d y d x = − x y. Substituting x = 2 gives y = ± 1 on the curve of intersection, so that d z  2  − 2  − 2  1 4 + 4          (0 . 2) +  ( ± 1)(0 . 2)  = 0 . 2 d t =          8 9 ± 1 9 = 5 36 cms / sec . 3

2. Determine the total derivative of u with respect to t in the case when u = xy + yz + zx, x = e t , y = e − t and z = x + y. Solution We use the formula d u d t = ∂u ∂x. d x d t + ∂u ∂y. d y d t + ∂u ∂z. d z d t, where ∂u ∂u ∂u ∂x = y + z, ∂y = z + x, ∂z = x + y and d x d t = e t = x, d y d t = − e − t = − y, d z d t = e t − e − t = x − y. Hence, d u d t = ( y + z ) x − ( z + x ) y + ( x + y )( x − y ) = − zy + zx + x 2 − y 2 = z ( x − y ) + ( x − y )( x + y ) . That is, d u d t = ( x − y )( x + y + z ) . 4

14.5.2 SEVERAL INDEPENDENT VARIABLES We may now extend the previous work to functions, f ( x, y.. ), of 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 = x 2 + y 2 , where x = r cos θ and y = r sin 2 θ. 5

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 = 2 x cos θ + 2 y sin 2 θ cos 2 θ + sin 2 2 θ � � = 2 r . ∂z ∂θ = 2 x ( − r sin θ ) + 2 y (2 r cos 2 θ ) = 2 r 2 (2 cos 2 θ sin 2 θ − 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 = x 2 + 2 y 2 + 2 z 2 , where x = u sin φ cos θ, y = u sin φ sin θ and z = u cos φ. 6

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 = 2 x sin φ cos θ + 4 y sin φ sin θ + 4 z cos φ = 2 u sin 2 φ cos 2 θ + 4 u sin 2 φ sin 2 θ + 4 u cos 2 φ, ∂w ∂θ = − 2 xu sin φ sin θ + 4 yu sin φ cos θ = − 2 u 2 sin 2 φ sin θ cos θ + 4 u 2 sin 2 φ sin θ cos θ = 2 u 2 sin 2 φ sin θ cos θ ∂w ∂φ = 2 xu cos φ cos θ + 4 yu cos φ sin θ − 4 zu sin φ = 2 u 2 sin φ cos φ cos 2 θ + 4 u 2 sin φ cos φ sin 2 θ − 4 u 2 sin φ cos φ = 2 u 2 sin φ cos φ cos 2 θ + 2sin 2 θ − 2 � � . 7