Two-Dimensional Calculus (2011)

Chapter 2. Differentiation

5. Continuity

The function f(x, y) = 2xy/(x² + y²) is defined everywhere except at the origin. Since it is the quotient of two polynomials, its partial derivatives f_x and f_y exist wherever the denominator is nonzero, which is everywhere except at the origin. Since the function is not defined at the origin, we may ask whether we cannot extend the function by suitably defining it at the origin so that its partial derivatives exist there too. We note that along the x; and y axes, where y = 0 or x = 0, f(x, y) = 0. Hence we make the definition

Then f(x, 0) ≡ 0 for all x, and hence f_x(x, 0) ≡ 0. In particular, f_x(0, 0) exists and is zero. Similarly f_y(0, 0) = 0. Thus the function defined by (5.1) has both partial derivatives f_x and f_y at every point.

For a function of one variable, the existence of a derivative at a point implies that the function is continuous at that point. On the other hand, for a function of two variables both partial derivatives may exist at every point, and the function may still not be continuous. In particular, the function defined by (5.1) is not continuous at the origin, since approaching the origin along the line y = x, we have f(x, y) ≡ 1 for (x, y) ≠ (0, 0), but at the origin f(x, y) suddenly jumps from 1 to 0 (see Fig. 3.4).

In order to express precisely our intuitive notions of continuity, we introduce the following notation:

Note that for each n, the quantity

is a fixed number, representing the distance of the point (x_n, y_n) from the point (x₀, y₀)· Thus condition (5.2) simply says that the ordinary sequence of numbers d_n tends to zero. Geometrically, the points (x_n, y_n) are tending to (x₀, y₀) although they may not be approaching from any fixed direction. Several examples are pictured in Fig. 5.1.

Note also that we have

FIGURE 5.1 Illustrations of “lim_{n → ∞} (x_n, y_n) = (x₀, y₀)”

Definition 5.1 f(x, y) tends to the limit L as (x, y) tends to (x₀, y₀) if

whenever

We adopt the notation

Definition 5.2 f(x, y) is continuous at (x₀, y₀) if

Thus the idea of continuity is essentially the same for functions of two variables as for one variable. We simply require that as a variable point approaches a fixed point, the value of the function should approach its value at that point.

Remark Although we only use the property of continuity given in the definition, there is an alternative characterization that may be helpful to those who are more accustomed to “∊, δ” definitions of continuity. Namely, f(x, y) is continuous at (x₀, y₀) if and only if for every ∊ > 0 there is a δ > 0 such that |f(x, y) − f(x₀, y₀)| < ∊ whenever .

It is important to note that it is not the same thing for a function f(x, y) to be continuous as a function of two variables, and to be continuous in each variable when the other is held fixed. It is an immediate consequence of our definition of continuity that if f(x, y) is continuous at (x₀, y₀), then f(x, y₀) is a continuous function of x at x = x₀, and f(x₀, y) is a continuous function of y at y = y₀. However, the converse is not true. For example, the function defined by Eq. (5.1) is continuous in each variable separately at the origin, since f(x, 0) ≡ 0 and f(0, y) ≡ 0; however, it is not continuous at the origin, since if we consider, for example, the sequence of points (1/n, 1/n), then

but

hence

So far, we have discussed only the notion of continuity at a point. In general, we are interested in continuity of a function in a domain.

Definition 5.3 A function f(x, y) defined in a domain D is continuous in D if it is continuous at each point of D.

Suppose now that f(x, y) is defined in D and that both partial derivatives f_x and f_y exist at each point of D. Then f_x and f_y are themselves functions defined throughout D, and we may ask whether they are continuous.

Definition 5.4 f(x, y) is called continuously differentiable in D if f_x and f_y exist at each point and if they are continuous functions in D.

By adding the hypothesis that f_x and f_y not only exist but are continuous, we rule out all “pathological” functions of the type (5.1). This is a consequence of the following lemma, which plays a key role in the theory of functions of two variables.

Lemma 5.1 Fundamental Lemma Let f(x, y) be continuously differentiable in a domain D, and let (x₀, y₀) be a point in D. Then there is a function h(x, y), defined in D, satisfying

and

where

Remark The function h(x, y) in Eq. (5.3) should be thought of as a remainder term. It is actually defined as the difference of the value of the function f(x, y) and the sum of the first three terms on the right-hand side of Eq. (5.3). The whole content of the lemma is given by Eq. (5.4) which says that as (x, y) → (x₀, y₀), the remainder term h(x, y) tends to zero faster than the distance d(x, y).

PROOF. Since Eq. (5.4) is concerned with the limit as (x, y) → (x₀, y₀), it is sufficient to consider only values of (x, y) in some circle

By our definition of a domain, (x₀, y₀) is an interior point of D, and hence if r is chosen sufficiently small all points of the circle (5.5) lie in D.

The proof is based on two observations. First, to evaluate the change in f as we move from (x₀, y₀) to (x, y), we may proceed in two steps, first fixing y and letting x vary and then fixing x and letting y vary. Second, we may apply to each step the mean-value theorem for a function of a single variable.¹ Thus, fixing y = y₀, the mean-value theorem asserts that there exists a value x_x between x₀ and x such that

(See Fig. 5.2.)

Similarly, we have

where y₁ lies between y₀ and y.

Adding the above equations, we find that

Substituting Eq. (5.8) into Eq. (5.3), we obtain

But as x → x₀ we have x₁ → x₀ and as y → y₀ we have y₁ → y₀. Thus, by the continuity of f_x and f_y at (x₀, y₀), we have

FIGURE 5.2 “Two-step” method of moving from (x₀, y₀) to (x, y)

Thus each of the expressions in the square brackets in Eq. (5.9) tends to zero as (x, y) → (x₀, y₀). On the other hand, it is evident that for all (x, y)

Thus, dividing through Eq. (5.9) by , we obtain the sum of two terms, each of which tends to zero as (x, y) → (x₀, y₀). This shows that Eq. (5.4) holds, and the lemma is proved.

In the next section we shall give a number of applications of this basic lemma. For now, we shall restrict ourselves to one immediate consequence and to a geometrical interpretation.

Corollary A continuously differentiable function is continuous.

PROOF. Using the lemma, at any point (x₀, y₀) we can express the function f(x, y) by Eq. (5.3). Then by Eq. (5.4) we have certainly

and hence from Eq. (5.3),

We turn now to the geometrical interpretation of Lemma 5.1. Given the function f(x, y) and a fixed point (x₀, y₀) we construct the function

L(x, y) is simply a linear function of the form Ax + By + C. Hence the equation z = L(x, y) defines a plane. Further, L(x₀, y₀) = f(x₀, y₀), so that this plane touches the surface z = f(x, y) above the point (x₀, y₀). Equation (5.3) can now be written in the form

Thus h(x,y) represents the difference between the height of the surface z = f(x, y) above the x, y plane, and the height of the plane z = L(x, y) (Fig. 5.3). For any plane passing through the same point, say

FIGURE 5.3 Geometric interpretation of the fundamental lemma: the tangent plane

the difference f(x, y) − l(x, y) would tend to zero as (x, y) → (x₀, y₀). The significance of Eq. (5.4) is that for the particular plane defined by Eq. (5.10), the difference h(x, y) tends to zero to a higher order, so that even the ratio

One can show that given a function f(x, y) and a point (x₀, y₀), there cannot be more than one plane z = l(x, y) for which

(See Ex. 5.4 below.) If there is a plane satisfying this condition, it is called the tangent plane to the surface z = f(x, y) at the point. Thus, the geometric content of Lemma 5.1 is the following: if f(x, y) is continuously differentiable, then it has a tangent plane at each point, and the equation of the tangent plane is z = L(x, y), where L(x, y) is defined by Eq. (5.10).

Example 5.1

Find the equation of the tangent plane to the surface

at the point (1,1).

We set

then

and

Hence,

and the tangent plane is

or

Exercises

5.1 Find the equation of the tangent plane to each of the following surfaces at the points indicated.

a. z = x⁴ − 3x²y³ + 2xy² + 3x − 4y + 5 at (0, 0)

b. z = 3x − 4y + 5 at (−1, 2)

c.

d.

e. z = 2xy/(x² + y²) at (2, −1)

f. z = ye^{x2 + y} at (0, 0)

5.2 Show that for the quadric surface ax² + by² + cz² = 1, where c > 0, the tangent plane at the point (x₀, y₀, z₀), where

has the equation

5.3 a. Find the equation of the tangent plane to the surface x^1/2 + y^1/2 + z^1/2 = 1 at an arbitrary point (x₀, y₀, z₀).

b. Find the intersection of the tangent plane with each of the axes.

c. Show that the sum of the intercepts does not depend on the choice of the point (x₀ y₀, z₀).

5.4 Fix a point (x₀, y₀) and let

Show that the following statements are true.

a. If g(x, y) is defined in some domain containing (x₀, y₀) and if

(Hint: g(x, y) = d(x, y)[g(x, y)/d(x, y)].)

b. If L(x, y) = Ax + By + C, and if

then

(Hint: First use part a, and then show that A = 0 and B = 0 by forming the limit first with a sequence such as (x₀ + 1/n, y₀) and then with the sequence (x₀, y₀ + 1/n).)

c. Let

If

then

d. Suppose f(x, y) is defined in a domain D containing (x₀, y₀). If

(Hint: |l₁ − l₂| ≤ |l₁ − f| + |f − l₂|.)

Note that Ex. 5.4d is the statement in the text that there cannot be more than one plane z = l(x, y) such that

If such a plane exists, it is called the tangent plane to the surface z = f(x, y) at the point. The following terminology is also frequently used. A function f(x, y) defined in a domain D is called differentiable at a point (x₀, y₀) in D if f(x, y) is continuous at (x₀, y₀) and if the surface z = f(x, y) has a tangent plane at the point. For functions of one variable the corresponding property (that is, existence of a tangent line) is equivalent to the existence of the derivative at the point. However, the situation in two variables is far more complicated. Since we make no use of this concept, we leave it to the interested reader to explore its basic properties in Exs. 5.5–5.8. These properties may be summarized as follows.

1. If f(x, y) is continuously differentiable in a domain D, then it is differentiable at each point of D (Lemma 5.1).

2. If f(x, y) is differentiable at a point (x₀, y₀), then f_x(x₀, y₀) and f_y(x₀, y₀) exist, and the equation of the tangent plane is

(Ex. 5.5).

3. A function may have both partial derivatives at a point but not be differentiable (Exs. 5.6 and 5.7).

4. A function may be differentiable, but not continuously differentiable (Ex. 5.8).

Thus differentiability implies more than mere existence of partial derivatives and less than existence and continuity of the partial derivatives. By property 2 above, differentiability is exactly the conclusion of Lemma 5.1, that is, that Eqs. (5.3) and (5.4) hold.

*5.5 Using the notation of Ex. 5.4, show that if

and if f(x, y) is continuous at (x₀, y₀), then f_x(x₀, y₀) and f_y(x₀, y₀) exist, and

(Hint: first apply the statement in Ex. 5.4a to the function g(x, y) = f(x, y) − L(x, y) to deduce L(x₀, y₀) = f(x₀, y₀). Then take the above limit, first fixing y = y₀ and then x = x₀.)

*5.6 Show that the function defined by Eq. (5.1) does not have a tangent plane at the origin, even though both partial derivatives exist at every point. (Hint: if there were a tangent plane, and if we took limits first along y = 0 and then along x = 0, the tangent plane would have to be the x, y plane. But taking limits along y = x gives a contradiction.)

*5.7 Show that the function f(x, y) = (x^1/3 + y^1/3)³ is continuous at (0, 0) and that both partial derivatives f_x(0, 0) and f_y(0, 0) exist, but that f(x, y) is not differentiable at (0, 0). (Hint: use the same reasoning as Ex. 5.6.)

5.8 Let f(x, y) = x³ sin (1/x) + y² for x ≠ 0, and f(0, y) = y².

a. Find the partial derivatives of f at any point (x₀, y₀) where x₀ ≠ 0.

b. Find the partial derivatives of f at the point (0, y₀) for any y₀. (Use the definition of f_x(0, y₀).)

c. Show that f(x, y) is continuous at (0, 0).

*d. Show that f_x(x, y) is not continuous at (0, 0).

*e. Show that f(x, y) is differentiable at (0, 0). In fact, show that the plane z = 0 is the tangent plane to the surface z = f(x, y) at the origin.

5.9 Show that the cone does not have a tangent plane at the origin. (Hint: assuming that a tangent plane z = L(x, y), where L(x, y) = Ax + By + C, does exist, then applying Ex. 5.4a to the condition

gives C = 0. Forming this limit with the sequence (1/n, 0) gives A = 1 and using (0, 1/n) gives B = 1. But using (1/n, 1/n) gives A + B = , a contradiction.)

5.10 Let f(x, y) be the function defined in Eq. (5.1).

a. Compute the partial derivatives of f with respect to x and y at any point other than (0, 0).

b. Show that f_x and f_y are not continuous at (0, 0).

5.11 Let f(x, y) be defined in a domain D containing the point (x₀, y₀). Let x = x₀ + at, y = y₀ + bt, define an arbitrary straight line through the point (x₀, y₀). Show that the function g(t) = f(x₀ + at, y₀ + bt) is a continuous function of t at t = 0.

Note. It was mentioned earlier that a continuous function of two variables reduces to a continuous function of each variable separately if the other variable is held fixed. Geometrically, this may be interpreted as saying that a continuous function of two variables is continuous along every horizontal and vertical line. What Ex. 5.11 states is that a continuous function of two variables is in fact continuous along every straight line. It may come as a surprise that the converse is not true. The following exercise gives an example of a function that is continuous when restricted to any straight line through the origin, but is not continuous at the origin when considered as a function of two variables.

5.12 Let

a. Show that if x = at, y = bt is any straight line through the origin, where a and b are not both zero, then

b. Show that at every point (1/n, 1/n²), f(x, y) is equal to 1. Conclude that f(x, y) is not continuous at (0, 0).

*c. Sketch the surface z = f(x, y) by observing that the parabolas y = ax² are all level curves of the function, and by seeing how the value of the function on each of these curves depends on the constant a. Try to visualize how the function can be continuous along every straight line through the origin, but still fail to be continuous if one approaches the origin along a parabola y = ax².

5.13 Let x(t), y(t) be continuous functions for a ≤ t ≤ b. Let x₀ = x(t₀) and y₀ = y(t₀). Let f(x, y) be defined in a domain containing the point (x₀, y₀) and let f(x, y) be continuous at (x₀, y₀). Show that the function g(t) = f(x(t), y(t)) is continuous at t = t₀.

Note. The property stated in Ex. 5.13 may be reformulated roughly in the following geometric form: a continuous function of two variables is continuous along every curve.

5.14 Show that if f(x, y) is continuous in a domain D, and if f(x, y) is never zero, then either f(x, y) > 0 throughout D, or else f(x, y) < 0 throughout D. (Hint: prove the contrapositive; if f(x₁, y₁) > 0 and f(x₂, y₂) < 0, then by connectedness (x₁, y₁) can be joined to (x₂, y₂) by a curve in D, and applying the intermediate-value theorem for functions of one variable to the function g(t) of Ex. 5.13 gives a point in D where f(x, y) = 0.)

*5.15 Prove the equivalence of Def. 5.2 of continuity and the , δ definition mentioned in the Remark following Def. 5.2.

5.16 Let f(x, y) be continuous in a domain D.

a. Prove that if f(x₀, y₀) = η > 0, then there exists a positive number r such that f(x, y) > η/2 for all points (x, y) satisfying (x − x₀)² + (y − y₀)² < r². (Hint: either use Ex. 5.15, or else prove by contradiction, choosing a sequence of circles whose radii tend to zero, in each of which there is a point (x_n, y_n), where f(x_n, y_n) ≤ η/2.)

b. Show that the set of points (x, y) in D, where f(x, y) > 0, is an open set.