Calculus 2nd Edition Rogawski

834 C H A P T E R 15 DIFFERENTIATION IN SEVERAL VARIABLES
EXAMPLE 4 Polar Coordinates Let f (x, y) be a function of two variables, and let
(r, θ) be polar coordinates.
(a) Express ∂f/∂θ in terms of ∂f/∂x and ∂f/∂y.
(b) Evaluate ∂f/∂θ at (x, y) = (1, 1) for f (x, y) = x 2 y.
Solution
(a) Since x = r cos θ and y = r sin θ,
By the Chain Rule,
∂x
∂θ
= −r sin θ,
∂y
∂θ = r cos θ
If you have studied quantum mechanics,
you may recognize the right-hand side of
Eq. (6) as the angular momentum operator
(with respect to the z-axis).
∂f
∂θ = ∂f ∂x
∂x ∂θ + ∂f ∂y
∂y ∂θ
∂f ∂f
= −r sin θ + r cos θ
∂x ∂y
Since x = r cos θ and y = r sin θ, we can write ∂f/∂θ in terms of x and y alone:
(b) Apply Eq. (6) to f (x, y) = x 2 y:
∂f
∂θ = x ∂f
∂y − y ∂f
∂x
∂f
∂θ = x ∂
∂y (x2 y) − y ∂
∂x (x2 y) = x 3 − 2xy 2
∂f
∂θ ∣ = 1 3 − 2(1)(1 2 ) = −1
(x,y)=(1,1)
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Implicit Differentiation
In single-variable calculus, we used implicit differentiation to compute dy/dx when y
is defined implicitly as a function of x through an equation f (x, y) = 0. This method
also works for functions of several variables. Suppose that z is defined implicitly by an
equation
F (x, y, z) = 0
Thus z = z(x, y) is a function of x and y. We may not be able to solve explicitly for
z(x, y), but we can treat F (x, y, z) as a composite function with x and y as independent
variables, and use the Chain Rule to differentiate with respect to x:
∂F ∂x
∂x ∂x + ∂F ∂y
∂y ∂x + ∂F ∂z
∂z ∂x = 0
We have ∂x/∂x = 1, and also ∂y/∂x = 0 since y does not depend on x. Thus
∂F
∂x + ∂F ∂z
∂z ∂x = F ∂z
x + F z
∂x = 0
If F z ̸= 0, we may solve for ∂z/∂x (we compute ∂z/∂y similarly):
∂z
∂x = −F x
F z
,
∂z
∂y = −F y
F z
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