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3.9. EXERCISES FOR CHAPTER 3 73 Example 3.27. In Example 3.20, we showed that the function given by f(x, y) = x3+y3 x2+y2 , which is defined for (x, y) ̸= (0, 0), could be made continuous at (0, 0) by setting f(0, 0) = 0. (Strictly speaking, this extended function is the one called ef in the definition.) Hence, by definition: lim (x,y)→(0,0) x3 + y3 x2 + y2 = 0. Similarly, in Example 3.19, for the function f(x, y) = xy x2+y2 , there is no value that can be assigned to f(0, 0) that makes f continuous at (0, 0). In this case: lim (x,y)→(0,0) xy x2 + y2 does not exist. Because of the close connection between the two concepts, properties of continuous functions have analogues for limits. Here is the version of Proposition 3.21 for limits. Proposition 3.28. Assume that limx→a f(x) and limx→a g(x) both exist. Then: 1. limx→a | Multivariable_Calculus_Shimamoto_Page_85_Chunk3301 |
74 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES (a) Sketch the level set corresponding to c = 0. (Hint: x3 −3xy2 = x(x + √ 3y)(x − √ 3y).) (b) Draw the region of the xy-plane in which f(x, y) > 0 and the region in which f(x, y) < 0. (Hint: Part (a) might help.) (c) Use the information from parts (a) and (b) to make a rough sketch of the monkey saddle. 1.9. It is possible to construct an origami approximation of the saddle surface z = x2 −y2 of Example 3.3 by folding a piece of paper. The resulting model is called a hypar. (The saddle surface itself is known as a hyperbolic paraboloid.) See Figure 3.28. Construct your own hypar by starting with a square piece of paper and following the instructions appended to the end of the chapter (page 80).3 Figure 3.28: A hypar Section 2 More surfaces in R3 In Exercises 2.1–2.4, sketch the surface in R3 described by the given equation. 2.1. x2 + y2 + z2 = 4 2.2. x2 + y2 = 4 2.3. x2 + z2 = 4 2.4. x2 + y2 9 + z2 4 = 1 In Exercises 2.5–2.9, sketch the level sets corresponding to the indicated values of c for the given function f(x, y, z) of three variables. Make a separate sketch for each individual level set. 2.5. f(x, y, z) = x2 + y2 + z2, c = 0, 1, 2 2.6. f(x, y, z) = x2 + y2, c = 0, 1, 2 2.7. f(x, y, z) = x2 + y2 −z, c = −1, 0, 1 2.8. f(x, y, z) = x2 + y2 −z2, c = −1, 0, 1 2.9. f(x, y, z) = x2 −y2 −z2, c = −1, 0, 1 2.10. Can the saddle surface z = x2 −y2 in R3 be described as a level set? If so, how? What about the parabola y = x2 in R2? 3For more information about hypars, see http://erikdemaine.org/hypar/. | Multivariable_Calculus_Shimamoto_Page_86_Chunk3302 |
3.9. EXERCISES FOR CHAPTER 3 75 Section 3 The equation of a plane 3.1. Find an equation of the plane through the point p = (1, 2, 3) with normal vector n = (4, 5, 6). 3.2. Find an equation of the plane through the point p = (2, −1, 3) with normal vector n = (1, −4, 5). 3.3. Find a point that lies on the plane x + 3y + 5z = 9 and a normal vector to the plane. 3.4. Find a point that lies on the plane x + z = 1 and a normal vector to the plane. 3.5. Are the planes x −y + z = 8 and 2x + y −z = −1 perpendicular? Are they parallel? 3.6. Are the planes x −y + z = 8 and 2x −2y + 2z = −1 perpendicular? Are they parallel? 3.7. Find an equation of the plane that contains the points p = (1, 0, 0), q = (0, 2, 0), and r = (0, 0, 3). 3.8. Find an equation of the plane that contains the points p = (1, 1, 1), q = (2, −1, 2), and r = (−1, 2, 0). 3.9. Find an equation of the plane that contains the point p = (1, 1, 1) and the line parametrized by α(t) = (1, 0, 1) + t(2, 3, −4). 3.10. Find a parametrization of the line of intersection of the planes x+y+z = 3 and x−y+z = 1. 3.11. What point on the plane x −y + 2z = 3 is closest to the point q = (2, 1, −1)? (Hint: Find a parametrization of an appropriate line through q.) 3.12. Two planes in R3 are perpendicular to each other. Their line of intersection is described by the parametric equations: x = 2 −t, y = −1 + t, z = −4 + 2t. If one of the planes has equation 2x + 4y −z = 4, find an equation for the other plane. 3.13. Consider the plane x + y + z = 1. (a) Sketch the cross-sections with the three coordinate planes. (b) Sketch and describe in words the portion of the plane that lies in the “first octant” of R3, that is, the part where x ≥0, y ≥0, and z ≥0. 3.14. Sketch and describe some of the level sets of the function f(x, y, z) = x + y + z. 3.15. Let π1 and π2 be parallel planes in R3 given by the equations: π1 : Ax + By + Cz = D1 and π2 : Ax + By + Cz = D2. (a) If p1 and p2 are points on π1 and π2, respectively, show that: n · (p2 −p1) = D2 −D1, where n = (A, B, C) is a normal vector to the two planes. | Multivariable_Calculus_Shimamoto_Page_87_Chunk3303 |
76 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES (b) Show that the perpendicular distance d between π1 and π2 is given by: d = |D2 −D1| √ A2 + B2 + C2 . (c) Find the perpendicular distance between the planes x + y + z = 1 and x + y + z = 5. 3.16. Let α be a path in R3 that has constant torsion τ = 0. You may assume that α′(t) ̸= 0 and T′(t) ̸= 0 for all t so that the Frenet vectors of α are always defined. (a) Prove that the binormal vector B is constant. (b) Fix a time t = t0, and let x0 = α(t0). Show that the function f(t) = B · (α(t) −x0) is a constant function. What is the value of the constant? (c) Prove that α(t) lies in a single plane for all t. In other words, curves that have constant torsion 0 must be planar. (Hint: Use part (b) to identify a point and a normal vector for the plane in which the path lies.) Section 4 Open sets In Exercises 4.1–4.7, let U be the set of points (x, y) in R2 that satisfy the given conditions. Sketch U, and determine whether it is an open set. Your arguments should be at a level of rigor comparable to those given in the text. 4.1. 0 < x < 1 and 0 < y < 1 4.2. |x| < 1 and |y| < 1 4.3. |x| ≤1 and |y| < 1 4.4. x2 + y2 = 1 4.5. 1 < x2 + y2 < 4 4.6. y > x 4.7. y ≥x 4.8. Is R2 −{(0, 0)}, the plane with the origin removed, an open set in R2? Justify your answer. 4.9. Prove that the union of any collection of open sets in Rn is open. That is, if {Uα} is a collection of open sets in Rn, then S α Uα is open. Section 5 Continuity 5.1. Let f : R2 →R be the function: f(x, y) = ( x3y x2+y2 if (x, y) ̸= (0, 0), 0 if (x, y) = (0, 0). (a) Let a be a positive real number. Show that, if ∥(x, y)∥< a, then |f(x, y)| < a2. | Multivariable_Calculus_Shimamoto_Page_88_Chunk3304 |
3.9. EXERCISES FOR CHAPTER 3 77 (b) Use the ϵδ-definition of continuity to determine whether f is continuous at (0, 0). 5.2. Consider the function f(x, y) defined by: f(x, y) = x2y4 (x2 + y4)2 if (x, y) ̸= (0, 0). Determine what happens to the value of f(x, y) as (x, y) approaches the origin along: (a) the x-axis, (b) the y-axis, (c) the line y = mx, (d) the parabola x = y2. (e) Is it possible to assign a value to f(0, 0) so that f is continuous at (0, 0)? Justify your answer using the ϵδ-definition of continuity. 5.3. Let f(x, y) = x2−y2 x2+y2 if (x, y) ̸= (0, 0). Is it possible to assign a value to f(0, 0) so that f is continuous at (0, 0)? Justify your answer using the ϵδ-definition. 5.4. Let f(x, y) = x3+2y3 x2+y2 if (x, y) ̸= (0, 0). Is it possible to assign a value to f(0, 0) so that f is continuous at (0, 0)? Justify your answer using the ϵδ-definition. Section 6 Some properties of continuous functions 6.1. Let a = (1, 2). Show that the function f : R2 →R given by f(x) = a · x is continuous. In Exercises 5.1–5.2, you determined the continuity of the following two functions at (0, 0). Now, consider points (x, y) other than (0, 0). Determine all such points at which f is continuous. 6.2. f(x, y) = x3y x2+y2 if (x, y) ̸= (0, 0), f(0, 0) = 0 6.3. f(x, y) = x2y4 (x2+y4)2 if (x, y) ̸= (0, 0) 6.4. Let U be an open set in Rn, and let f : U →R be a function that is continuous at a point a of U. (a) If f(a) > 0, show that there exists an open ball B = B(a, r) centered at a such that f(x) > f(a) 2 for all x in B. (Hint: Let ϵ = f(a) 2 .) (b) Similarly, if f(a) < 0, show that there exists an open ball B = B(a, r) such that f(x) < f(a) 2 for all x in B. In particular, if f(a) ̸= 0, there is an open ball B centered at a throughout which f(x) has the same sign as f(a). 6.5. Let U be the set of all points (x, y) in R2 such that y > sin x, that is: U = {(x, y) ∈R2 : y > sin x}. Prove that U is an open set in R2. (Hint: Apply the previous exercise to an appropriate continuous function.) | Multivariable_Calculus_Shimamoto_Page_89_Chunk3305 |
78 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES Section 7 The Cauchy-Schwarz and triangle inequalities 7.1. Let v and w be vectors in Rn. (a) Show that ∥v∥−∥w∥≤∥v −w∥. (Hint: v = (v −w) + w.) (b) Show that ∥v∥−∥w∥ ≤∥v −w∥. 7.2. (a) Let a be a point of Rn. If r > 0, show that B(a, r) ⊂B(0, ∥a∥+r). Draw a picture that illustrates the result in R2. (Hint: To prove a set inclusion of this type, you must show that, if x ∈B(a, r), then x ∈B(0, ∥a∥+r), that is, if ∥x−a∥< r, then ∥x−0∥< ∥a∥+r.) (b) In R2, if a ∈B((0, 0), 1) and r = 1 −∥a∥, show that B(a, r) ⊂B((0, 0), 1). (Recall that this inclusion was used in Example 3.12 to prove that B((0, 0), 1) is an open set.) 7.3. Let a be a point of Rn, and let r be a positive real number. Prove that the open ball B(a, r) is an open set in Rn. Exercises 7.4–7.7 provide the missing proofs of parts of Proposition 3.21 about properties of continuous functions. Here, U is an open set in Rn, f, g: U →R are real-valued functions defined on U, and a is a point of U. 7.4. If f is continuous at a and c is a scalar, prove that cf is continuous at a as well. (Hint: One approach is to consider the two cases c = 0 and c ̸= 0 separately. Another is to start by showing that |cf(x) −cf(a)| ≤(|c| + 1) |f(x) −f(a)|.) 7.5. If f is continuous at a, prove that there exists a δ > 0 such that, if x ∈B(a, δ), then |f(x)| < |f(a)| + 1. (Hint: Use Exercise 7.1.) 7.6. If f and g are continuous at a, prove that their product fg is continuous at a. (Hint: Use an “add zero” trick and the previous exercise to show that there is a δ1 > 0 such that, if x ∈B(a, δ1), then f(x)g(x)−f(a)g(a) ≤ | Multivariable_Calculus_Shimamoto_Page_90_Chunk3306 |
3.9. EXERCISES FOR CHAPTER 3 79 Section 8 Limits 8.1. Prove parts 1 and 3 of Proposition 3.28 about limits of sums and products. (Hint: By using the corresponding properties of continuous functions in Proposition 3.21, you should be able to avoid ϵ’s and δ’s entirely.) 8.2. Let U be an open set in Rn, x0 a point of U, and f a real-valued function defined on U, except possibly at x0, such that limx→x0 f(x) = L. Let I be an open interval in R, and let α: I →U be a continuous path in U that passes through x0, i.e., α(t0) = x0 for some t0 in I. Consider the function g: I →R given by: g(t) = ( f(α(t)) if α(t) ̸= x0, L if α(t) = x0. Show that limt→t0 g(t) = L. 8.3. (Sandwich principles.) Let U be an open set in Rn, and let a be a point of U. (a) Let f, g, h: U →R be real-valued functions such that f(x) ≤g(x) ≤h(x) for all x in U. If f(a) = h(a)—let’s call the common value c—and if f and h are continuous at a, prove that g(a) = c and that g is continuous at a as well. (b) Let f, g, h be real-valued functions defined on U, except possibly at the point a, such that f(x) ≤g(x) ≤h(x) for all x in U, except possibly when x = a. If limx→a f(x) = limx→a h(x) = L, prove that limx→a g(x) = L, too. | Multivariable_Calculus_Shimamoto_Page_91_Chunk3307 |
80 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES Addendum: Folding a hypar (Exercise 1.9) 1. Take your square piece of paper, and fold and unfold along each of the diagonals. Step 1 2. Turn the paper over. 3. Fold the bottom edge of the paper to the center point, but only crease the part between the diagonals. Unfold back to the square. Step 3 Step 4 4. Repeat for the other three sides. 5. Now, fold the bottom edge up to the upper crease line that you just made, 3/4 of the way up, again only creasing the part between the diagonals. This is a fairly short crease. Unfold back to the square. Step 5 Step 6 6. Fold the bottom edge to the lower crease line (the one you made in step 3), once again only creasing between the diagonals. Unfold back to the square. 7. Repeat for all four sides. At this point, the square should be divided into four concentric square rings. See the figure on the next page. 8. Turn the paper over. 9. The goal now is to create additional creases halfway between the ones constructed so far, for a total of eight concentric square rings. You can do this by folding the bottom edge up to previously constructed folds, but use only every other fold. That is, fold the bottom edge up to the existing creases 7/8, 5/8, 3/8, and 1/8 of the way to the top, again only creasing between the diagonals. Do this for all four sides. | Multivariable_Calculus_Shimamoto_Page_92_Chunk3308 |
3.9. EXERCISES FOR CHAPTER 3 81 Step 7 Step 9 10. Turn the paper over. The creases should alternate as you move from one concentric square to the next: mountain, valley, mountain, valley, . . . . 11. Now, try to fold all the creases. It’s easiest to start from the outer ring and work your way in, pinching in at the corners and trying to compress the faces of adjacent rings together as you move towards the center. Half of the short diagonal creases between the rings need to be reversed so that the rings nest inside each other, though the paper can be coaxed to do this without too much trouble. In the end, you should get something shaped like an X, or a four-armed starfish. 12. Open up the arms of the starfish. The paper will naturally assume the hypar shape. | Multivariable_Calculus_Shimamoto_Page_93_Chunk3309 |
82 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES | Multivariable_Calculus_Shimamoto_Page_94_Chunk3310 |
Chapter 4 Real-valued functions: differentiation We now extend the notion of derivative from real-valued functions of one variable, as in first-year calculus, to real-valued functions of n variables f : U →R, where U is an open set in Rn. The most natural choice would be to copy the one-variable definition verbatim and let the derivative at a point a be limx→a f(x)−f(a) x−a . Unfortunately, this involves dividing by a vector, which does not make sense. One potential remedy would be to consider limx→a f(x)−f(a) ∥x−a∥ instead. Here, it turns out that the limit may fail to exist even for very simple functions. For example, for the function f(x) = x of one variable, this would be limx→a x−a |x−a| which approaches +1 from the right and −1 from the left. So we must look harder to find a formulation of the one-variable derivative that can be generalized. 4.1 The first-order approximation In first-year calculus, we think of a function f as being differentiable at a point a if it has a good tangent line approximation: f(x) ≈ℓ(x) when x is near a, where y = ℓ(x) = mx + b is the tangent line to the graph of f at the point (a, f(a)). See Figure 4.1. Figure 4.1: Approximating a differentiable function f(x) by one of its tangent lines ℓ(x) In fact, the tangent line is given by ℓ(x) = f(a) + f′(a) · (x −a), (4.1) so m = f′(a) and b = f(a) −f′(a) · a. 83 | Multivariable_Calculus_Shimamoto_Page_95_Chunk3311 |
84 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION By definition, f′(a) = limx→a f(x)−f(a) x−a , assuming the limit exists. This can be rewritten as 0 = limx→a | Multivariable_Calculus_Shimamoto_Page_96_Chunk3312 |
4.1. THE FIRST-ORDER APPROXIMATION 85 If h < 0, then the denominator in equation (4.4) is |h| = −h. The minus sign can be factored out and canceled, so equation (4.5) still holds. This can be solved for T(e1): m1 = T(e1) = lim h→0 f(a + he1) −f(a) h = lim h→0 f(a1 + h, a2, . . . , an) −f(a1, a2, . . . , an) h . This limit of a difference quotient is the definition of the ordinary one-variable derivative where x2, . . . , xn are held fixed at x2 = a2, . . . , xn = an and only x1 varies. It is called the partial derivative ∂f ∂x1 (a). Similarly, by approaching a in the x2, . . . , xn directions, we find that m2 = ∂f ∂x2 (a), . . . , mn = ∂f ∂xn (a) and hence that A = h ∂f ∂x1 (a) ∂f ∂x2 (a) . . . ∂f ∂xn (a) i . Definition. For a real-valued function f : U →R defined on an open set U in Rn and a point a of U: 1. If j = 1, 2, . . . , n, the partial derivative of f at a with respect to xj is defined by: ∂f ∂xj (a) = lim h→0 f(a + hej) −f(a) h . Note that a+hej = (a1, . . . , aj +h, . . . , an), so a+hej and a differ only in the jth coordinate. Thus the partial derivative is defined by the one-variable difference quotient for the derivative with variable xj. Other common notations for the partial derivative are fxj(a) and (Djf)(a). 2. Df(a) is defined to be the 1 by n matrix Df(a) = h ∂f ∂x1 (a) ∂f ∂x2 (a) . . . ∂f ∂xn (a) i . The corresponding vector ∇f(a) = | Multivariable_Calculus_Shimamoto_Page_97_Chunk3313 |
86 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION Definition. Let U be an open set in Rn, f : U →R a real-valued function, and a a point of U. Then f is said to be differentiable at a if: lim x→a f(x) −f(a) −Df(a) · (x −a) ∥x −a∥ = 0. When this happens, the matrix Df(a) is called the derivative of f at a. It is also known as the Jacobian matrix. The affine function ℓ(x) = f(a)+Df(a)·(x−a) is called the first-order affine approximation of f at a. (Note the resemblance to the one-variable tangent line approximation (4.1).) The product in the numerator of the definition is the matrix product: Df(a) · (x −a) = h ∂f ∂x1 (a) ∂f ∂x2 (a) . . . ∂f ∂xn (a) i x1 −a1 x2 −a2 ... xn −an . It could also be written as a dot product: ∇f(a) · (x −a). The definition generalizes equation (4.2) for one variable in that it says that f is differentiable at a if the error in using the first-order approximation ℓgoes to 0 faster than ∥x −a∥as x approaches a. In particular, the first-order approximation is a good approximation. This is worth emphasizing: the existence of a good first-order approximation is often the most important way to think of what it means for a function to be differentiable. Note that the derivative Df(a) itself is no longer just a number the way it is in first-year calculus, but rather a matrix or, even better, the linear part of a good affine approximation of f near a. Example 4.2. Let f : R2 →R be f(x, y) = x2 + y2, and let a = (1, 2). Is f differentiable at a? We calculate the various elements that go into the definition. First, ∂f ∂x = 2x and ∂f ∂y = 2y, so Df(x, y) = 2x 2y and Df(a) = Df(1, 2) = 2 4 . Also, f(a) = 5, and x −a = x −1 y −2 . Thus we need to check if: lim (x,y)→(1,2) x2 + y2 −5 − 2 4 x −1 y −2 p (x −1)2 + (y −2)2 = 0. Let’s simplify the numerator, completing the square at an appropriate stage: x2 + y2 −5 − 2 4 x −1 y −2 = x2 + y2 −5 − | Multivariable_Calculus_Shimamoto_Page_98_Chunk3314 |
4.1. THE FIRST-ORDER APPROXIMATION 87 The function in this last expression is continuous, as it is a composition of sums and products of continuous pieces. Thus the value of the limit is simply the value of the function at the point (1, 2). In other words: lim (x,y)→(1,2) x2 + y2 −5 − 2 4 x −1 y −2 p (x −1)2 + (y −2)2 = p (1 −1)2 + (2 −2)2 = 0. This shows that the answer is: yes, f is differentiable at a = (1, 2). In addition, the preceding calculations show that the first-order approximation of f(x, y) = x2 + y2 at a = (1, 2) is: ℓ(x, y) = f(a) + Df(a) · (x −a) = 5 + 2 4 x −1 y −2 = 2x + 4y −5. (4.6) For instance, (x, y) = (1.05, 1.95) is near a, and f(1.05, 1.95) = 1.052 + 1.952 = 4.905 while ℓ(1.05, 1.95) = 2(1.05) + 4(1.95) −5 = 4.9. The values are pretty close. The same sort of reasoning can be used to show that, more generally, the function f(x, y) = x2 + y2 is differentiable at every point of R2. This is Exercise 1.16. Example 4.3. Let f(x, y) = ( xy x2+y2 if (x, y) ̸= (0, 0), 0 if (x, y) = (0, 0) . Its graph is shown in Figure 4.2. Is f differentiable at a = (0, 0)? Figure 4.2: The graph z = xy x2+y2 Using the definition requires knowing the partial derivatives at (0, 0). To calculate them, we could apply the quotient rule to the formula that defines f, but the results would be valid only when (x, y) ̸= (0, 0), which is precisely what we don’t need. So instead we go back to the basic idea behind partial derivatives, namely, that they are one-variable derivatives where all variables but one are held constant. For example, to find ∂f ∂x(0, 0), fix y = 0 in f(x, y) and look at the resulting function of x. By the formula for f, f(x, 0) = x·0 x2+02 = 0 if x ̸= 0. This expression also holds when x = 0: by definition, f(0, 0) = 0. Thus f(x, 0) = 0 for all x. Differentiating this with respect to x gives ∂f ∂x(x, 0) = d dx | Multivariable_Calculus_Shimamoto_Page_99_Chunk3315 |
88 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION Moreover, x −a = x −0 y −0 = x y , and ∥x −a∥= p x2 + y2. Thus, according to the definition of differentiability, we need to check if: lim (x,y)→(0,0) xy x2+y2 −0 − 0 0 x y p x2 + y2 = 0, i.e., if lim(x,y)→(0,0) xy (x2+y2)3/2 = 0. For this, suppose we approach the origin along the line y = x. Then we are looking at the behavior of x·x (x2+x2)3/2 = x2 23/2|x|3 = 1 23/2|x|. This blows up as x goes to 0. It certainly doesn’t approach 0. Hence there’s no way that lim(x,y)→(0,0) xy (x2+y2)3/2 can equal 0 (in fact, the limit does not exist), so f is not differentiable at (0, 0). This example shows that a function of more than one variable can fail to be differentiable at a point even though all of its partial derivatives exist there. This differs from the one-variable case. 4.2 Conditions for differentiability We now look for ways to tell if a function is differentiable without having to do all the work of going through the definition. As we saw in the last example, calculating the partial derivatives is not enough. That’s bad news, because calculating partial derivatives is pretty easy. The good news is that we are about to obtain results that show that the examples in the last section could have been solved much more quickly using means other than the definition. For the remainder of this section: U denotes an open set in Rn, f : U →R is a real-valued function defined on U, and a is a point of U. We won’t keep repeating this. For instance, in Example 4.3 above, we asked if the function f(x, y) = ( xy x2+y2 if (x, y) ̸= (0, 0), 0 if (x, y) = (0, 0) (4.7) were differentiable at (0, 0). In fact, we had shown before that f is not continuous at (0, 0) (see Example 3.19 in Chapter 3). In first-year calculus, discontinuous functions cannot be differentiable. If the same were true for functions of more than one variable, we could have saved ourselves a lot of work. We state the result in contrapositive form. Proposition 4.4. If f is differentiable at a, then f is continuous at a. Proof. We just outline the idea and leave the details for the exercises (Exercise 2.3). If f is differentiable at a, then limx→a(f(x) −ℓ(x)) = 0, where ℓis the first-order approximation at a. In other words: lim x→a(f(x) −f(a) −Df(a) · (x −a)) = 0. The matrix product term Df(a) · (x −a) goes to 0 as x approaches a (it’s a continuous function of x), so the last line becomes limx→a(f(x) −f(a) −0) = 0. Hence limx→a f(x) = f(a). As a result, f is continuous at a. | Multivariable_Calculus_Shimamoto_Page_100_Chunk3316 |
4.3. THE MEAN VALUE THEOREM 89 Example 4.5. Returning to function (4.7) above, we could have simply said from the start that f is not continuous at (0, 0) and therefore is not differentiable there either, avoiding the definition of differentiability entirely. Conversely, we know from first-year calculus that a function can be continuous without being differentiable. The absolute value function f(x) = |x| is the standard example. It is continuous, but, because its graph has a corner, it is not differentiable at the origin. Sometimes, we can use the failure of a one-variable derivative to tell us about the differentiability of a function of more than one variable. Proposition 4.6. If some partial derivative ∂f ∂xj does not exist at a, then f is not differentiable at a. Proof. This follows simply because if ∂f ∂xj (a) doesn’t exist, then neither does Df(a), so the condition for differentiability cannot be satisfied. Example 4.7. One analogue of the absolute value function is the function f : R2 →R, f(x, y) = ∥(x, y)∥= p x2 + y2. We have seen that its graph is a cone with vertex at the origin (Example 3.2). To find ∂f ∂x at the origin, we fix y = 0 to get f(x, 0) = √ x2 = |x|. The derivative with respect to x does not exist at x = 0, so, by definition, ∂f ∂x(0, 0) does not exist either. Thus f is not differentiable at (0, 0) (though it is continuous there). We have saved the most useful criterion for differentiability for last. To understand why it is true, however, we take a slight detour. 4.3 The mean value theorem We devote this section to reviewing an important, and possibly underappreciated, result from first-year calculus. Theorem 4.8 (Mean Value Theorem). Let f : [a, b] →R be a continuous function that is differen- tiable on the open interval (a, b). Then there exists a point c in (a, b) such that: f(b) −f(a) = f′(c) · (b −a). The theorem relates the difference in function values f(b) −f(a) to the difference in domain values b −a. The relation can be useful even if one cannot completely pin down the location of the point c (in fact, one rarely tries). For instance, if f(x) = sin x, then |f′(c)| = | cos c| ≤1 regardless of what c is. Hence the mean value theorem gives | sin b −sin a| ≤|b −a| for all a, b. By the same reasoning, | cos b −cos a| ≤|b −a| as well. There is an n-variable version of the mean value theorem that follows from results we prove later (see Exercise 5.5), but for now we show how the one-variable theorem can be used to study a difference in function values for a function of two variables by allowing only one variable to change at a time and using partial derivatives. Example 4.9. Let f(x, y) be a continuous real-valued function of two variables that is defined on an open ball B in R2 and whose partial derivatives ∂f ∂x and ∂f ∂y exist at every point of B. Let a = (a1, a2) and b = (b1, b2) be points of B. We show that there exist points p and q in B such that ∥p −a∥≤∥b −a∥, ∥q −a∥≤∥b −a∥, and: f(b) −f(a) = ∂f ∂x(p) · (b1 −a1) + ∂f ∂y (q) · (b2 −a2). (4.8) | Multivariable_Calculus_Shimamoto_Page_101_Chunk3317 |
90 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION Figure 4.3: Going from a to b by way of c See Figure 4.3. To prove this, we introduce the intermediate point c = (b1, a2) and think of going from a to b by first going from a to c and then from c to b, as in Figure 4.3. Only one variable varies along each of the segments, so by writing f(b) −f(a) = (f(b) −f(c)) + (f(c) −f(a)), (4.9) the mean value theorem applies to each of the differences on the right. That is, in the second difference, using x as variable and y = a2 fixed, the mean value theorem implies that there exists a point p = (p1, a2) on the line segment between a and c such that: f(c) −f(a) = ∂f ∂x(p) · (b1 −a1). (4.10) Similarly, for the first difference, using y as variable and x = b1 fixed, there exists a point q = (b1, q2) on the line segment between c and b such that: f(b) −f(c) = ∂f ∂y (q) · (b2 −a2). (4.11) Substituting (4.10) and (4.11) into equation (4.9) results in f(b)−f(a) = ∂f ∂y (q)·(b2 −a2)+ ∂f ∂x(p)· (b1 −a1), which is the desired conclusion (4.8). 4.4 The C1 test We present the result that often gives the simplest way to show that a function is differentiable. It uses partial derivatives. Definition. Let U be an open set in Rn. A real-valued function f : U →R all of whose partial derivatives ∂f ∂x1 , ∂f ∂x2 , . . . , ∂f ∂xn are continuous on U is said to be of class C1, or continuously differentiable, on U. Theorem 4.10 (The C1 test). If f is of class C1 on U, then f is differentiable at every point of U. Proof. We give the proof in the case of two variables, that is, U ⊂R2. This contains the main new idea, and we leave for the reader the generalization to n variables, which is a relatively straightfor- ward extension. | Multivariable_Calculus_Shimamoto_Page_102_Chunk3318 |
4.4. THE C1 TEST 91 We use the definition of differentiability. Let a be a point of U. We first analyze the error f(x) −f(a) −Df(a) · (x −a) in the first-order approximation of f at a. Since U is an open set and we are interested only in what happens when x is near a, we may assume that x lies in an open ball about a that is contained in U. Say a = (c, d) and x = (x, y). By the result we just obtained in Example 4.9, there are points p and q such that ∥p −a∥≤∥x −a∥, ∥q −a∥≤∥x −a∥, and: f(x) −f(a) = ∂f ∂x(p) · (x −c) + ∂f ∂y (q) · (y −d). Also: Df(a) · (x −a) = h ∂f ∂x(a) ∂f ∂y (a) i x −c y −d = ∂f ∂x(a) · (x −c) + ∂f ∂y (a) · (y −d). Hence: f(x) −f(a) −Df(a) · (x −a) = | Multivariable_Calculus_Shimamoto_Page_103_Chunk3319 |
92 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION of (4.7). We saw that f is not differentiable at the origin. On the other hand, at points other than the origin, the partial derivatives can be found using the quotient rule. This gives: ∂f ∂x = (x2 + y2) · y −xy · 2x (x2 + y2)2 = y3 −x2y (x2 + y2)2 = y(y2 −x2) (x2 + y2)2 and ∂f ∂y = (x2 + y2) · x −xy · 2y (x2 + y2)2 = x(x2 −y2) (x2 + y2)2 . As algebraic combinations of x and y, both of these partial derivatives are continuous on the open set R2 −{(0, 0)}. As a result, the C1 test says that f is differentiable at all points other than the origin. 4.5 The Little Chain Rule In one-variable calculus, the derivative measures the rate of change of a function. When there is more than one variable, the derivative is no longer just a number, but it still has something to say about rates of change. Let U be an open set in Rn, and let f : U →R be a real-valued function defined on U. Suppose that α: I →U is a path in U defined on an interval I in R. The composition f ◦α: I →R is a real-valued function of one variable that describes how f behaves along the path. We examine the rate of change of this composition. Let t0 be a point of I, and let a = α(t0). If f is differentiable at a, then its first-order approximation applies: f(x) ≈f(a) + ∇f(a) · (x −a), or: f(x) −f(a) ≈∇f(a) · (x −a) when x is near a. Here, we use the gradient form of the derivative. Assume in addition that α is differentiable at t0. Then α is continuous at t0, so, when t is near t0, α(t) is near α(t0). Substituting x = α(t) and a = α(t0) into the previous approximation gives: f(α(t)) −f(α(t0)) ≈∇f(α(t0)) · (α(t) −α(t0)) when t is near t0. Thus f(α(t))−f(α(t0)) t−t0 ≈∇f(α(t0))· α(t)−α(t0) t−t0 . In the limit as t approaches t0, the left side approaches the one-variable derivative (f ◦α)′(t0) and the right approaches ∇f(α(t0)) · α′(t0). Writing t in place of t0, this gives the following result. Theorem 4.13 (Little Chain Rule). Let U be an open set in Rn, α: I →U a path in U defined on an open interval I in R, and f : U →R a real-valued function. If α is differentiable at t and f is differentiable at α(t), then f ◦α is differentiable at t and: (f ◦α)′(t) = ∇f(α(t)) · α′(t). In words, the derivative of the composition is “gradient dot velocity.” While we outlined the idea why this is true, a complete proof of the result as stated entails a more careful treatment of the approximations involved. The details appear in the exercises (see Exercise 5.6). 4.6 Directional derivatives In this section and the next, we present some interpretations of the gradient that follow from the Little Chain Rule. The first concerns the rate of change of a function of n variables in a given | Multivariable_Calculus_Shimamoto_Page_104_Chunk3320 |
4.6. DIRECTIONAL DERIVATIVES 93 direction. Let a be a point of an open set U in Rn. To represent a direction going away from a, we choose a unit vector u pointing in that direction. We travel in that direction on the line through a and parallel to u, which is parametrized by α(t) = a + tu. The parametrization has velocity α′(t) = u and hence constant speed ∥u∥= 1. If f : U →R is a real-valued function, then the derivative (f ◦α)′(t) is the rate of change of f along this line. In particular, since α(0) = a, we think of (f ◦α)′(0) as the rate of change at a in the direction of u. Definition. Given the input described above, (f ◦α)′(0) is called the directional derivative of f at a in the direction of the unit vector u, denoted by (Duf)(a). If v is any nonzero vector in Rn, we define the directional derivative in the direction of v to be (Duf)(a), where u = 1 ∥v∥v is the unit vector in the direction of v. Directional derivatives are easy to compute, for by the Little Chain Rule: (f ◦α)′(0) = ∇f(α(0)) · α′(0) = ∇f(a) · u. In other words: Proposition 4.14. If f is differentiable at a and u is a unit vector, then the directional derivative is given by: (Duf)(a) = ∇f(a) · u. As a result, (Duf)(a) = ∥∇f(a)∥∥u∥cos θ = ∥∇f(a)∥cos θ, where θ is the angle between ∇f(a) and the direction vector u. Keeping a fixed but varying the direction, we see that (Duf)(a) is maximized when θ = 0, i.e., when u points in the same direction as ∇f(a), and that, in this direction, (Duf)(a) = ∥∇f(a)∥. Corollary 4.15. The maximum directional derivative of f at a: • occurs in the direction of ∇f(a) and • has a value of ∥∇f(a)∥. Sometimes, weather reports mention the “temperature gradient.” Usually, this does not refer to the gradient vector that we have been studying but rather to one of the aspects of the gradient in the corollary, that is, the magnitude and/or direction of the maximum rate of temperature increase. Example 4.16. Let f : R3 →R be f(x, y, z) = x2 + y2 + z2, and let a = (1, 2, 3). (a) Find the directional derivative of f at a in the direction of v = (2, −1, 1). (b) Find the direction and rate of maximum increase at a. (a) First, ∇f = (2x, 2y, 2z), so ∇f(a) = (2, 4, 6). Next, the unit vector in the direction of v is u = 1 ∥v∥v = 1 √ 6(2, −1, 1). Thus: (Duf)(a) = ∇f(a) · u = (2, 4, 6) · 1 √ 6(2, −1, 1) = 1 √ 6(4 −4 + 6) = 6 √ 6 = √ 6. (b) The direction of maximum increase is the direction of ∇f(a) = (2, 4, 6), that is, the direction of the unit vector u = 1 √4+16+36(2, 4, 6) = 1 √ 56(2, 4, 6) = 1 √ 14(1, 2, 3). The rate of maximum increase is ∥∇f(a)∥= √ 56 = 2 √ 14. | Multivariable_Calculus_Shimamoto_Page_105_Chunk3321 |
94 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION 4.7 ∇f as normal vector Let f : U →R be a differentiable real-valued function of n variables defined on an open set U in Rn, and let S be the level set of f corresponding to f = c. Let a be a point of S, and let v be a vector tangent to S at a. We take this to mean that v can be realized as the velocity vector of a path in S, that is, there is a differentiable path α: I →Rn that: • lies in S, i.e., α(t) ∈S for all t, • passes through a, say a = α(t0) for some t0 in I, and • has velocity v at a, i.e., α′(t0) = v. This is illustrated in Figure 4.4. Then f(α(t)) = c for all t, so (f ◦α)′(t) = c′ = 0. By the Figure 4.4: A level set S of f and a vector v tangent to it Little Chain Rule, this can be written ∇f(α(t)) · α′(t) = 0. In particular, when t = t0, we obtain ∇f(a)·v = 0. This is true for all vectors v tangent to S at a, which has the following consequence. Proposition 4.17. If a is a point on a level set S of f, then ∇f(a) is a normal vector to S at a. Example 4.18. Find an equation for the plane tangent to the graph z = x2 + y2 at the point (1, 2, 5). As with the equation of any plane, it suffices to find a point p on the plane and a normal vector n. As a point, we can use the point of tangency, p = (1, 2, 5). To find a normal vector, we rewrite the graph as a level set. That is, rewrite z = x2 +y2 as x2 +y2 −z = 0. Then the graph is the level set of the function of three variables F(x, y, z) = x2 + y2 −z corresponding to c = 0. In particular, ∇F(p) is a normal vector to the level set, that is, to the graph, at p, so it is normal to the tangent plane as well. See Figure 4.5. The rest is calculation: ∇F = (2x, 2y, −1), so n = ∇F(p) = (2, 4, −1) is our normal vector. Substituting into the equation n · x = n · p for a plane gives: (2, 4, −1) · (x, y, z) = (2, 4, −1) · (1, 2, 5) = 2 + 8 −5 = 5, or 2x + 4y −z = 5. Note that we just studied the graph of the function f(x, y) = x2 + y2 of two variables, and we considered the point p = (1, 2, 5) = (a, f(a)) on the graph, where a = (1, 2). In the remarks following Example 4.2, we saw that the first-order approximation of f at a is given by ℓ(x, y) = 2x+4y−5 (see equation (4.6)). Thus the graph of the approximation ℓhas equation z = 2x+4y−5, which is exactly the tangent plane we just found. | Multivariable_Calculus_Shimamoto_Page_106_Chunk3322 |
4.7. ∇f AS NORMAL VECTOR 95 Figure 4.5: The tangent plane to z = x2 + y2 at p = (1, 2, 5) In other words, geometrically, the approximation we get by using the tangent plane at (a, f(a)) to approximate the graph coincides with the first-order approximation at a. This extends a familiar idea from first-year calculus, where tangent lines are used to obtain the first-order approximation of a differentiable function. In fact, it is the idea we used to motivate our multivariable definition of differentiability in the first place. Example 4.19. Suppose that T(x, y) represents the temperature at the point (x, y) in an open set U of the plane. There are two natural families of curves associated with this function. The first is the curves along which the temperature is constant. They are called isotherms and are the level sets of T: all points on a single isotherm have the same temperature. See Figure 4.6.4 Figure 4.6: Isotherms are level curves of temperature. Heat travels along curves that are always orthogonal to the isotherms. The second is the curves along which heat flows. The principle here is that, at every point, heat travels in the direction in which the temperature is decreasing as rapidly as possible, from hot to cold. From our work on directional derivatives, we know that the maximum rate of decrease occurs in the direction of −∇T. And we’ve just seen that ∇T is always normal to the level sets, the isotherms. In other words, the isotherms and the curves of heat flow are mutually orthogonal families of curves. If you have a map of the isotherms, you can visualize the flow of heat: always move at right angles to the isotherms. 4Figure taken from https://www.noaa.gov/jetstream/surface-temperature-map by National Weather Service, public domain. | Multivariable_Calculus_Shimamoto_Page_107_Chunk3323 |
96 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION 4.8 Higher-order partial derivatives Once they start partial differentiating, most people can’t stop. Example 4.20. Let f(x, y) = x3 −3x2y4 + exy. Then: ∂f ∂x = 3x2 −6xy4 + yexy and ∂f ∂y = −12x2y3 + xexy. Both of these are again real-valued functions of x and y, so they have partial derivatives of their own. In the case of ∂f ∂x: ∂2f ∂x2 = ∂ ∂x ∂f ∂x = 6x −6y4 + y2exy and ∂2f ∂y ∂x = ∂ ∂y ∂f ∂x = −24xy3 + exy + xyexy. Similarly, for ∂f ∂y : ∂2f ∂x ∂y = ∂ ∂x ∂f ∂y = −24xy3 + exy + xyexy and ∂2f ∂y2 = ∂ ∂y ∂f ∂y = −36x2y2 + x2exy. These are the second-order partial derivatives. They too have partial derivatives, and the process could continue indefinitely to obtain partial derivatives of higher and higher order. Perhaps the most striking part of the preceding calculation is that the two partial derivatives ∂2f ∂y ∂x and ∂2f ∂x ∂y turned out to be equal. They are called the mixed partials. To see whether they might be equal in general, we return to the definition of a partial derivative as a limit of difference quotients and try to describe the second-order partials in terms of the original function f. In fact, what we are about to discuss is a method one might use to numerically approximate the second-order partials. Let a = (c, d). First, consider ∂2f ∂x ∂y = ∂ ∂x | Multivariable_Calculus_Shimamoto_Page_108_Chunk3324 |
4.8. HIGHER-ORDER PARTIAL DERIVATIVES 97 This expression is symmetric enough in h and k that perhaps you are willing to believe that you would get the same thing when the order of differentiation is reversed. But if you are nervous about it, here is the parallel calculation for ∂2f ∂y ∂x: ∂2f ∂y ∂x(c, d) = ∂ ∂y | Multivariable_Calculus_Shimamoto_Page_109_Chunk3325 |
98 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION Example 4.22. Find ∂4f ∂x ∂y2 ∂z if f(x, y, z) = | Multivariable_Calculus_Shimamoto_Page_110_Chunk3326 |
4.10. MAX/MIN: CRITICAL POINTS 99 this for functions of more than one variable. This is a big subject that overflows with applications. Our treatment is not comprehensive. Instead, we have selected a couple of topics that illustrate some of the similarities and differences that pertain to moving to the multivariable case. Definition. Let f : U →R be a function defined on an open set U in Rn. f is said to have a local maximum at a point a of U if there is an open ball B(a, r) centered at a such that f(a) ≥f(x) for all x in B(a, r). It is said to have a local minimum at a if instead f(a) ≤f(x) for all x in B(a, r). See Figure 4.7. It has a global maximum or global minimum at a if the associated inequalities are true for all x in U, not just in B(a, r). Figure 4.7: Local maximum (left) and local minimum (right) at a If f is smooth and has a local maximum or local minimum at a, then it has a local maximum or local minimum in every direction, in particular, in each of the coordinate directions. The derivatives in these directions are the partial derivatives. They are essentially one-variable derivatives, so from first-year calculus, at a local maximum or minimum, all of them must be 0, that is, ∂f ∂xj (a) = 0 for j = 1, 2, . . . , n. Definition. A point a is called a critical point of f if ∂f ∂xj (a) = 0 for j = 1, 2, . . . , n. Equivalently, Df(a) = 0 0 . . . 0 , or ∇f(a) = 0. The preceding discussion proves the following. Proposition 4.23. Let U be an open set in Rn, and let f : U →R be a smooth function. If f has a local maximum or a local minimum at a, then a is a critical point of f. Hence, if one is looking for local maxima or minima of a smooth function on an open set, the critical points are the only possible candidates. This is just like the one-variable case. Conversely, as in the one-variable case, not every critical point is necessarily a local max or min, though there are new ways in which this failure can occur. A good example to keep in mind are functions whose graphs are saddles, such as f(x, y) = x2 −y2 from Example 3.3 in Chapter 3. At a saddle point, the function has a local maximum in one direction and a local minimum in another, so there is no open ball about the point in which it is exclusively one or the other. Even so, in first-year calculus, a given critical point can often be classified as a local maximum or minimum using the second derivative. There is an analogue of this for more than one variable that is similar in spirit while reflecting the greater number of possibilities. We discuss the situation for two variables. Let U be an open set in R2, and let f : U →R be a smooth function. If a ∈U, consider the matrix of second-order partial derivatives: H(a) = " ∂2f ∂x2 (a) ∂2f ∂y ∂x(a) ∂2f ∂x ∂y(a) ∂2f ∂y2 (a) # . | Multivariable_Calculus_Shimamoto_Page_111_Chunk3327 |
100 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION It is called the Hessian matrix of f at a. By the equality of mixed partials, the two off-diagonal terms, ∂2f ∂x ∂y(a) and ∂2f ∂y ∂x(a), are actually equal. Theorem 4.24 (Second derivative test for functions of two variables). Let a be a critical point of f such that det H(a) ̸= 0. Under this assumption, a is called a nondegenerate critical point. 1. If det H(a) > 0 and: (a) if ∂2f ∂x2 (a) > 0, then f has a local minimum at a, (b) if ∂2f ∂x2 (a) < 0, then f has a local maximum at a. 2. If det H(a) < 0, then f has a saddle point at a. If det H(a) = 0, the test is inconclusive. We discuss the ideas behind the test in the next section, but, before that, we present some examples. The first involves three simple functions for which we already know the answers, but it is meant to illustrate how the test works and confirm that it gives the right results. In a way, these three functions are prototypes of the general situation. Example 4.25. Find all critical points of the following functions and, if possible, classify each as a local maximum, local minimum, or saddle point. (a) f(x, y) = x2 + y2 (b) f(x, y) = −x2 −y2 (c) f(x, y) = x2 −y2 Intuitively, it’s clear that the first two of these attain their minimum and maximum values, respectively, at the origin, while, as noted earlier, the third has a saddle point there. The graphs of the functions are shown in Figure 4.8. Figure 4.8: Prototypical nondegenerate critical points at (0, 0): f(x, y) = x2 + y2 (left), f(x, y) = −x2 −y2 (middle), f(x, y) = x2 −y2 (right) The second derivative test reaches these conclusions as follows. (a) To find the critical points, we set ∂f ∂x = 2x = 0 and ∂f ∂y = 2y = 0. The only solution is x = 0 and y = 0. Hence the only critical point is (0, 0). To classify its type, we look at the second-order partials: ∂2f ∂x2 = 2, ∂2f ∂x ∂y = ∂2f ∂y ∂x = 0, and ∂2f ∂y2 = 2. Therefore: H(0, 0) = 2 0 0 2 . Since det H(0, 0) = 4 > 0 and ∂2f ∂x2 (0, 0) = 2 > 0, the second derivative test implies that f has a local minimum at (0, 0), as expected. | Multivariable_Calculus_Shimamoto_Page_112_Chunk3328 |
4.10. MAX/MIN: CRITICAL POINTS 101 (b) By the same calculation, (0, 0) is the only critical point, only this time: H(0, 0) = −2 0 0 −2 . Hence det H(0, 0) = 4 > 0 and ∂2f ∂x2 (0, 0) = −2 < 0, so f has a local maximum at (0, 0). (c) Again, (0, 0) is the only critical point, but: H(0, 0) = 2 0 0 −2 . Since det H(0, 0) = −4 < 0, f has a saddle point at (0, 0). Example 4.26. We repeat for the function f(x, y) = x3 + 8y3 −3xy. A portion of the graph of f is shown in Figure 4.9. Figure 4.9: The graph of f(x, y) = x3 + 8y3 −3xy for −1 ≤x ≤1, −1 ≤y ≤1 First, to find the critical points, we set: (∂f ∂x = 3x2 −3y = 0 ∂f ∂y = 24y2 −3x = 0. The first equation implies that y = x2, and substituting this into the second gives 24x4 −3x = 0, or 3x(8x3 −1) = 0. Hence x = 0, in which case y = 02 = 0, or x3 = 1 8, i.e., x = 1 2, in which case y = | Multivariable_Calculus_Shimamoto_Page_113_Chunk3329 |
102 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION Figure 4.10: The graph of f(x, y) = x3 + 8y3 −3xy near (0, 0) (left) and near (1 2, 1 4) (right) 4.11 Classifying nondegenerate critical points We now say a few words about why the second derivative test for functions of two variables works (Theorem 4.24 of the preceding section). We won’t give a complete proof, but, since the test may appear to come out of nowhere, a word or two of explanation is warranted. We begin by reviewing the corresponding situation for functions of one variable. Let f : I →R be a smooth function defined on an open interval I in R. The critical points of f are the points a where f′(a) = 0. At such points, the usual first-order approximation f(x) ≈f(a) + f′(a) · (x −a) degenerates to f(x) ≈f(a), which gives us little information about the behavior of f(x) when x is near a. As a refinement, we add an additional term by considering the second-order Taylor approxima- tion at a: f(x) ≈f(a) + f′(a) · (x −a) + f′′(a) 2 · (x −a)2. (4.16) If a is a critical point, this reduces to f(x) ≈f(a)+ f′′(a) 2 ·(x−a)2, a parabola whose general shape is determined by the sign of the coefficient f′′(a) 2 . If f′′(a) > 0, then the approximating parabola is concave up and f has a local minimum at a. If f′′(a) < 0, the parabola is concave down and f has a local maximum. If f′′(a) = 0, the approximation degenerates again. Further information is required to determine the type of critical point in this case. 4.11.1 The second-order approximation Let f : U →R be a smooth function of two variables defined on an open set U in R2, and let a be a point of U. The situation here is that the second-order approximation of f(x) for x near a is: f(x) ≈f(a) + Df(a) · (x −a) + 1 2(x −a)t · H(a) · (x −a). (4.17) The various dots are matrix products, where, writing x = (x, y) and a = (c, d): • H(a) = " ∂2f ∂x2 (a) ∂2f ∂x ∂y(a) ∂2f ∂x ∂y(a) ∂2f ∂y2 (a) # is the Hessian matrix at a, • x −a = x −c y −d , and • (x −a)t = x −c y −d , the transpose of x −a. To justify the approximation, let x be a point near a, and let v = x −a. Consider the real- valued function g defined by g(t) = f(a + tv). Then g(0) = f(a) and g(1) = f(a + v) = f(x), so | Multivariable_Calculus_Shimamoto_Page_114_Chunk3330 |
4.11. CLASSIFYING NONDEGENERATE CRITICAL POINTS 103 finding an approximation for f(x) is the same as approximating g(1). As a real-valued function of one variable, g(t) has a second-order Taylor approximation (4.16) at 0, which for g(1) gives: f(x) = g(1) ≈g(0) + g′(0) · (1 −0) + g′′(0) 2 · (1 −0)2 ≈f(a) + g′(0) + g′′(0) 2 . (4.18) To find the derivatives of g, note that, by construction, g is the composition g(t) = f(α(t)), where α(t) = a + tv. Hence, by the Little Chain Rule: g′(t) = ∇f(α(t)) · α′(t) = ∇f(a + tv) · v. (4.19) In particular, g′(0) = ∇f(a) · v, or in matrix form, g′(0) = Df(a) · v. Substituting this into (4.18) and recalling that v = x −a gives: f(x) ≈f(a) + Df(a) · (x −a) + g′′(0) 2 . (4.20) So far, we have succeeded only in reconstructing the first-order approximation of f at a. The really new information comes from determining g′′(0). From equation (4.19), note that g′ is also a composition, namely, g′(t) = u(α(t)), where u(x) = ∇f(x) · v and α(t) = a + tv as before. Therefore, by the same Little Chain Rule calculation: g′′(0) = (u ◦α)′(0) = ∇u(a) · v. (4.21) To find ∇u, we let v = (h, k) and write out u(x) = ∇f(x)·v = | Multivariable_Calculus_Shimamoto_Page_115_Chunk3331 |
104 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION We remark that one can continue in this way to obtain higher-order approximations of f by using higher-order Taylor approximations of g. The coefficients involve the higher-order derivatives g(n)(0) of g. The idea is that, by the Little Chain Rule, each additional derivative of g corresponds to applying the “operator” v·∇= (h, k)·( ∂ ∂x, ∂ ∂y) = h ∂ ∂x +k ∂ ∂y. Thus g(n)(0) = | Multivariable_Calculus_Shimamoto_Page_116_Chunk3332 |
4.12. MAX/MIN: LAGRANGE MULTIPLIERS 105 about the signs of the coefficients A and AC−B2 A then gives the second-derivative test as stated in Theorem 4.24. The key idea of the argument was to complete the square to turn f into one of the prototypes of being a sum and/or difference of squares, at least up to second order. This is not as ad hoc as it might seem. A theorem in linear algebra states that, if M is a symmetric matrix in the sense that Mt = M, then it is always possible to complete the square and convert the expression vt ·M ·v into a sum and/or difference of squares. This is one interpretation of a powerful result known as the spectral theorem. Thanks to the equality of mixed partials, the Hessian matrix H(a) is symmetric, so the spectral theorem applies. In fact, the spectral theorem is true for n by n symmetric matrices for all n. Using similar ideas, this leads to a corresponding classification of nondegenerate critical points for real-valued functions of n variables when n > 2. 4.12 Max/min: Lagrange multipliers We now consider the problem of maximizing or minimizing a function subject to a “constraint.” The constraint is often what gives an optimization problem substance. For instance, trying to find the maximum value of a function like f(x, y, z) = x + 2y + 3z is silly: the function can be made arbitrarily large by going far away from the origin where x, y, and z are all large. But if (x, y, z) is confined to a limited region of space, then there’s something to think about. We illustrate with a specific example. Example 4.27. Let S be the surface described by x2 4 + y2 9 + z2 = 1 in R3. It is an ellipsoid, as sketched in Figure 4.11. If f(x, y, z) = x + 2y + 3z, at what points of S does f attain its maximum and minimum values, and what are those values? Figure 4.11: The ellipsoid x2 4 + y2 9 + z2 = 1 Here, we want to maximize and minimize f = x + 2y + 3z subject to the constraint x2 4 + y2 9 + z2 = 1. One approach is to convert this to a two-variable problem, for instance, writing z = ± q 1 −x2 4 −y2 9 on S, so f = x + 2y + 3 q 1 −x2 4 −y2 9 or f = x + 2y −3 q 1 −x2 4 −y2 9 . These are functions of two variables, so we could find the critical points, i.e., set the partial derivatives with respect to x and y equal to 0, and proceed as in the previous two sections. Instead, we try a new approach based on organizing the values of f on S according to the level sets of f. Each level set f = x + 2y + 3z = c is a plane that intersects S typically in a curve, if at all. As c varies, we get a family of level curves on S with f constant on each. See Figure 4.12. Let’s say we want to maximize f on S. Then, starting at some point, we move on S in the direction of increasing c. We continue doing this until we reach a point a where we can’t go any farther. Let cmax denote the value of f at this point a. Observe that at a: | Multivariable_Calculus_Shimamoto_Page_117_Chunk3333 |
106 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION Figure 4.12: Intersecting S with a level set f = c (left) and a family of intersections (right) • The level set f = cmax must be tangent to S. Otherwise we would be able to cross the level curve at a and make the value of f larger while remaining on S. • ∇f(a) is a normal vector to the level set f = cmax. This is because gradients are always normal to level sets (Proposition 4.17). • Similarly, the surface S is itself a level set, namely, the set where: g(x, y, z) = x2 4 + y2 9 + z2 = 1. Hence ∇g(a) is a normal vector to S. The situation is shown in Figure 4.13. Figure 4.13: The level set f = cmax and the ellipsoid S are tangent at a. Since the level set f = cmax and the surface S are tangent at a, their normal vectors are scalar multiples of each other. In other words, there is a scalar λ such that ∇f(a) = λ∇g(a). This is the idea behind the following general principle. Proposition 4.28 (Method of Lagrange multipliers). Let U be an open set in Rn, and suppose that you want to maximize or minimize a smooth function f : U →R subject to the constraint g(x) = c for some smooth function g: U →R and some constant c. If f has a local maximum or local minimum at a point a and if ∇g(a) ̸= 0, then there is a scalar λ such that: ∇f(a) = λ∇g(a). The scalar λ is called a Lagrange multiplier. We return to Example 4.27 and find the maximum and minimum of f = x + 2y + 3z subject to the constraint g = x2 4 + y2 9 +z2 = 1. According to the method of Lagrange multipliers, the solutions | Multivariable_Calculus_Shimamoto_Page_118_Chunk3334 |
4.12. MAX/MIN: LAGRANGE MULTIPLIERS 107 occur at points where ∇f = λ∇g, that is, where (1, 2, 3) = λ(1 2x, 2 9y, 2z). This gives a system of equations: 1 = 1 2λx 2 = 2 9λy 3 = 2λz. There are four unknowns and only three equations, which may seem like not enough information, but we must remember that the constraint provides a fourth equation that can be included as part of the system. In this case, the equations in the system imply that λ ̸= 0, so we can solve for each of x, y, and z in terms of λ, then substitute into the constraint: x = 2 λ, y = 9 λ, z = 3 2λ, and: | Multivariable_Calculus_Shimamoto_Page_119_Chunk3335 |
108 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION It’s worth stepping back to review this calculation, as there is more here than meets the eye. A partial derivative can be viewed as a one-variable rate of change. For instance, in the previous example, ∂f ∂x is the rate at which the output changes per dollar spent on labor. This is called the marginal productivity with respect to labor. There are similar interpretations of ∂f ∂y and ∂f ∂z with respect to equipment and research, respectively. The Lagrange multiplier condition ∇f = λ∇g with g = x + y + z says that (∂f ∂x, ∂f ∂y , ∂f ∂z ) = λ(1, 1, 1) = (λ, λ, λ), or: ∂f ∂x = ∂f ∂y = ∂f ∂z = λ. (4.29) In other words, the marginal productivities with respect to labor, equipment, and research are equal. This part of the analysis applies to any production function f, not just the Cobb-Douglas model of the example. Thus, at the optimal level of production, an extra dollar spent on labor, equipment, or research all increase production by the same amount. This makes sense since, if there were an advantage to increasing the amount spent on one of them, then the manufacturer could increase production by spending more on that one and less on the others. At the optimal level, no such advantage can exist. Moreover, according to (4.29), the common value of these marginal productivities is the La- grange multiplier λ. In the example, by (4.25): λ = 10y1/3z1/2 x5/6 = 10 | Multivariable_Calculus_Shimamoto_Page_120_Chunk3336 |
4.13. EXERCISES FOR CHAPTER 4 109 (a) Find ∂f ∂x(a), ∂f ∂y (a), and ∂f ∂z (a). (b) Find Df(a) and ∇f(a). (c) Find the first-order approximation ℓ(x, y, z) of f(x, y, z) at a. (You may assume that f is differentiable at a.) (d) Compare the values of f(1.05, −1.1, 0.95) and ℓ(1.05, −1.1, 0.95). In Exercises 1.3–1.8, find (a) the partial derivatives ∂f ∂x and ∂f ∂y and (b) the matrix Df(x, y). 1.3. f(x, y) = x3 −2x2y + 3xy2 −4y3 1.4. f(x, y) = sin x sin 2y 1.5. f(x, y) = xexy 1.6. f(x, y) = x2−y2 x2+y2 1.7. f(x, y) = 1 x2+y2 1.8. f(x, y) = xy In Exercises 1.9–1.13, find (a) the partial derivatives ∂f ∂x, ∂f ∂y , and ∂f ∂z and (b) the gradient ∇f(x, y, z). 1.9. f(x, y, z) = x + 2y + 3z + 4 1.10. f(x, y, z) = xy + xz + yz −xyz 1.11. f(x, y, z) = e−x2−y2−z2 sin(x + 2y) 1.12. f(x, y, z) = x+y y+z 1.13. f(x, y, z) = ln(x2 + y2) 1.14. Let: f(x, y) = xyxyxy sin(xy) + x3y + x4 y2 + xy + 1 arctan ln x x5 + y6 . Evaluate ∂f ∂y (1, π). (Hint: Do not try to find a general formula for ∂f ∂y . Ever.) 1.15. Let f(x, y) = x + 2y. Use the definition of differentiability to prove that f is differentiable at every point a = (c, d) of R2. 1.16. Use the definition of differentiability to show that the function f(x, y) = x2 + y2 is differen- tiable at every point a = (c, d) of R2. 1.17. Let f : R2 →R be the function: f(x, y) = p x4 + y4. (a) Find formulas for the partial derivatives ∂f ∂x and ∂f ∂y at all points (x, y) other than the origin. (b) Find the values of the partial derivatives ∂f ∂x(0, 0) and ∂f ∂y (0, 0). (Note that the formulas you found in part (a) probably do not apply at (0, 0).) | Multivariable_Calculus_Shimamoto_Page_121_Chunk3337 |
110 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION (c) Use the definition of differentiability to determine whether f is differentiable at (0, 0). (Hint: Polar coordinates might be useful.) 1.18. Consider the function f : R2 →R defined by: f(x, y) = ( x3+2y3 x2+y2 if (x, y) ̸= (0, 0), 0 if (x, y) = (0, 0). (a) Find the values of the partial derivatives ∂f ∂x(0, 0) and ∂f ∂y (0, 0). (This shouldn’t require much calculation.) (b) Use the definition of differentiability to determine whether f is differentiable at (0, 0). 1.19. Let f and g be differentiable real-valued functions defined on an open set U in Rn. Prove that: ∇(fg) = f ∇g + g ∇f. (Hint: Partial derivatives are one-variable derivatives.) 1.20. Let f, g: U →R be real-valued functions defined on an open set U in Rn, and let a be a point of U. Use the definition of differentiability to prove the following statements. (a) If f and g are differentiable at a, so is f + g. (It’s reasonably clear that D(f + g)(a) = Df(a) + Dg(a), so the problem is to show that the limit in the definition of differentia- bility for f + g is 0.) (b) If f is differentiable at a, so is cf for any scalar c. Section 2 Conditions for differentiability 2.1. Let f : R2 →R be the function defined by: f(x, y) = ( x2 −y2 if (x, y) ̸= (0, 0), π if (x, y) = (0, 0). Is f differentiable at (0, 0)? If so, find Df(0, 0). 2.2. Is the function f(x, y) = x2/3 + y2/3 differentiable at (0, 0)? If so, find Df(0, 0). 2.3. This exercise gives the proof of Proposition 4.4, that differentiability implies continuity. Let U be an open set in Rn, f : U →R a real-valued function, and a a point of U. Define a function Q: U →R by: Q(x) = ( f(x)−f(a)−∇f(a)·(x−a) ∥x−a∥ if x ̸= a, 0 if x = a. Note that the quotient is the one that appears in the definition of differentiability using the gradient form of the derivative. (a) If f is differentiable at a, show that Q is continuous at a. (b) If f is differentiable at a, show that there exists a δ > 0 such that, if x ∈B(a, δ), then |Q(x)| < 1. | Multivariable_Calculus_Shimamoto_Page_122_Chunk3338 |
4.13. EXERCISES FOR CHAPTER 4 111 (c) If f is differentiable at a, prove that f is continuous at a. (Hint: Use the triangle inequality and the Cauchy-Schwarz inequality to show that: |f(x) −f(a)| ≤|Q(x)|∥x −a∥+ ∥∇f(a)∥∥x −a∥.) Section 3 The mean value theorem 3.1. Let B = B(a, r) be an open ball in R2 centered at the point a = (c, d), and let f : B →R be a differentiable function such that: Df(x) = 0 0 for all x in B. Prove that f(x) = f(a) for all x in B. In other words, if the derivative of f is zero everywhere, then f is a constant function. Section 4 The C1 test 4.1. We showed in Example 4.7 that the function f(x, y) = p x2 + y2 is not differentiable at (0, 0). Is f differentiable at points other than the origin? If so, which points? In Exercises 4.2–4.6, use the various criteria for differentiability discussed so far to determine the points at which the function is differentiable and the points at which it is not. Your reasons may be brief as long as they are clear and precise. It should not be necessary to use the definition of differentiability. 4.2. f : R2 →R, f(x, y) = sin(x + y) 4.3. f : R2 →R, f(x, y) = ( x2−y2 x2+y2 if (x, y) ̸= (0, 0), 1 if (x, y) = (0, 0) 4.4. f : R2 →R, f(x, y) = |x| + |y| 4.5. f : R3 →R, f(x, y, z) = x4 + 2x2yz −y3z3 4.6. f : R4 →R, f(w, x, y, z) = det w x y z 4.7. Let f(x, y) = 3p x3 + 8y3. (a) Find formulas for the partial derivatives ∂f ∂x and ∂f ∂y at all points (x, y) where x3+8y3 ̸= 0. (b) Do ∂f ∂x(0, 0) and ∂f ∂y (0, 0) exist? If so, what are their values? If not, why not? (c) Is f differentiable at (0, 0)? (d) More generally, at which points of R2 is f differentiable? (Hint: If a = (c, d) is a point other than the origin where c3 + 8d3 = 0, note that 3√ x3 + 8d3 = 3√ x3 −c3 = 3p (x −c)(x2 + cx + c2) = 3√x −c · 3√ x2 + cx + c2 where x2 + cx + c2 ̸= 0 when x = c.) Section 5 The Little Chain Rule 5.1. Let f(x, y) = x2 + y2, and let α(t) = (t2, t3). Calculate (f ◦α)′(1) in two different ways: | Multivariable_Calculus_Shimamoto_Page_123_Chunk3339 |
112 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION (a) by using the Little Chain Rule, (b) by substituting for x and y in terms of t in the formula for f to obtain (f ◦α)(t) directly and differentiating the result. 5.2. Let f(x, y, z) = xyz, and let α(t) = (cos t, sin t, t). Calculate (f ◦α)′(π 6 ) in two different ways: (a) by using the Little Chain Rule, (b) by substituting for x, y, and z in terms of t in the formula for f to obtain (f ◦α)(t) directly and differentiating the result. 5.3. Let f(x, y, z) be a differentiable real-valued function of three variables, and let α(t) = (x(t), y(t), z(t)) be a differentiable path in R3. If w = f(α(t)), use the Little Chain Rule to find a formula for dw dt in terms of the partial derivatives of f and the derivatives with respect to t of x, y, and z. 5.4. The Little Chain Rule often lurks in the background of a type of first-year calculus problem known as “related rates.” For instance, as an (admittedly artificial) example, suppose that the length ℓand width w of a rectangular region in the plane are changing as functions of time t, so that the area A = ℓw also changes. (a) Find a formula for dA dt in terms of ℓ, w, and their derivatives by differentiating the formula A = ℓw directly with respect to t. (b) Obtain the same result using the Little Chain Rule. (Hint: Let α(t) = (ℓ(t), w(t)).) (c) Suppose that, at the instant that the length of the region is 100 inches and its width is 40 inches, the length is increasing at a rate of 2 inches/second and the width is increasing at a rate of 3 inches/second. How fast is the area changing? 5.5. Let a be a point of Rn, and let B = B(a, r) be an open ball centered at a. Let f : B →R be a differentiable function. If b is any point of B, show that there exists a point c on the line segment connecting a and b such that: f(b) −f(a) = ∇f(c) · (b −a). Note that this is a generalization of the mean value theorem to real-valued functions of more than one variable. (Hint: Explain why the line segment can be parametrized by α(t) = a + t(b −a), 0 ≤t ≤1. Then, note that the composition f ◦α is a real-valued function of one variable, so the one-variable mean value theorem applies.) 5.6. In this exercise, we prove the Little Chain Rule (Theorem 4.13). Let U be an open set in Rn, α: I →U a path in U defined on an open interval I, and f : U →R a real-valued function. Let t0 be a point of I. Assume that α is differentiable at t0 and f is differentiable at α(t0). Let a = α(t0), and let Q: U →R be the function defined in Exercise 2.3: Q(x) = ( f(x)−f(a)−∇f(a)·(x−a) ∥x−a∥ if x ̸= a, 0 if x = a. (a) Prove that limt→t0 f(α(t))−f(α(t0))−∇f(α(t0))·(α(t)−α(t0)) t−t0 = 0. (Hint: f(x)−f(a)−∇f(a)· (x −a) = ∥x −a∥Q(x). Or see Exercise 8.2 in Chapter 3.) (b) Prove the Little Chain Rule: (f ◦α)′(t0) = ∇f(α(t0)) · α′(t0). | Multivariable_Calculus_Shimamoto_Page_124_Chunk3340 |
4.13. EXERCISES FOR CHAPTER 4 113 5.7. Let f : R3 →R be a differentiable function with the property that ∇f(x) points in the same direction as x for all nonzero x in R3. If a > 0, prove that f is constant on the sphere x2 +y2 +z2 = a2. (Hint: If p and q are any two points on the sphere, there is a differentiable path α on the sphere from p to q.) Section 6 Directional derivatives In Exercises 6.1–6.4, find the directional derivative of f at the point a in the direction of the vector v. 6.1. f(x, y) = e−2xy3, a = (0, 1), v = (4, 1) 6.2. f(x, y) = x2 + y2, a = (1, 2), v = (−2, 1) 6.3. f(x, y, z) = x3 −2x2yz + xz −3, a = (1, 0, −1), v = (1, −1, 2) 6.4. f(x, y, z) = sin x sin y cos z, a = (0, π 2 , π), v = (π, 2π, 2π) 6.5. For a certain differentiable real-valued function f : R2 →R, the maximum directional deriva- tive at the point a = (0, 0) has a value of 6 and occurs in the direction from a to the point b = (4, 1). Find ∇f(a). 6.6. Let f(x, y) = x2 −y2. (a) At the point a = ( √ 2, 1), find a unit vector u that points in the direction in which f is increasing most rapidly. What is the rate of increase in this direction? (b) Describe the set of all points a = (x, y) in R2 such that f increases most rapidly at a in the direction that points directly towards the origin. 6.7. A certain function has the form: f(x, y) = rx2 + sy2, where r and s are constant. Find values of r and s so that the maximum directional derivative of f at the point a = (1, 2) has a value of 10 and occurs in the direction from a to the point b = (2, 3). 6.8. Let f(x, y) = xy, and let a = (5, −5). (a) Find all directions in which the directional derivative of f at a is equal to 1. (b) Is there a direction in which the directional derivative at a is equal to 10? 6.9. The temperature in a certain region of space, in degrees Celsius, is modeled by the function T(x, y, z) = 20e−x2−2y2−4z2, where x, y, z are measured in meters. At the point a = (2, −1, 3): (a) In which direction is the temperature increasing most rapidly? (b) In which direction is it decreasing most rapidly? (c) If you travel in the direction described in part (a) at a speed of 10 meters/second, how fast is the observed temperature changing at a in degrees Celsius per second? 6.10. Let c = (c1, c2, . . . , cn) be a vector in Rn, and define f : Rn →R by f(x) = c · x. (a) Find the derivative Df(x). | Multivariable_Calculus_Shimamoto_Page_125_Chunk3341 |
114 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION (b) If u is a unit vector in Rn, find a formula for the directional derivative (Duf)(x). 6.11. You discover that your happiness is a function of your location in the plane. At the point (x, y), your happiness is given by the formula: H(x, y) = x3e2y happs, where “happ” is a unit of happiness. For instance, at the point (1, 0), you are H(1, 0) = 1 happ happy. (a) At the point (1, 0), in which direction is your happiness increasing most rapidly? What is the value of the directional derivative in this direction? (b) Still at the point (1, 0), in which direction is your happiness decreasing most rapidly? What is the value of the directional derivative in this direction? (c) Suppose that, starting at (1, 0), you follow a curve so that you are always traveling in the direction in which your happiness increases most rapidly. Find an equation for the curve, and sketch the curve. An equation describing the relationship between x and y along the curve is fine. No further parametrization is necessary. (Hint: What can you say about the slope at each point of the curve?) (d) Suppose instead that, starting at (1, 0), you travel along a curve such that the directional derivative in the tangent direction is always equal to zero. Find an equation describing this curve, and sketch the curve. 6.12. Does there exist a differentiable function f : U →R defined on an open subset U of Rn with the property that, for some point a in U, the directional derivatives at a satisfy (Duf)(a) > 0 for all unit vectors u in Rn, i.e., the directional derivative at a is positive in every direction? Either find such a function and point a, or explain why none exists. Section 7 ∇f as normal vector 7.1. Find an equation of the tangent plane to the surface z = x2 −y2 at the point a = (1, 2, −3). 7.2. Find an equation of the tangent plane to the surface x2+y2−z2 = 1 at the point a = (1, −1, 1). 7.3. The surface in R3 given by (9x2 + y2 + z2 −1)3 −y2z3 −2 5x2z3 = 0 was featured in an article in the February 14, 2019 issue of The New York Times.5 See Figure 4.14. Nowhere in the article was the tangent plane at the point (0, 1, 1) mentioned. Scoop the Times by finding an equation of the tangent plane. 7.4. Let S be the ellipsoid x2 + y2 2 + z2 4 = 4. Let a = (1, 2, 2), and let n be the unit normal vector to S at a that points outward from S. If f(x, y, z) = xyz, find the directional derivative (Dnf)(a). 7.5. The saddle surface y2 −x2 −z = 0 and the sphere x2 + y2 + z2 = 14 intersect in a curve C that passes through the point a = (1, 2, 3). Find a parametrization of the line tangent to C at a. 5https://www.nytimes.com/2019/02/14/science/math-algorithm-valentine.html. Actually, the article con- cerns a family of such surfaces, parametrized by two real numbers. The surface in the exercise is a specific example. | Multivariable_Calculus_Shimamoto_Page_126_Chunk3342 |
4.13. EXERCISES FOR CHAPTER 4 115 Figure 4.14: The surface (9x2 + y2 + z2 −1)3 −y2z3 −2 5x2z3 = 0 7.6. Find all values of c such that, at every point of intersection of the spheres (x −c)2 + y2 + z2 = 3 and x2 + (y −1)2 + z2 = 1, the respective tangent planes are perpendicular to one another. Section 8 Higher-order partial derivatives In Exercises 8.1–8.4, find all four second-order partial derivatives of the given function f. 8.1. f(x, y) = x4 −2x3y + 3x2y2 −4xy3 + 5y4 8.2. f(x, y) = sin x cos y 8.3. f(x, y) = e−x2−y2 8.4. f(x, y) = y x+y 8.5. Let f(x, y) = x4y3. Find i and j so that the (i + j)th-order partial derivative ∂i+jf ∂xi ∂yj (0, 0) is nonzero. What is the value of this partial derivative? 8.6. Consider the function f : R2 →R defined by: f(x, y) = ( x3y x2+y2 if (x, y) ̸= (0, 0), 0 if (x, y) = (0, 0). (a) Find formulas for the partial derivatives ∂f ∂x(x, y) and ∂f ∂y (x, y) if (x, y) ̸= (0, 0). (b) Find the values of ∂f ∂x(0, 0) and ∂f ∂y (0, 0). (c) Use your answers to parts (a) and (b) to evaluate the second-order partial derivatives: ∂2f ∂y ∂x(0, 0) = ∂ ∂y | Multivariable_Calculus_Shimamoto_Page_127_Chunk3343 |
116 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION Figure 4.15: The rectangle with vertices a, u, v, w 8.7. This exercise gives the details of a full proof of the equality of mixed partials (Theorem 4.21). Let a = (c, d) be a point of R2, and let f : B →R be a real-valued function defined on an open ball B = B(a, r) centered at a. Assume that the first and second-order partial derivatives of f are continuous on B. Let R be the rectangle whose vertices are a = (c, d), u = (c + h, d), v = (c + h, d + k), and w = (c, d + k), where h and k are nonzero real numbers small enough that R ⊂B. See Figure 4.15. Let: ∆= f(v) −f(u) −f(w) + f(a) = f(c + h, d + k) −f(c + h, d) −f(c, d + k) + f(c, d). (a) Show that ∆= | Multivariable_Calculus_Shimamoto_Page_128_Chunk3344 |
4.13. EXERCISES FOR CHAPTER 4 117 10.8. Let f(x, y) = 3x4 −4x2y + y2. (a) Make a sketch in R2 indicating the set of points (x, y) where f(x, y) > 0 and the set where f(x, y) < 0. (Hint: Factor f.) (b) Explain why f does not have a local maximum or minimum at (0, 0). (c) On the other hand, show that f does have a local minimum at (0, 0) when restricted to any line y = mx. 10.9. (a) Given a point a in Rn, define f : Rn →R by f(x) = ∥x −a∥2. Show that ∇f(x) = 2(x −a). (b) Let p1, p2, . . . , pm be m distinct points in Rn, and let F(x) = Pm k=1 ∥x−pk∥2. Assuming that F has a global minimum, show that it must occur when: x = 1 m m X k=1 pk (the “centroid”). 10.10. Let (x1, y1), (x2, y2), . . . , (xn, yn) be a collection of n distinct points in R2, where n ≥2. We think of the points as “data.” The line y = mx + b that best fits the data in the sense of least squares is defined to be the one that minimizes the quantity: E = n X i=1 (mxi + b −yi)2. Note that using a line y = mx + b to predict the data value y = yi when x = xi gives an error of magnitude |mxi + b −yi|. Consequently E is called the total squared error. It depends on the line, which we identify here by its slope m and y-intercept b. Thus E is a function of the two variables m and b, and we can apply the methods we have been studying to find where it assumes its minimum. Calculate the partial derivatives ∂E ∂m and ∂E ∂b , and show that the critical points (m, b) of E are the solutions of the system of equations: | Multivariable_Calculus_Shimamoto_Page_129_Chunk3345 |
118 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION (b) Let (x1, y1), (x2, y2), . . . , (xn, yn) be given data points. Show that the Hessian of the total squared error E is given by: H(m, b) = 2 | Multivariable_Calculus_Shimamoto_Page_130_Chunk3346 |
4.13. EXERCISES FOR CHAPTER 4 119 (b) Show that the quadratic term Ah2 + 2Bhk + Ck2 = 2Bhk + Ck2 in the second-order approximation can be written as a difference of two perfect squares. (Hence, based on our prototypical models, we predict a to be a saddle point. Hint: In the case that C = 0, too, consider the expansions of (a ± b)2.) Section 12 Max/min: Lagrange multipliers 12.1. Consider the problem of finding the maximum and minimum values of the function f(x, y) = xy subject to the constraint x + 2y = 1. (a) Explain why a minimum value does not exist. (b) On the other hand, you may assume that a maximum does exist. Use Lagrange multi- pliers to find the maximum value and the point at which it is attained. 12.2. (a) Find the points on the curve x2 −xy + y2 = 4 at which the function f(x, y) = x2 + y2 attains its maximum and minimum values. You may assume that these maximum and minimum values exist. (b) Use your answer to part (a) to help sketch the curve x2 −xy + y2 = 4. (Hint: Note that f(x, y) = x2 + y2 represents the square of the distance from the origin.) 12.3. At what points on the unit sphere x2 +y2 +z2 = 1 does the function f(x, y, z) = 2x−3y +4z attain its maximum and minimum values? 12.4. Find the minimum value of the function f(x, y, z) = x2 + y2 + z2 subject to the constraint 2x −3y + 4z = 1. At what point is the minimum attained? 12.5. Find the positive values of x, y, and z for which the function f(x, y, z) = xy2z3 attains its maximum value subject to the constraint 3x+4y+5z = 12. You may assume that a maximum exists. 12.6. (a) Use the method of Lagrange multipliers to find the positive values of x, y, and z such that xyz = 1 and x + y + z is as small as possible. What is the minimum value? You may assume that a minimum exists. (b) Use part (a) to show that, for all positive real numbers a, b, c: a + b + c 3 ≥ 3√ abc. This is called the inequality of the arithmetic and geometric means. (Hint: Let k = 3√ abc, and consider a/k, b/k, and c/k.) 12.7. Let A be a 3 by 3 symmetric matrix, A = a b c b d e c e f , and consider the quadratic polynomial: Q(x) = xtAx = x y z a b c b d e c e f x y z , where x = x y z . Use Lagrange multipliers to show that, when Q is restricted to the unit sphere x2+y2+z2 = 1, any point x at which it attains its maximum or minimum value satisfies | Multivariable_Calculus_Shimamoto_Page_131_Chunk3347 |
120 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION Ax = λx for some scalar λ and that the maximum or minimum value is the corresponding value of λ. (In the language of linear algebra, x is called an eigenvector of A and λ is called the corresponding eigenvalue.) In fact, maximum and minimum values do exist, so the conclusion is not an empty one. 12.8. You discover that your happiness is a function of your daily routine. After a great deal of soul-searching, you decide to focus on three activities to which you will devote your entire day: if x, y, and z are the number of hours a day that you spend eating, sleeping, and studying multivariable calculus, respectively, then, based on data provided by the federal government, your happiness is given by the function: h(x, y, z) = 1000 −23x2 −23y2 −46y −z2 happs.6 There is a mix of eating, sleeping, and studying that leads to the greatest possible daily happiness (you may assume this). What is it? Answer in two ways, as follows. (a) Substitute for z in terms of x and y in the formula for h, and find the critical points of the resulting function of x and y. (b) Use Lagrange multipliers. 12.9. After conducting extensive market research, a manufacturer of monkey saddles discovers that, if it produces a saddle consisting of x ounces of leather, y ounces of copper, and z ounces of premium bananas, the monkey rider experiences a total of: S = 2xy + 3xz + 4yz units of satisfaction. Leather costs $1 per ounce, copper $2 per ounce, and premium bananas $3 per ounce, and the manufacturer is willing to spend at most $1000 per saddle. What combination of ingredients yields the most satisfied monkey? You may assume that a maximum exists. 6See Exercise 6.11, page 114. | Multivariable_Calculus_Shimamoto_Page_132_Chunk3348 |
Chapter 5 Real-valued functions: integration 5.1 Volume and iterated integrals We now study the integral of real-valued functions of more than one variable. Most of our time is devoted to functions of two variables. In this section, we introduce the integral informally by thinking about how to visualize it. This uses an idea likely to be familiar from first-year calculus. It won’t be until the next section that we define what the integral actually is. Let D be a subset of R2. At this point, we won’t be too careful about putting any conditions on D, but it’s best to think of it as some sort of two-dimensional blob, as opposed to a set of discrete points or a curve. Let f : A →R be a real-valued function whose domain A is a subset of R2 that contains D. For the moment, assume that f(x) ≥0 for all x in D. Let W be the region in R3 that lies below the graph z = f(x, y) and above D, as shown in Figure 5.1. That is: W = {(x, y, z) ∈R3 : (x, y) ∈D and 0 ≤z ≤f(x, y)}. Figure 5.1: The region W under the graph z = f(x, y) and above D Consider the volume of W, which we denote variously by RR D f(x, y) dA, RR D f(x, y) dx dy, or just RR D f dA. To calculate it, we apply the following principle from first-year calculus: Volume = Integral of cross-sectional area. For instance, in first-year calculus, the cross-sections may turn out to be disks, washers, or, in a slight variant, cylinders. Example 5.1. Find the volume of the region under the plane z = 4 −x + 2y and above the square 1 ≤x ≤3, 2 ≤y ≤4, in the xy-plane. See Figure 5.2. 121 | Multivariable_Calculus_Shimamoto_Page_133_Chunk3349 |
122 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION Figure 5.2: The region under z = 4 −x + 2y and above the square 1 ≤x ≤3, 2 ≤y ≤4 Here, f(x, y) = 4 −x + 2y, and, if D denotes the specified square, we are looking for RR D(4 − x + 2y) dA. As above, let W denote the region in question, below the plane and above the square. We consider first cross-sections perpendicular to the x-axis, that is, intersections of W with planes of the form x = constant. There is such a cross-section for each value of x from x = 1 to x = 3, so: Volume = Z 3 1 A(x) dx, (5.1) where A(x) is the area of the cross-section at x. A typical cross-section is shown in Figure 5.3. The Figure 5.3: A cross-section perpendicular to the x-axis base is a line segment from y = 2 to y = 4, and the top is a curve lying in the graph z = 4−x+2y. (In fact, the cross-sections in this example are trapezoids, but let’s not worry about that.) In other words, A(x) is an area under a curve, so it too is an integral: A(x) = R 4 2 (4 −x + 2y) dy, where, for the given cross-section, x is fixed. Integrating the cross-sectional area as in equation (5.1) and evaluating gives: Volume = Z 3 1 Z 4 2 | Multivariable_Calculus_Shimamoto_Page_134_Chunk3350 |
5.1. VOLUME AND ITERATED INTEGRALS 123 Alternatively, we could have used cross-sections perpendicular to the y-axis, in which case the cross-sections occur from y = 2 to y = 4 (Figure 5.4): Volume = Z 4 2 A(y) dy. (5.2) Figure 5.4: A cross-section perpendicular to the y-axis Each cross-section lies below the graph and above a line segment that goes from x = 1 to x = 3, so A(y) = R 3 1 (4 −x + 2y) dx. This time, the integral of the cross-sectional area (5.2) gives: Volume = Z 4 2 Z 3 1 | Multivariable_Calculus_Shimamoto_Page_135_Chunk3351 |
124 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION Figure 5.5: The cylinders x2 + z2 = 1 (left), y2 + z2 = 1 (middle), and the region contained in both (right) Figure 5.6: The portion in the first octant (left) and the portion under only x2 + z2 = 1 (right) The volume of the original solid is 8 times the volume of the portion over D1. This lies below x2 + z2 = 1, i.e., below the graph z = √ 1 −x2. Hence: Volume = 8 ZZ D1 p 1 −x2 dA. To calculate the volume over D1, let’s try cross-sections perpendicular to the x-axis. These exist from x = 0 to x = 1: Volume = 8 Z 1 0 A(x) dx. The base of each cross-section is a line segment perpendicular to the x-axis within D1. The lower endpoint is always at y = 0, but the upper endpoint depends on x. Indeed, the upper endpoint lies on the line y = x. See Figure 5.7. Figure 5.7: The base of the cross-section at x | Multivariable_Calculus_Shimamoto_Page_136_Chunk3352 |
5.1. VOLUME AND ITERATED INTEGRALS 125 Hence A(x) = R x 0 √ 1 −x2 dy, so: Volume = 8 Z 1 0 Z x 0 p 1 −x2 dy dx = 8 Z 1 0 p 1 −x2 · y y=x y=0 dx = 8 Z 1 0 x p 1 −x2 dx (let u = 1 −x2, du = −2x dx) = 8 −1 2 · 2 3(1 −x2)3/2 1 0 = −8 3(03/2 −13/2) = 8 3. Alternatively, suppose we had used cross-sections perpendicular to the y-axis. The cross-sections exist from y = 0 to y = 1, and, for each such y, the base of the cross-section is a line segment in the x-direction that goes from x = y to x = 1, as illustrated in Figure 5.8. Figure 5.8: The base of the cross-section at y Consequently: Volume = 8 Z 1 0 A(y) dy = 8 Z 1 0 Z 1 y p 1 −x2 dx dy. While not impossible, it is a little less obvious how to do the first partial antidifferentiation with respect to x by hand. In other words, our original order of antidifferentiation might be preferable. Example 5.3. Consider the iterated integral Z 2 0 Z 4 x2 x3ey3 dy dx. (a) Sketch the domain of integration D in the xy-plane. (b) Evaluate the integral. (a) The domain of integration can be reconstructed from the endpoints of the integrals. The outermost integral says that x goes from x = 0 to x = 2. Geometrically, we are integrating areas of cross-sections perpendicular to the x-axis. Then, the inner integral says that, for each x, y goes from y = x2 to y = 4. Figure 5.9 shows the relevant input. Hence D is described by the conditions: D = {(x, y) ∈R2 : 0 ≤x ≤2, x2 ≤y ≤4}. This is the region in the first quadrant bounded by the parabola y = x2, the line y = 4, and the y-axis. | Multivariable_Calculus_Shimamoto_Page_137_Chunk3353 |
126 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION Figure 5.9: The domain of integration: cross-sections perpendicular to the x-axis (b) To evaluate the integral as presented, we would antidifferentiate first with respect to y, treating x as constant: Z 2 0 Z 4 x2 x3ey3 dy dx = Z 2 0 x3 Z 4 x2 ey3 dy dx. The innermost antiderivative looks hard. So, having nothing better to do, we try switching the order of antidifferentiation. Changing to cross-sections perpendicular to the y-axis, we see from the description of D that y goes from y = 0 to y = 4, and, for each y, x goes from x = 0 to x = √y. The thinking behind the switched order is illustrated in Figure 5.10. Figure 5.10: Reversing the order of antidifferentiation Therefore: Z 2 0 Z 4 x2 x3ey3 dy dx = Z 4 0 Z √y 0 x3ey3 dx dy = Z 4 0 1 4x4ey3 x=√y x=0 dy = Z 4 0 | Multivariable_Calculus_Shimamoto_Page_138_Chunk3354 |
5.2. THE DOUBLE INTEGRAL 127 • D is a bounded subset of R2: this means that there is a rectangle R in R2 such that D ⊂R. • f is a bounded function: this means that there is a scalar M such that |f(x)| ≤M for all x in D. To define the integral, we take as a guiding principle that it is a limit of weighted sums. This is often what gives the integral its power. We begin by reviewing the one-variable case. Let f : [a, b] →R be defined on a closed interval [a, b]. We subdivide the interval into n subintervals of widths △x1, △x2, . . . , △xn and choose a point pi in the ith subinterval for i = 1, 2, . . . , n. See Figure 5.11. We then form the sum P i f(pi) △xi. This is called a Riemann sum, and we refer to the pi as sample points. We think of the Figure 5.11: Subdividing [a, b] into n subintervals of widths △x1, △x2, . . . , △xn with sample points p1, p2, . . . , pn Riemann sum as the sum of the values f(pi) at the sample points weighted by the width of the subintervals (though it may be equally appropriate on occasion to think of it as the widths of the subintervals weighted by the function values). The integral is defined as a limit of the Riemann sums: Z b a f(x) dx = lim △xi→0 X i f(pi) △xi. The limit is a different type from the ones we encountered before. Here, it means that, given any ϵ > 0, there exists a δ > 0 such that, for all Riemann sums P i f(pi) △xi based on subdivisions of [a, b] with △xi < δ for all i, we have | P i f(pi) △xi− R b a f(x) dx| < ϵ. This is extremely cumbersome to work with. Moving on to functions of two variables f : D →R, where D ⊂R2, we do our best to mimic the one-variable case and proceed in two steps. Step 1. Integrals over rectangles. We first introduce some notation that makes it easier to specify the kind of rectangles we have in mind. Definition. If X and Y are sets, then the Cartesian product X × Y is defined to be the set of all ordered pairs: X × Y = {(x, y) : x ∈X, y ∈Y }. For instance, the product of two closed intervals R = [a, b] × [c, d] = {(x, y) : x ∈[a, b], y ∈ [c, d]} = {(x, y) : a ≤x ≤b, c ≤y ≤d} is a rectangle in R2 whose sides are parallel to the coordinate axes (Figure 5.12). We define how to integrate a function f(x, y) over such a rectangle R. We chop up R into a grid of subrectangles with sides parallel to the axes and of dimensions △xi by △yj. A simple example is shown in Figure 5.13. We choose a sample point pij in each subrectangle and consider the Riemann sum P i,j f(pij) △xi △yj. This is a weighted sum in which the function values at the sample points are weighted by the areas of the subrectangles. Now, let △xi and △yj go to 0. | Multivariable_Calculus_Shimamoto_Page_139_Chunk3355 |
128 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION Figure 5.12: The rectangle R = [a, b] × [c, d] in R2 Figure 5.13: A subdivision of a rectangle into 4 subrectangles Definition. Let R = [a, b] × [c, d] be a rectangle in R2. The integral of a bounded function f : R →R is defined by: ZZ R f(x, y) dA = lim △xi→0 △yj→0 X i,j f(pij) △xi △yj, assuming that the limit exists. When it does, we say that f is integrable on R. There are other versions of integrals, too, and the one defined here is known technically as the Riemann integral. Integrals of functions of two variables are called double integrals. We begin with a standard somewhat artificial example that is constructed to make a point. Example 5.4. Let R = [0, 1] × [0, 1], and define f : R →R by: f(x, y) = ( 0 if x and y are both rational, 1 otherwise. For instance, f(1 2, 3 5) = 0, while f(1 2, √ 2 2 ) = 1 and f(π 6 , √ 2 2 ) = 1. The graph z = f(x, y) consists of two 1 by 1 squares, one at height z = 0 and one at height z = 1, both with lots of missing points. To evaluate the integral RR R f(x, y) dA using the definition, consider a subdivision of R into subrectangles. Regardless of the subdivision, each subrectangle contains points where f = 0 as well as points where f = 1. Thus, depending on the choice of sample points, some Riemann sums have the form P i.j 0·△xi △yj = 0, and some have the form P i,j 1·△xi △yj = Area of R = 1. Still others have values in between. This is true no matter what △xi and △yj are, so P i,j f(pij) △xi △yj does not approach a limit as △xi and △yj go to 0. Consequently f is not integrable on R. This raises the question of whether it’s possible to tell when a function is integrable. For the preceding function, in some sense the problem is that it is too discontinuous. | Multivariable_Calculus_Shimamoto_Page_140_Chunk3356 |
5.2. THE DOUBLE INTEGRAL 129 Theorem 5.5. Let f : R →R be a bounded function defined on a rectangle R = [a, b] × [c, d]. If f is continuous on R, except possibly at a finite number of points or on a finite number of smooth curves, then f is integrable on R. This is actually a special case of a more general theorem that characterizes precisely when a function is Riemann integrable. Unfortunately, the proof uses concepts that are best left for a course in real analysis, so we do not go into it further. One consequence, however, is the useful fact that continuous functions are integrable. This is worth remembering. It covers most of the functions that we work with. A second question is that, once a function is known to be integrable, is there a practical way to evaluate the integral? Here again, we only indicate why the answer is plausible without attempting anything like a correct proof. The intuition is that, after you subdivide a rectangle and choose sample points, you can organize the Riemann sum P i,j f(pij) △xi △yj either: • column-by-column: P i | Multivariable_Calculus_Shimamoto_Page_141_Chunk3357 |
130 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION An example of a graph of ef is shown in Figure 5.15. Figure 5.15: The general case: the graph z = ef(x, y) The integral of ef, as a function defined on a rectangle, has been covered in Step 1. Definition. We say that f : D →R is integrable on D if ef is integrable on R. If so, we define RR D f(x, y) dA = RR R ef(x, y) dA. Thus the integral of f over D is defined as a limit of Riemann sums for ef, P i,j ef(pij) △xi △yj, based on subdivisions of the larger rectangle R into subrectangles. Since ef = 0 away from D, however, only those subrectangles that intersect D contribute to the Riemann sum. Hence we might write: ZZ D f(x, y) dA = lim △xi→0 △yj→0 X subrectangles that intersect D f(pij) △xi △yj. The subdivision is indicated in Figure 5.16. Figure 5.16: The general case: subdividing a rectangle containing D into subrectangles. Only those subrectangles that intersect D may contribute to a Riemann sum for ef. As Figure 5.15 indicates, even if f is continuous on D, the function ef might well be discontinuous on the boundary of D. Nevertheless, by Theorem 5.5, ef will still be integrable on R as long as the boundary of D consists of finitely many smooth curves. In other words, the integral can tolerate a certain amount of bad behavior on the boundary. Having this latitude is something we take advantage of later on. Here are some basic properties of the integral. They are quite reasonable and can be proven using the definition of the integral as a limit of sums. Proposition 5.7. Let f and g be integrable functions on a bounded set D in R2. Then: 1. RR D(f + g) dA = RR D f dA + RR D g dA. | Multivariable_Calculus_Shimamoto_Page_142_Chunk3358 |
5.3. INTERPRETATIONS OF THE DOUBLE INTEGRAL 131 2. RR D cf dA = c RR D f dA for any scalar c. 3. If f(x, y) ≤g(x, y) for all (x, y) in D, then RR D f dA ≤ RR D g dA. 5.3 Interpretations of the double integral We have defined the integral of a bounded real-valued function f(x, y) of two variables over a bounded subset D of R2 as: ZZ D f(x, y) dA = lim △xi→0 △yj→0 X subrectangles that intersect D f(pij) △xi △yj, where the subrectangles come from subdividing a rectangle R that contains D. Now, we go about trying to understand why this might be useful. First, the integral is a limit of sums. Therefore, if a typical summand f(pij) △xi △yj represents “something,” then RR D f(x, y) dA represents the total “something.” Also, we have noted before that a Riemann sum is the sum of the sample values f(pij) weighted by the area of the subrectangles. We consider a few situations where such a weighted sum might arise. Example 5.8. The first is when f(x, y) represents some sort of density, or quantity per unit area, at the point (x, y), for instance, mass density or population density. Then f(pij) △xi △yj is approximately the quantity on a typical subrectangle, and: ZZ D (density) dA = total quantity in D, for instance, the total mass or total population. Example 5.9. Suppose that f is the constant function f(x, y) = 1 for all (x, y) in D. Then f(pij) △xi △yj = 1 · △xi △yj is the area of a subrectangle, so: ZZ D 1 dA = Area (D). Example 5.10. Let’s return to the graph of a function z = f(x, y), where f(x, y) ≥0 for all (x, y) in D. Then f(pij) is the height of the graph above the sample point pij, and f(pij) △xi △yj is approximately the volume below the graph and above a typical subrectangle (Figure 5.17). Thus: ZZ D f(x, y) dA = volume below the graph of f and above D. As a result, volume can be regarded either as an iterated integral, as in Section 5.1, or as a double integral, i.e., a limit of Riemann sums. That these two approaches give the same thing is another way of understanding Fubini’s theorem. Example 5.11. Average value of a function. To average a list of numbers, say 2, 4, 7, 4, 4, we find the sum and divide by the number of entries: average = 2 + 4 + 7 + 4 + 4 5 = 21 5 = 4.2. | Multivariable_Calculus_Shimamoto_Page_143_Chunk3359 |
132 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION Figure 5.17: Visualizing a Riemann sum as approximating the volume under a graph This could also be written as: 2 · 1 5 + 4 · 1 5 + 7 · 1 5 + 4 · 1 5 + 4 · 1 5 or 2 · 1 5 + 4 · 3 5 + 7 · 1 5. In other words, it’s a sum in which each value is weighted by its relative frequency. Similarly, to find the average value of a function f(x, y) over a region D, we take a sampling of points pij in D and consider the sum in which each function value f(pij) is weighted by the fraction of D that it represents. If the sampling comes from a subdivision into subrectangles and we think of each sample point as representing its subrectangle, this gives: Average value ≈ X f(pij) △xi △yj Area (D) = 1 Area (D) X f(pij) △xi △yj. Taking the limit as △xi and △yj go to 0 leads to the following definition. Definition. 1. The average value of a bounded function f(x, y) on a bounded subset D of R2, denoted by f, is defined to be: f = 1 Area (D) ZZ D f dA. 2. For example, the average x and y-coordinates on D are: x = 1 Area (D) ZZ D x dA and y = 1 Area (D) ZZ D y dA. The point (x, y) is called the centroid of D. Example 5.12. Let D be the triangular region in R2 with vertices (0, 0), (2, 2), and (−2, 2). (a) Find the average value of f(x, y) = y2 −x2 on D. The graph of f is in Figure 5.18. (b) Find the centroid of D. (a) Here, f = 1 Area (D) RR D(y2 −x2) dA, where the region D is shown in Figure 5.19. We could find the area of D using Area (D) = RR D 1 dA as in Example 5.9, but, since D is a triangle, we can use geometry instead: Area (D) = 1 2(base)(height) = 1 2 · 4 · 2 = 4. | Multivariable_Calculus_Shimamoto_Page_144_Chunk3360 |
5.3. INTERPRETATIONS OF THE DOUBLE INTEGRAL 133 Figure 5.18: The graph of f(x, y) = y2 −x2 over the triangle D Figure 5.19: The triangle D To set up the double integral RR D(y2 −x2) dA as an iterated integral, note that D is bounded on the left by y = −x, on the right by y = x, and on top by y = 2. We use cross-sections perpendicular to the y-axis. These cross-sections exist from y = 0 to y = 2, and, for each y, x goes from x = −y to x = y. Hence: ZZ D (y2 −x2) dA = Z 2 0 Z y −y (y2 −x2) dx dy = Z 2 0 | Multivariable_Calculus_Shimamoto_Page_145_Chunk3361 |
134 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION Figure 5.20: Reversing the order of integration to integrate over D Actually, we could try to take advantage of symmetry properties of D and f to avoid having to evaluate both halves. We take up this idea more fully when we discuss the change of variables theorem in Chapter 7. (b) Because of the symmetry of D in the y-axis, it’s intuitively clear that the average x-coordinate is 0: x = 0. Again, we justify this more carefully later (see Example 7.7 in Chapter 7). For the average y-coordinate, by definition, y = 1 Area (D) RR D y dA = 1 4 RR D y dA. We describe D using the same limits of integration as in part (a) with cross-sections perpendicular to the y-axis: ZZ D y dA = Z 2 0 Z y −y y dx dy = Z 2 0 yx x=y x=−y dy = Z 2 0 (y2 −(−y2)) dy = Z 2 0 2y2 dy = 2 3y3 2 0 = 16 3 . Thus y = 1 4 · 16 3 = 4 3, and the centroid is the point (x, y) = (0, 4 3). 5.4 Parametrizations of surfaces Thus far, we have studied surfaces that arise as: • graphs z = f(x, y) (or, analogously, y = g(x, z) and x = h(y, z)) or • level sets f(x, y, z) = c. We next describe surfaces by how they are traced out. This is like the way we described curves using parametrizations, except that now, since surfaces are two-dimensional, we need two parameters. In other words, surfaces will be described by functions σ: D →Rn, where D ⊂R2, as illustrated in Figure 5.21. We want the domain of the parametrization to be two-dimensional, so we assume that D is an open set U in R2 together with all of its boundary points. The technical term is that D is the closure of U. Actually, the theory carries over if some of the boundary points are missing, | Multivariable_Calculus_Shimamoto_Page_146_Chunk3362 |
5.4. PARAMETRIZATIONS OF SURFACES 135 but, in our examples, D will often be a rectangle or a closed disk in R2, in which case all boundary points are included. The coordinates in D are the parameters. There are often natural choices for what to call them depending on the surface, though generically we call them s and t. For every (s, t) in D, σ(s, t) is a point in Rn, so we may write σ(s, t) = (x1(s, t), x2(s, t), . . . , xn(s, t)). We shall focus almost exclusively on surfaces in R3, in which case σ(s, t) = (x(s, t), y(s, t), z(s, t)). Figure 5.21: A parametrization σ of a surface S in R3 The reason for introducing the topic now is that this is a good time to talk about how to integrate real-valued functions over surfaces. Because surfaces are two-dimensional objects, in principle, these integrals should be extensions of the double integrals we are in the midst of studying. For example, the interpretations of limits of weighted sums that apply to integrals over regions in the plane should then carry over to integrals over surfaces as well. In the remainder of this section, we look at several examples of parametrizations of surfaces and, in particular, some common geometric parameters. Integrals we leave for the next section. Example 5.13. Consider the portion of the cone z = p x2 + y2 in R3 where 0 ≤z ≤2. See Figure 5.22. The surface is already described in terms of x and y, so we use (x, y) as parameters with z = p x2 + y2. That is, let: σ(x, y) = (x, y, p x2 + y2). The domain D of σ is the set of all (x, y) such that 0 ≤ p x2 + y2 ≤2, or x2 + y2 ≤4. This is a disk of radius 2 in the xy-plane. As (x, y) ranges over D, σ(x, y) sweeps out the given portion of the cone. Figure 5.22: A parametrization of a cone using x and y as parameters | Multivariable_Calculus_Shimamoto_Page_147_Chunk3363 |
136 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION In this example, we used the x and y coordinates of the xyz-coordinate system as parameters for the surface. This is always an option for surfaces that are graphs z = f(x, y) of functions of two variables. We next introduce some other coordinate systems that can be used to locate points in the plane or in three-dimensional space. These coordinate systems provide the most convenient way to approach a variety of situations, though, for the time being, we just show how two of the coordinates are natural parameters for certain common surfaces. 5.4.1 Polar coordinates (r, θ) in R2 We referred to polar coordinates in Chapter 3, but perhaps it is best to introduce them formally. To plot a point in R2 with polar coordinates (r, θ), go out r units along the positive x-axis, then rotate counterclockwise about the origin by the angle θ (Figure 5.23). This brings you to the Figure 5.23: Polar coordinates (r, θ) point (r cos θ, r sin θ) in rectangular coordinates. Thus the conversions from polar to rectangular coordinates are: ( x = r cos θ y = r sin θ. Also, by the Pythagorean theorem, r = p x2 + y2. All of R2 can be described by choosing r and θ in the ranges r ≥0, 0 ≤θ ≤2π. Actually, θ = 0 and θ = 2π represent the same points, along the positive x-axis, but it is convenient to allow this degree of redundancy. For example, the disk D of radius 2 about the origin in Example 5.13 can be described in polar coordinates by the conditions 0 ≤r ≤2, 0 ≤θ ≤2π, that is, (r, θ) ∈[0, 2] × [0, 2π]. We think of imposing polar coordinates on D as a transformation T : [0, 2] × [0, 2π] →R2 from the rθ-plane to the xy-plane, where T(r, θ) = (r cos θ, r sin θ). Horizontal segments 0 ≤r ≤2, θ = constant, in the rθ-plane get mapped to radial segments emanating from the origin in the xy-plane, and vertical segments r = constant, 0 ≤θ ≤2π, get mapped to circles centered at the origin. This is illustrated in Figure 5.24. Example 5.14. We return to the cone z = p x2 + y2, 0 ≤z ≤2, from Example 5.13. In polar coordinates, the equation of the cone is z = r, and combining this with the polar description of the disk D gives a parametrization eσ of the cone with r and θ as parameters: eσ: [0, 2] × [0, 2π] →R3, eσ(r, θ) = (r cos θ, r sin θ, r). See Figure 5.25. The actual cone is the same one as before, but one advantage of this parametriza- tion is that, for the purposes of integration, rectangles are nicer domains than disks. | Multivariable_Calculus_Shimamoto_Page_148_Chunk3364 |
5.4. PARAMETRIZATIONS OF SURFACES 137 Figure 5.24: The polar coordinate transformation from the rθ-plane to the xy-plane Figure 5.25: A parametrization of a cone using polar coordinates r and θ as parameters 5.4.2 Cylindrical coordinates (r, θ, z) in R3 Cylindrical coordinates are polar coordinates in the xy-plane together with the usual z-coordinate. To plot a point with cylindrical coordinates (r, θ, z), go out r units along the positive x-axis, rotate counterclockwise about the positive z-axis by the angle θ, and then go vertically z units. These coordinates are shown in Figure 5.26. This brings you to the point (r cos θ, r sin θ, z) in xyz-space. Figure 5.26: Cylindrical coordinates (r, θ, z) Thus the conversions from cylindrical to rectangular coordinates are: x = r cos θ y = r sin θ z = z. | Multivariable_Calculus_Shimamoto_Page_149_Chunk3365 |
138 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION Again r = p x2 + y2. All of R3 is covered by choosing r ≥0, 0 ≤θ ≤2π, −∞< z < ∞. Example 5.15 (Circular cylinders). We parametrize the circular cylinder of radius a and height h given by the conditions: x2 + y2 = a2, 0 ≤z ≤h. The axis of the cylinder is the z-axis. In cylindrical coordinates, along the cylinder, r is fixed at the radius a, but θ and z can vary independently. The surface is traced out by setting r = a and letting θ and z range over 0 ≤θ ≤2π, 0 ≤z ≤h. In other words, we can parametrize the cylinder using θ and z as parameters. With r = a fixed, the conversions from cylindrical to rectangular coordinates give the parametrization: σ: [0, 2π] × [0, h] →R3, σ(θ, z) = (a cos θ, a sin θ, z). See Figure 5.27. Figure 5.27: A parametrization of a cylinder using cylindrical coordinates θ and z as parameters It is instructive to think about how σ transforms the rectangle D = [0, 2π] × [0, h] to become the cylinder S. In polar/cylindrical coordinates, θ = 0 and θ = 2π represent the same thing in xyz-space. Thus σ maps the points (0, z) and (2π, z) in the θz-parameter plane to the same point in R3. Apart from this, distinct points in D get mapped to distinct points in xyz-space. One way to construct a cylinder is to take a rectangle like D and glue together, or identify, points (0, z) on the left edge with points (2π, z) on the right. It is clear geometrically that the result is a cylinder, and σ gives a formula that carries it out. 5.4.3 Spherical coordinates (ρ, ϕ, θ) in R3 The three spherical coordinates of a point in R3 are described geometrically as follows. • ρ is the distance from the origin to the point. • ϕ is the angle from the positive z-axis to the point. It is like an angle of latitude, except measured down from the positive z-axis rather than up or down from the equatorial plane. • θ is the usual polar/cylindrical angle. To plot a point in R3 with spherical coordinates (ρ, ϕ, θ), first go out ρ units along the positive z- axis. This brings you to the point (0, 0, ρ) in rectangular coordinates. Then, rotate counterclockwise about the positive y-axis by the angle ϕ, bringing you to the point (ρ sin ϕ, 0, ρ cos ϕ) in the xz- plane, a distance ρ from the origin and r = ρ sin ϕ from the z-axis. Finally, rotate counterclockwise | Multivariable_Calculus_Shimamoto_Page_150_Chunk3366 |
5.4. PARAMETRIZATIONS OF SURFACES 139 Figure 5.28: Spherical coordinates (ρ, ϕ, θ) about the positive z-axis by the angle θ. This lands at the point (ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ). The coordinates are labeled in Figure 5.28. Thus the conversions from spherical to rectangular coordinates are: x = ρ sin ϕ cos θ y = ρ sin ϕ sin θ z = ρ cos ϕ. From the Pythagorean theorem, the distance from the origin is ρ = p x2 + y2 + z2. All of R3 is covered by spherical coordinates in the intervals ρ ≥0, 0 ≤ϕ ≤π, 0 ≤θ ≤2π. (Note that the angular interval 0 ≤ϕ ≤π ranges from the direction of the positive z-axis to the direction of the negative z-axis and the interval 0 ≤θ ≤2π sweeps all the way around the z-axis. This is why values of ϕ in the interval π < ϕ ≤2π are not needed—they would duplicate points already covered.) Example 5.16 (Spheres). We parametrize the sphere of radius a centered at the origin: x2 + y2 + z2 = a2. In spherical coordinates, the sphere is traced out by keeping the value of ρ fixed at the radius a, while ϕ and θ vary independently. More precisely, set ρ = a, and let ϕ and θ range over 0 ≤ϕ ≤π, 0 ≤θ ≤2π. Hence we use ϕ and θ as parameters, and, from the spherical to rectangular conversions with ρ = a fixed, obtain the parametrization: σ: [0, π] × [0, 2π] →R3, σ(ϕ, θ) = (a sin ϕ cos θ, a sin ϕ sin θ, a cos ϕ). This is illustrated in Figure 5.29. Again the parametrization gives precise instructions for how to transform the rectangle D = [0, π] × [0, 2π] into a sphere. Here, all points where ϕ = 0 get mapped to the north pole. These are the points along the left edge of D. Similarly, the right edge of D, where ϕ = π, gets mapped to the south pole. Finally, points where θ = 0 and θ = 2π, that is, along the top and bottom edges, are identified in pairs along a meridian on the sphere in the xz-plane. Thus, to construct a sphere from a rectangle, collapse each of the left and right sides to points and zip together the remaining two sides to close up the surface. Example 5.17. We return one last time to the cone z = p x2 + y2, 0 ≤z ≤2, which we have parametrized twice already, once using rectangular coordinates (x, y) and once using po- lar/cylindrical coordinates (r, θ) as parameters. It can be parametrized using spherical coordinates as well. | Multivariable_Calculus_Shimamoto_Page_151_Chunk3367 |
140 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION Figure 5.29: A parametrization of a sphere using spherical coordinates ϕ and θ as parameters Here, along the cone, the angle ϕ down from the positive z-axis is fixed at π 4 . To see this, note, for instance, that the cross-section with the yz-plane is z = |y|. On the other hand, ρ and θ can vary independently. For example, ρ ranges from 0 to 2 √ 2. (To determine the upper endpoint, one can either use trigonometry or solve for ρ using z = ρ cos ϕ with z = 2 and ϕ = π 4 .) Also, θ rotates all the way around from 0 to 2π. From the spherical conversions to rectangular coordinates, we obtain a parametrization: bσ: [0, 2 √ 2] × [0, 2π] →R3, bσ(ρ, θ) = | Multivariable_Calculus_Shimamoto_Page_152_Chunk3368 |
5.5. INTEGRALS WITH RESPECT TO SURFACE AREA 141 two-dimensional partition of small pieces of surface area △Sij, choose a sample point pij in each piece, form a sum P i,j f(pij) △Sij, and take the limit as the size of the pieces goes to zero. We use a parametrization of S to convert the calculation into a double integral over a region in the plane. The approach is similar to what we did when we defined integrals over curves with respect to arclength, though that was a long time ago (Section 2.3 to be precise). Thus let σ: D →R3 be a parametrization of S, where D ⊂R2. We write σ(s, t) = (x(s, t), y(s, t), z(s, t)). We subdivide D (or, really, a rectangle that contains D) into small subrectangles of dimensions △si by △tj, and choose a sample point aij in each subrectangle. Intuitively, the expectation is that σ transforms each subrectangle into a small “curvy quadrilateral” in S. This is one of the small pieces of S alluded to earlier. We denote its area by △Sij and take pij = σ(aij) as the corresponding sample point. This is depicted in Figure 5.31. Figure 5.31: The integral with respect to surface area Let ∂σ ∂s and ∂σ ∂t be the vectors given by: ∂σ ∂s = ∂x ∂s , ∂y ∂s, ∂z ∂s and ∂σ ∂t = ∂x ∂t , ∂y ∂t , ∂z ∂t . We think of these vectors as velocities with respect to s and t, respectively. Then σ transforms the △si side of a typical subrectangle of D approximately to the vector ∂σ ∂s △si and the △tj side approximately to ∂σ ∂t △tj. This gives an approximation of the area of the curvy quadrilaterals: △Sij ≈area of the parallelogram determined by ∂σ ∂s △si and ∂σ ∂t △tj ≈ ∂σ ∂s △si × ∂σ ∂t △tj (by an old property about the cross product and areas, page 36) ≈ ∂σ ∂s × ∂σ ∂t △si △tj. See Figure 5.32. Hence P i,j f(pij) △Sij ≈P i,j f(σ(aij)) ∥∂σ ∂s × ∂σ ∂t ∥△si △tj. In the limit as △si and △tj go to zero, this becomes the double integral RR D f(σ(s, t)) ∥∂σ ∂s × ∂σ ∂t ∥ds dt. With this as motivation, we adopt the following definition. Definition. Let S be a surface in R3, and let f : S →R be a continuous real-valued function. Then the integral of f with respect to surface area, denoted by RR S f dS, is defined to be: ZZ S f dS = ZZ D f(σ(s, t)) ∂σ ∂s × ∂σ ∂t ds dt, where σ: D →R3 is a parametrization of S. | Multivariable_Calculus_Shimamoto_Page_153_Chunk3369 |
142 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION Figure 5.32: A blowup of the previous figure: σ sends a small △si by △tj subrectangle to a curvy quadrilateral whose area △Sij is approximately ∂σ ∂s × ∂σ ∂t △si △tj. We should acknowledge a point that was also glossed over in connection with integrals with re- spect to arclength. Namely, we motivated the surface integral using Riemann sums P i,j f(pij) △Sij based on the surface, but the actual definition relies on a parametrization σ. Perhaps a better nota- tion for the integral as defined would be RR σ f dS. The issue is whether the choice of parametrization affects the value of the integral. Once we have a few more tools available, we shall see that there is no need for concern: using the definition of the integral with two different parametrizations of the same surface gives the same value. The effect of parametrizations on surface integrals is discussed at the end of Chapter 10. Example 5.18. If f(x, y, z) = 1 for all (x, y, z) in S, then P i,j f(pij) △Sij = P i,j 1 · △Sij, the sum of the small pieces of surface area. This is the total area of S, that is: Surface area of S = ZZ S 1 dS. (5.3) Example 5.19. Let S be the sphere x2 + y2 + z2 = a2 of radius a. (a) Find the surface area of S. (b) Find RR S z2 dS. (c) Find RR S (x2 + y2 + z2) dS. We use the parametrization of a sphere derived in Example 5.16 with spherical coordinates (ϕ, θ) as parameters and ρ = a fixed: σ: [0, π] × [0, 2π] →R3, σ(ϕ, θ) = (a sin ϕ cos θ, a sin ϕ sin θ, a cos ϕ). Let D denote the domain of the parametrization, D = [0, π] × [0, 2π]. (a) Applying the previous example, we integrate the function f = 1. Then, using the definition of the integral: Area (S) = ZZ S 1 dS = ZZ D 1 · ∂σ ∂ϕ × ∂σ ∂θ dϕ dθ. We calculate ∂σ ∂ϕ and ∂σ ∂θ and place them in the rows of the determinant for a cross product: ∂σ ∂ϕ × ∂σ ∂θ = det i j k a cos ϕ cos θ a cos ϕ sin θ −a sin ϕ −a sin ϕ sin θ a sin ϕ cos θ 0 = | Multivariable_Calculus_Shimamoto_Page_154_Chunk3370 |
5.5. INTEGRALS WITH RESPECT TO SURFACE AREA 143 so: ∂σ ∂ϕ × ∂σ ∂θ = a2 sin ϕ q sin2 ϕ cos2 θ + sin2 ϕ sin2 θ + cos2 ϕ = a2 sin ϕ q sin2 ϕ + cos2 ϕ = a2 sin ϕ. (5.4) Hence: Area (S) = ZZ D a2 sin ϕ dϕ dθ = Z 2π 0 Z π 0 a2 sin ϕ dϕ dθ = Z 2π 0 −a2 cos ϕ ϕ=π ϕ=0 dθ = Z 2π 0 | Multivariable_Calculus_Shimamoto_Page_155_Chunk3371 |
144 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION (c) This time, f(x, y, z) = x2 + y2 + z2. We use a trick! On the sphere S, x2 + y2 + z2 = a2, so RR S (x2 + y2 + z2) dS = RR S a2 dS = a2 RR S 1 dS = a2Area (S). Using the formula for the area from part (a) then gives: ZZ S (x2 + y2 + z2) dS = a2 · 4πa2 = 4πa4. The trick enabled us to evaluate the integral with almost no calculation. This is a highly desirable situation. In retrospect, we can get even more mileage out of it by going back and using the result of part (c) to redo RR S z2 dS in part (b). For the sphere S is symmetric in x, y, and z, whence RR S x2 dS = RR S y2 dS = RR S z2 dS. Therefore RR S (x2 + y2 + z2) dS = 3 RR S z2 dS, or: ZZ S z2 dS = 1 3 ZZ S (x2 + y2 + z2) dS = 1 3 · 4πa4 = 4 3πa4, where the next-to-last step used part (c). Happily, this is the same as the answer we obtained in part (b) originally by grinding it out with a parametrization. 5.6 Triple integrals and beyond Integrals of real-valued functions of three or more variables are conceptually the same as double integrals, though the notation is more encumbered. We merely outline the definition. Given: • a bounded subset W of Rn (that is, there is an n-dimensional box R = [a1, b1] × [a2, b2] × · · · × [an, bn] such that W ⊂R) and • a bounded real-valued function f : W →R (that is, there is a scalar M such that |f(x)| ≤M for all x in W), one subdivides W into small n-dimensional subboxes of dimensions △x1 by △x2 by · · · by △xn, chooses a sample point p in each subbox, and considers sums of the form P f(p) △x1 △x2 · · · △xn, where the summation is over all the subboxes. The integral is the limit of these sums as the dimensions of the subboxes go to zero. It is denoted by: ZZ · · · Z W f dV or ZZ · · · Z W f dx1 dx2 · · · dxn. For instance, if W ⊂R3 and f = f(x, y, z) is a function of three variables, then RRR W f dV is called a triple integral. The higher-dimensional integral has the same sort of weighted sum intepretations as the double integral, such as: • n-dimensional volume: Vol (W) = RR · · · R W 1 dV , • average value: f = 1 Vol (W) RR · · · R W f dV . For example, if W ⊂R3, the average x-coordinate on W is x = 1 Vol (W) RRR W x dV , similarly for y and z, and the point (x, y, z) is called the centroid of W. To calculate the integral, one usually evaluates a sequence of n one-dimensional partial integrals. This can be complicated, because it may be difficult to determine the nested limits of integration that describe the domain W. This is true even in three dimensions, where, in principle, it is still possible to visualize what W looks like. We illustrate this by examining in detail a single example of a triple integral. | Multivariable_Calculus_Shimamoto_Page_156_Chunk3372 |
5.6. TRIPLE INTEGRALS AND BEYOND 145 Figure 5.33: Three of the planes that bound W: z = y (left), y = x (middle), and x = 1 (right). The fourth is z = 0, the xy-plane. Example 5.20. Find RRR W (y + z) dV if W is the solid region in R3 bounded by the planes z = y, y = x, x = 1, and z = 0. See Figure 5.33. The planes in question slice through one another in such a way that the solid W that they bound is a tetrahedron. Each of its four triangular faces is contained in one of the planes. Looking at W towards the origin from the first octant, the z = y face is the top, z = 0 is the bottom, x = 1 is the front, and y = x is the back. This is shown on the left of Figure 5.34. Figure 5.34: The region of integration W (left) and its projection D on the xy-plane (right) To find the endpoints that describe W in an iterated integral, our approach is to focus on choosing which variable to integrate with respect to first. Once that variable is integrated out of the picture, what remains is a double integral, which by now is a much more familiar situation. For instance, say we decide to integrate first with respect to z, treating x and y as constant. We need to know the values of x and y for which this process makes sense, that is, the pairs (x, y) for which there is at least one z such that (x, y, z) ∈W. These pairs are precisely the projection of W on the xy-plane. In this example, this is the triangle D at the base of the tetrahedron, which is shown on the right of Figure 5.34. In the xy-plane, it is bounded by the x-axis and the lines x = 1 and y = x. For each (x, y) in D, the values of z for which (x, y, z) ∈W go from z = 0 to z = y. Thus: ZZZ W (y + z) dV = ZZ D Z y 0 (y + z) dz dx dy. The remaining limits of integration come from describing D as a double integral in the usual way, for instance, using the order of integration dy dx. After that, the integral can be evaluated one | Multivariable_Calculus_Shimamoto_Page_157_Chunk3373 |
146 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION antidifferentiation at a time. Thus: ZZZ W (y + z) dV = Z 1 0 Z x 0 Z y 0 (y + z) dz dy dx = Z 1 0 Z x 0 (yz + 1 2z2) z=y z=0 dy dx = Z 1 0 Z x 0 3 2y2 dy dx = Z 1 0 1 2y3 y=x y=0 dx = Z 1 0 1 2x3 dx = 1 8x4 1 0 = 1 8. Example 5.21. Express the same integral RRR W (y+z) dV from the previous example as an iterated integral in which: (a) the first integration is with respect to x, (b) the first integration is with respect to y. (a) To integrate first with respect to x, we treat y and z as constant. Proceeding as in the previous example, the relevant values of (y, z) are the points in the projection D of W on the yz-plane. As before, this is a triangle, though no longer one of the faces of the tetrahedron. The triangle is bounded in the yz-plane by the y-axis and the lines y = 1 and z = y. See Figure 5.35 on the left. For each (y, z) in D, the point (x, y, z) is in W from x = y to x = 1. Hence: ZZZ W (y + z) dV = ZZ D Z 1 y (y + z) dx dy dz. Describing the remaining double integral over D using the order of integration, say dz dy, gives: ZZZ W (y + z) dV = Z 1 0 Z y 0 Z 1 y (y + z) dx dz dy. We won’t carry out the iterated antidifferentiation, though we are optimistic that the final answer is again 1 8. (b) This time, the integral will be a double integral of an integral with respect to y over the projection of W on the xz-plane . Again, the projection D is a triangle. Two sides of the triangle are contained in the x-axis and the line x = 1 in the xz-plane. The third side is the projection of the line where the planes z = y and y = x intersect. This line of intersection in R3 is described by z = y = x, and its projection on the xz-plane is z = x. This is the remaining side of D. See Figure 5.35 on the right. For each (x, z) in D, y goes from y = z to y = x. Therefore: ZZZ W (y + z) dV = ZZ D Z x z (y + z) dy dx dz. Finally, integrating over D using the order dz dx results in: ZZZ W (y + z) dV = Z 1 0 Z x 0 Z x z (y + z) dy dz dx. | Multivariable_Calculus_Shimamoto_Page_158_Chunk3374 |
5.7. EXERCISES FOR CHAPTER 5 147 Figure 5.35: Integrating first with respect to x (left) or y (right) 5.7 Exercises for Chapter 5 Section 1 Volume and iterated integrals In Exercises 1.1–1.4, evaluate the given iterated integral. 1.1. Z 2 1 Z 4 3 x2y dy dx 1.2. Z 1 0 Z x x2 (x + y) dy dx 1.3. Z π/2 0 Z sin y 0 (x cos y + 1) dx dy 1.4. Z 2 1 Z y −y sin(3x + 2y) dx dy 1.5. Evaluate RR D x2 y dA if D is the rectangle described by 1 ≤x ≤3, 2 ≤y ≤4. 1.6. Evaluate RR D(x + y) dA if D is the quarter-disk x2 + y2 ≤1, x ≥0, y ≥0. 1.7. Evaluate RR D ex+y dA if D is the triangular region with vertices (0, 0), (0, 2), (1, 0). 1.8. Evaluate RR D xy dA if D is the triangular region with vertices (0, 0), (1, 1), (2, 0). In Exercises 1.9–1.14, (a) sketch the domain of integration D in the xy-plane and (b) write an equivalent expression with the order of integration reversed. 1.9. Z 2 1 Z 4 3 f(x, y) dy dx 1.10. Z 1 0 Z e ex f(x, y) dy dx 1.11. Z 1 0 Z y y2 f(x, y) dx dy 1.12. Z π/2 0 Z sin y 0 f(x, y) dx dy | Multivariable_Calculus_Shimamoto_Page_159_Chunk3375 |
148 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION 1.13. Z 1 −1 Z √ 1−x2 0 f(x, y) dy dx 1.14. Z 6 −3 Z −1+√3+y 1 3 y f(x, y) dx dy 1.15. Consider the iterated integral Z 1 1 4 Z 4 1 x yexy dy dx. (a) Sketch the domain of integration D in the xy-plane. (b) Write an equivalent expression with the order of integration reversed. (c) Evaluate the integral using whatever method seems best. 1.16. Evaluate the integral Z 6 0 Z 3 y 2 sin(x2) dx dy. 1.17. Evaluate the integral ZZ D p 1 −x2 dA, where D is the quarter-disk in the first quadrant described by x2 + y2 ≤1, x ≥0, and y ≥0. 1.18. Find the volume of the wedge-shaped solid that lies above the xy-plane, below the plane z = y, and inside the cylinder x2 + y2 = 1. 1.19. Find the volume of the pyramid-shaped solid in the first octant bounded by the three coor- dinate planes and the planes x + z = 1 and y + 2z = 2. 1.20. (a) If c is a positive constant, find the volume of the tetrahedron in the first octant bounded by the plane x + y + z = c and the three coordinate planes. (b) Consider the tetrahedron W in the first octant bounded by the plane x + y + z = 1 and the three coordinate planes. Suppose that you want to divide W into three pieces of equal volume by slicing it with two planes parallel to x + y + z = 1, i.e., with planes of the form x + y + z = c. How should the slices be made? 1.21. Let D be the unit square 0 ≤x ≤1, 0 ≤y ≤1, and let f : D →R be the function given by f(x, y) = min{x, y}. Find RR D f(x, y) dx dy. Section 2 The double integral 2.1. Let R = [0, 1] × [0, 1], and let f : R →R be the function given by: f(x, y) = ( 1 if (x, y) = (1 2, 1 2), 0 otherwise. (a) Let R be subdivided into a 3 by 3 grid of subrectangles of equal size (Figure 5.36). What are the possible values of Riemann sums based on this subdivision? (Different choices of sample points may give different values.) (b) Repeat for a 4 by 4 grid of subrectangles of equal size. (c) How about for an n by n grid of subrectangles of equal size? | Multivariable_Calculus_Shimamoto_Page_160_Chunk3376 |
5.7. EXERCISES FOR CHAPTER 5 149 Figure 5.36: A 3 by 3 subdivision of R = [0, 1] × [0, 1] (d) Let δ be a positive real number. If R is subdivided into a grid of subrectangles of dimensions △xi by △yj, where △xi < δ and △yj < δ for all i, j, show that any Riemann sum based on the subdivision lies in the range 0 ≤P i,j f(pij) △xi △yj < 4δ2. (e) Show that RR R f(x, y) dA = 0, i.e., that lim △xi→0 △yj→0 P i,j f(pij) △xi △yj = 0. Technically, this means you need to show that, given any ϵ > 0, there exists a δ > 0 such that, for any subdivision of R into subrectangles with △xi < δ and △yj < δ for all i, j, every Riemann sum based on the subdivision satisfies P i,j f(pij) △xi △yj −0 < ϵ. 2.2. Let R = [0, 1] × [0, 1], and let f : R →R be the function given by: f(x, y) = ( 1 if x = 0 or y = 0, 0 otherwise. (a) Let R be subdivided into a 3 by 3 grid of subrectangles of equal size (Figure 5.36). What are the possible values of Riemann sums based on this subdivision? (b) Repeat for a 4 by 4 grid of subrectangles of equal size. (c) How about for an n by n grid of subrectangles of equal size? (d) Is f integrable on R? If so, what is the value of RR R f(x, y) dA? 2.3. Here is a result that will come in handy increasingly as we calculate more integrals. Let R = [a, b] × [c, d] be a rectangle, and let F : R →R be a real-valued function such that the variables x and y separate into two continuous factors, that is, F(x, y) = f(x)g(y), where f : [a, b] →R and g: [c, d] →R are continuous. Show that: ZZ R f(x)g(y) dx dy = Z b a f(x) dx Z d c g(y) dy . 2.4. Let f : U →R be a continuous function defined on an open set U in R2, and let a be a point of U. (a) If f(a) > 0, prove that there exists a rectangle R containing a such that: ZZ R f dA > 0. (Hint: Use Exercise 6.4 of Chapter 3.) | Multivariable_Calculus_Shimamoto_Page_161_Chunk3377 |
150 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION (b) Let g: U →R be another continuous function on U. If f(a) > g(a), prove that there exists a rectangle R containing a such that: ZZ R f dA > ZZ R g dA. (Hint: Consider f −g.) 2.5. In this exercise, we use the double integral to give another proof of the equality of mixed partial derivatives. Let f(x, y) be a real-valued function defined on an open set U in R2 whose first and second-order partial derivatives are continuous on U. (a) Let R = [a, b] × [c, d] be a rectangle that is contained in U. Show that: ZZ R ∂2f ∂x ∂y(x, y) dA = ZZ R ∂2f ∂y ∂x(x, y) dA. (Hint: Use Fubini’s theorem, with appropriate orders of integration, and the fundamen- tal theorem of calculus to show that both sides are equal to f(b, d) −f(b, c) −f(a, d) + f(a, c).) (b) Show that ∂2f ∂x ∂y(a) = ∂2f ∂y ∂x(a) for all a in U. (Hint: Use Exercise 2.4 to show that something goes wrong if they are not equal.) Section 3 Interpretations of the double integral 3.1. The population density of birds in a wildlife refuge decreases at a uniform rate with the distance from a river. If the river is modeled as the x-axis in the plane, then the density at the point (x, y) is given by f(x, y) = 5 −|y| hundred birds per square mile, where x and y are measured in miles. Find the total number of birds in the rectangle R = [−2, 2] × [0, 1]. 3.2. A small cookie has the shape of the region in the first quadrant bounded by the curves y = x2 and x = y2, where x and y are measured in inches. Chocolate is poured unevenly on top of the cookie in such a way that the density of chocolate at the point (x, y) is given by f(x, y) = 100(x + y) grams per square inch. Find the total mass of chocolate on the cookie. 3.3. The eye of a tornado is positioned directly over the origin in the plane. Suppose that the wind speed on the ground at the point (x, y) is given by v(x, y) = 30(x2 + y2) miles per hour. (a) Find the average wind speed on the square R = [0, 2] × [0, 2]. (b) Find all points (x, y) of R at which the wind speed equals the average. 3.4. Find the centroid of the half-disk of radius a in R2 described by x2 + y2 ≤a2, y ≥0. 3.5. Find the centroid of the triangular region in R2 with vertices (0, 0), (1, 2), and (1, 3). 3.6. Find the centroid of the triangular region in R2 with vertices (0, 0), (2, 2), and (1, 3). 3.7. Let D be the fudgsicle-shaped region of R2 that consists of the rectangle [−1, 1] × [0, h] of height h topped by a half-disk of radius 1, as shown in Figure 5.37. Find the value of h such that the point (0, h) is the centroid of D. | Multivariable_Calculus_Shimamoto_Page_162_Chunk3378 |
5.7. EXERCISES FOR CHAPTER 5 151 Figure 5.37: The fudgsicle 3.8. Let R be a rectangle in R2, and let f : R →R be an integrable function. If f is continuous, then it is true that there is a point of R at which f assumes its average value, that is, there exists a point c of R such that f(c) = f. The proof is not hard, but it depends on subtle properties of the real numbers that are beyond what we cover here. The same conclusion need not hold, however, for discontinuous functions. Find an example of a rectangle R and an integrable function f : R →R such that there is no point c of R at which f(c) = f. Section 4 Parametrization of surfaces 4.1. Find a parametrization of the upper hemisphere of radius a described by x2 + y2 + z2 = a2, z ≥0, using the following coordinates as parameters. Be sure to state what the domain of your parametrization is. (a) the rectangular coordinates x, y (b) the polar/cylindrical coordinates r, θ (c) the spherical coordinates ϕ, θ 4.2. Let S be the surface described in cylindrical coordinates by z = θ, 0 ≤r ≤1, and 0 ≤θ ≤4π. It is called a helicoid. Converting to rectangular coordinates gives a parametrization: σ: [0, 1] × [0, 4π] →R3, σ(r, θ) = (r cos θ, r sin θ, θ). (a) Describe the curve that is traced out by σ if r = 1 2 is held fixed and θ varies, that is, the curve parametrized by the path α(θ) = σ(1 2, θ), 0 ≤θ ≤4π. (b) Similarly, describe the curve that is traced out by σ if θ = π 4 is held fixed and r varies, that is, the curve parametrized by β(r) = σ(r, π 4 ), 0 ≤r ≤1. (c) Describe S in a few words, and draw a sketch. Section 5 Integrals with respect to surface area 5.1. Let S be the triangular surface in R3 whose vertices are (1, 0, 0), (0, 1, 0), and (0, 0, 1). (a) Find a parametrization of S using x and y as parameters. What is the domain of your parametrization? (Hint: The triangle is contained in a plane.) (b) Use your parametrization and formula (5.3) to find the area of S. Check that your answer agrees with what you would get using the formula: Area = 1 2(base)(height). (c) Find RR S 7 dS. (Hint: Use your answer to part (b).) | Multivariable_Calculus_Shimamoto_Page_163_Chunk3379 |
152 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION 5.2. Let S be the cylindrical surface x2 + y2 = 4, 0 ≤z ≤3. Find the integral RR S z2 dS. 5.3. Consider the graph z = f(x, y) of a smooth real-valued function f defined on a bounded subset D of R2. Show that the surface area of the graph is given by the formula: Surface area = ZZ D s 1 + ∂f ∂x 2 + ∂f ∂y 2 dx dy. 5.4. Let f : [a, b] →R be a smooth real-valued function of one variable, where a ≥0, and let S be the surface of revolution that is swept out when the curve z = f(x) in the xz-plane is rotated all the way around the z-axis. Show that the surface area of S is given by the formula: Surface area = 2π Z b a x p 1 + f′(x)2 dx. (Hint: Parametrize S using cylindrical coordinates.) 5.5. Let W be the solid bounded on top by the plane z = y + 5, on the sides by the cylinder x2 + y2 = 4, and on the bottom by the plane z = 0. (a) Sketch W. (b) Let S be the surface that bounds W (top, sides, and bottom). Find the surface area of S. (c) Find RR S z dS, where S is the surface in part (b). 5.6. Let S be the helicoid from Exercise 4.2, parametrized by: σ: [0, 1] × [0, 4π] →R3, σ(r, θ) = (r cos θ, r sin θ, θ). Find RR S z p x2 + y2 dS. 5.7. Let W be the solid region of Example 5.2 that lies inside the cylinders x2 + z2 = 1 and y2 +z2 = 1 and above the xy-plane, and let S be the exposed part of the surface that bounds W, that is, the cylindrical surfaces but not the base. (a) Find the surface area of S. (b) Find RR S z dS. 5.8. The sphere x2 + y2 + z2 = a2 of radius a sits inscribed in the circular cylinder x2 + y2 = a2. Given real numbers c and d, where −a ≤c < d ≤a, show that the portions of the sphere and the cylinder lying between the planes z = c and z = d have equal surface area. (Hint: Parametrize both surfaces using the cylindrical coordinates θ and z as parameters.) Section 6 Triple integrals and beyond In Exercises 6.1–6.2, evaluate the given iterated integral. 6.1. Z 1 0 Z x 0 Z x y (x + y −z) dz dy dx 6.2. Z 1 0 Z 2x x Z xy 0 x3y2z dz dy dx | Multivariable_Calculus_Shimamoto_Page_164_Chunk3380 |
5.7. EXERCISES FOR CHAPTER 5 153 6.3. Find RRR W xyz dV if W is the solid region in R3 below the surface z = x2 + y2 and above the square 0 ≤x ≤1, 0 ≤y ≤1. 6.4. Find RRR W (xz + y) dV if W is the solid region in R3 above the xy-plane bounded by the surface y = x2 and the planes z = y, y = 1, and z = 0. 6.5. Find RRR W x dV if W is the region in the first octant bounded by the saddle surface z = x2−y2 and the planes x = 2, y = 0, and z = 0. 6.6. Find RRR W (1 + x + y + z)2 dV if W is the solid tetrahedron in the first octant bounded by the plane x + y + z = 1 and the three coordinate planes. 6.7. Let a be a positive real number, and let W = [0, a] × [0, a] × [0, a] in R3. Find the average value of f(x, y, z) = x2 + y2 + z2 on W. For each of the triple integrals in Exercises 6.8–6.11, (a) sketch the domain of integration in R3, (b) write an equivalent expression in which the first integration is with respect to x, and (c) write an equivalent expression in which the first integration is with respect to y. 6.8. Z 2 −2 Z q 2−x2 2 0 Z q 1−x2 4 −y2 2 0 f(x, y, z) dz dy dx 6.9. Z 1 −1 Z √ 1−x2 0 Z 1 0 f(x, y, z) dz dy dx 6.10. Z 1 0 Z 1 x Z x 0 f(x, y, z) dz dy dx 6.11. Z 1 0 Z 1 x Z y 0 f(x, y, z) dz dy dx | Multivariable_Calculus_Shimamoto_Page_165_Chunk3381 |
154 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION | Multivariable_Calculus_Shimamoto_Page_166_Chunk3382 |
Part IV Vector-valued functions 155 | Multivariable_Calculus_Shimamoto_Page_167_Chunk3383 |
Chapter 6 Differentiability and the chain rule So far, we have studied functions of which at least one of the domain or codomain is a subset of R, in other words: • vector-valued functions of one variable, also known as paths, α: I →Rn, where I ⊂R, or • real-valued functions of n variables f : U →R, where U ⊂Rn. Now, we consider the general case of vector-valued functions of n variables, that is, functions of the form f : U →Rm, where U ⊂Rn and both n and m are allowed to be greater than 1. Given such a vector-valued function, we can leverage what we know about real-valued functions because, if x ∈U, then, as an element of Rm, f(x) has m coordinates: f(x) = (f1(x), f2(x), . . . , fm(x)), where each fi(x) is a real number. In other words, each component is a real-valued function fi : U →R. Thus a vector-valued function is also a sequence of real-valued functions. We use the notation above for the components consistently from now on. For example, if f : R2 →R2 is the polar coordinate transformation f(r, θ) = (r cos θ, r sin θ), then f1(r, θ) = r cos θ and f2(r, θ) = r sin θ. Also, as with paths, we denote f in plainface type, reserving boldface for a special kind of vector-valued function that we study beginning in Chapter 8. We treated the case of real-valued functions fairly rigorously, and we shall see that much of the theory carries over to vector-valued functions without incident. On the other hand, we discussed paths back in Chapter 2 more informally, so what we are about to do here in the vector-valued case can be taken as establishing the theory behind what we did then. 6.1 Continuity revisited The motivation and definition for continuity of vector-valued functions carry over almost verbatim from the real-valued case. Definition. Let U be an open set in Rn, and let f : U →Rm be a function. We say that f is continuous at a point a of U if, given any open ball B(f(a), ϵ) about f(a), there exists an open ball B(a, δ) about a such that: f(B(a, δ)) ⊂B(f(a), ϵ). In other words, if x ∈B(a, δ), then f(x) ∈B(f(a), ϵ). See Figure 6.1. Equivalently, given any ϵ > 0, there exists a δ > 0 such that: if ∥x −a∥< δ, then ∥f(x) −f(a)∥< ϵ. 157 | Multivariable_Calculus_Shimamoto_Page_169_Chunk3384 |
158 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE Figure 6.1: Continuity of a vector-valued function: given any ball B(f(a), ϵ) about f(a), there is a ball B(a, δ) about a such that f(B(a, δ)) ⊂B(f(a), ϵ). We say that f is a continuous function if it is continuous at every point of its domain. Proposition 6.1. If f(x) = (f1(x), f2(x), . . . , fm(x)), then f is continuous at a if and only if f1, f2, . . . , fm are all continuous at a. Proof. Intuitively, this is because f(x)−f(a) can be made small if and only if all of its components fi(x)−fi(a) can be made small, but this needs to be made more precise. There are two implications that need to be proven here, that each condition implies the other. We give the details for one of the implications and leave the other for the exercises. Namely, we show that, if f1, f2, . . . , fm are continuous at a, then so is f. Let ϵ > 0 be given. We want to ensure that ∥f(x) −f(a)∥< ϵ. Note that: ∥f(x) −f(a)∥= q | Multivariable_Calculus_Shimamoto_Page_170_Chunk3385 |
6.2. DIFFERENTIABILITY REVISITED 159 is continuous at a. When this happens, we write limx→a f(x) = L. As before, it is immediate that f : U →Rm is continuous at a if and only if limx→a f(x) = f(a). One consequence of this is that Proposition 6.1 then implies that a vector-valued limit can also be viewed as a sequence of real-valued limits. Proposition 6.2. If f(x) = (f1(x), f2(x), . . . , fm(x)) and L = (L1, L2, . . . , Lm), then limx→a f(x) = L if and only if limx→a fi(x) = Li for all i = 1, 2, . . . , m. 6.2 Differentiability revisited Recall that, in the real-valued case, f : U →R is differentiable at a if it has a first-order approxi- mation f(x) ≈ℓ(x) = f(a) + ∇f(a) · (x −a) such that limx→a f(x)−ℓ(x) ∥x−a∥ = 0. For a vector-valued function f : U →Rm, we find an approximation, again denoted by ℓ, by compiling the approxima- tions of the component functions f1, f2, . . . , fm into one vector: ℓ(x) = ℓ1(x) ℓ2(x) ... ℓm(a) = f1(a) + ∇f1(a) · (x −a) f2(a) + ∇f2(a) · (x −a) ... fm(a) + ∇fm(a) · (x −a) = f1(a) f2(a) ... fm(a) + ∇f1(a) · (x −a) ∇f2(a) · (x −a) ... ∇fm(a) · (x −a) = f(a) + ∇f1(a) ∇f2(a) ... ∇fm(a) · (x −a) = f(a) + Df(a) · (x −a), where, in the last two lines, Df(a) = ∇f1(a) ∇f2(a) ... ∇fm(a) is the matrix whose rows are the gradients of the component functions and x −a is a column vector. We consider the error f(x) −ℓ(x) = f(x) −f(a) −Df(a) · (x −a) in using the approximation and proceed as in the real-valued case. Definition. Let U be an open set in Rn, and let f : U →Rm be a function. If a ∈U, consider the matrix: Df(a) = ∂f1 ∂x1 (a) · · · ∂f1 ∂xn (a) ... ... ... ∂fm ∂x1 (a) · · · ∂fm ∂xn (a) . We say that f is differentiable at a if: lim x→a f(x) −f(a) −Df(a) · (x −a) ∥x −a∥ = 0. (6.1) When this happens, Df(a) is called the derivative, or Jacobian matrix, of f at a. | Multivariable_Calculus_Shimamoto_Page_171_Chunk3386 |
160 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE To repeat a point made in the real-valued case, the derivative is not a single number, but rather a matrix or, even better, the linear part of a good affine approximation of f near a. We begin by calculating a few simple examples of Df(a). In general, it is an m by n matrix whose entries are various partial derivatives. The ith row is the gradient ∇fi, and the jth column is the “velocity” ∂ ∂xj of f with respect to xj. Example 6.3. Let f : R2 →R2 be f(r, θ) = (r cos θ, r sin θ). Then: Df(r, θ) = cos θ −r sin θ sin θ r cos θ . The formula above gives what the matrix looks like at a general point a = (r, θ). At a specific point, such as a = (2, π 4 ), we have Df(a) = Df(2, π 4 ) = " √ 2 2 −2 · √ 2 2 √ 2 2 2 · √ 2 2 # = " √ 2 2 − √ 2 √ 2 2 √ 2 # . Example 6.4. Let g: R3 →R2 be the projection of xyz-space onto the xy-plane, g(x, y, z) = (x, y). Then: Dg(x, y, z) = 1 0 0 0 1 0 . In other words, Dg(a) = 1 0 0 0 1 0 for all a in R3. Example 6.5. Let h: R2 →R3 be h(x, y) = (x2 + y3, x4y5, e6x+7y). Then, if a = (x, y): Dh(a) = Dh(x, y) = 2x 3y2 4x3y5 5x4y4 6e6x+7y 7e6x+7y . To determine whether a function is differentiable without going through the definition, we may use criteria similar to those in the real-valued case. For instance, the same reasoning as before shows that a differentiable function must be continuous and that it is necessary for all the partial derivatives ∂fi ∂xj to exist in order for f to be differentiable. In addition, the condition for differentiability in the definition (6.1) requires that a certain limit be the zero vector. This can happen if and only if the limit of each of the components is zero, too. This follows from Proposition 6.2. In other words, we have the following componentwise criterion for differentiability. Proposition 6.6. If f(x) = (f1(x), f2(x), . . . , fm(x)), then f is differentiable at a if and only if f1, f2, . . . , fm are all differentiable at a. Example 6.7. Let I be an open interval in R, and let α: I →Rn be a path in Rn, where α(t) = (x1(t), x2(t), . . . , xn(t)). By the last proposition, α is differentiable if and only if each of its components x1, x2, . . . , xn is differentiable. The latter condition is essentially how we defined differentiability of paths in Chapter 2, so what we did there is supported by the theory we now have in place. Moreover, by definition of the derivative matrix: Dα(t) = x′ 1(t) x′ 2(t) ... x′ n(t) . This agrees with α′(t), the velocity, under the usual identification of a vector in Rn with a column matrix. | Multivariable_Calculus_Shimamoto_Page_172_Chunk3387 |
6.3. THE CHAIN RULE: A CONCEPTUAL APPROACH 161 According to Proposition 6.6, we may apply the C1 test for real-valued functions to each of the components of a vector-valued function to obtain the following criterion. Theorem 6.8 (The C1 test). If all the partial derivatives ∂fi ∂xj (i.e., the entries of Df(x)) exist and are continuous on U, then f is differentiable at every point of U. To illustrate this, in each of Examples 6.3–6.5 above, all entries of the D matrices are continuous, so those three functions f, g, h are differentiable at every point of their domains. A function f is called smooth if each of its component functions f1, f2, . . . , fm is smooth in the sense we have discussed previously (see Section 4.9), that is, they have continuous partial derivatives of all orders. As before, we generally state our results about differentiability for smooth functions without worrying about whether this is a stronger assumption than necessary. 6.3 The chain rule: a conceptual approach We now discuss the all-important rule for the derivative of a composition. We first take a fairly high-level approach to understand where the rule comes from. Let U and V be open sets in Rn and Rm, respectively, and let f : U →V and g: V →Rp be functions. The composition g ◦f : U →Rp is then defined, and by definition, if a ∈U, the composition is differentiable at a if there is a good first-order approximation of the form: g(f(x)) ≈g(f(a)) + D(g ◦f)(a) · (x −a) (6.2) when x is near a. The challenge is to figure out the expression that goes into the box for D(g◦f)(a). The idea is that the first-order approximation of the composition g◦f should be the composition of the first-order approximations of g and f individually. Let b = f(a), and assume that f is differentiable at a and g is differentable at b. Then there are good first-order approximations: • for f : f(x) ≈ℓf(x) ≈f(a) + Df(a) · (x −a) when x is near a, (6.3) • for g : g(y) ≈ℓg(y) ≈g(b) + Dg(b) · (y −b) when y is near b ≈g(f(a)) + Dg(f(a)) · (y −f(a)). (6.4) Note that, by continuity, when x is near a, f(x) is near f(a) = b, so substituting y = f(x) in (6.4) gives: g(f(x)) ≈g(f(a)) + Dg(f(a)) · (f(x) −f(a)). Then, using (6.3) to substitute f(x) −f(a) ≈Df(a) · (x −a) yields: g(f(x)) ≈g(f(a)) + Dg(f(a)) · Df(a) · (x −a). If you like, you can check that the right side of this last approximation is also the composition ℓg(ℓf(x)) of the first-order approximations of g and f. In any case, the upshot is that we have identified something that fits naturally in the box in equation (6.2). Theorem 6.9 (Chain rule). Let U and V be open sets in Rn and Rm, respectively, and let f : U →V and g: V →Rp be functions. Let a be a point of U. If f is differentiable at a and g is differentiable at f(a), then g ◦f is differentiable at a and: D(g ◦f)(a) = Dg(f(a)) · Df(a), where the product on the right is matrix multiplication. | Multivariable_Calculus_Shimamoto_Page_173_Chunk3388 |
162 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE In terms of the sizes of the various parties involved, the left side of the chain rule is a p by n matrix, while the right side is a product (p by m) × (m by n). This need not be memorized. In practice, it usually takes care of itself. The chain rule says that the derivative of a composition is the product of the derivatives, or, to be even more sophisticated, the composition of the linear parts of the associated first-order approximations. Though the objects involved are more complicated, this is actually the same as what happens back in first-year calculus. There, if y is a function of u, say y = g(u), and u is a function of x, say u = f(x), then y becomes a function of x via composition: y = g(f(x)). The one-variable chain rule says: dy dx = dy du du dx. Again, the derivative of a composition is the product of the derivatives of the composed steps. The line of reasoning we used to obtain the chain rule is natural and reasonably straightforward, but a proper proof requires greater attention to the various approximations involved. The letter ϵ is likely to appear. The arguments are a little like those needed for a correct proof of the Little Chain Rule back in Exercise 5.6 of Chapter 4, though the details here are more complicated. We shall say more about the proof of the chain rule in the next section. The following example is meant to illustrate how the pieces of the chain rule fit together. Example 6.10. Let f : R2 →R2 be given by f(r, θ) = (r cos θ, r sin θ) and g: R2 →R3 by g(x, y) = (x2 + y2, 2x + 3y, xy2). Find D(g ◦f)(2, π 2 ). By the chain rule, D(g ◦f)(2, π 2 ) = Dg | Multivariable_Calculus_Shimamoto_Page_174_Chunk3389 |
6.4. THE CHAIN RULE: A COMPUTATIONAL APPROACH 163 dot product of the ith row and jth column: ... · · · (i, j) · · · ... = ∗ ∗ · · · ∗ ∗ ∗ ∗ ... ∗ ∗ . There is one such dot product for each entry of D(g ◦f)(a). We shall write out these entries in a few cases in a moment and discover that we have seen this type of product before. You can think of the discovery either as an alternative way to derive the chain rule or as confirmation of the consistency of the theory. We take a simple situation that we have studied previously and then tweak it a couple of times. Example 6.11. Let α: R →R3 be a smooth path in R3, α(t) = (x(t), y(t), z(t)), and let g: R3 →R be a smooth real-valued function of three variables, w = g(x, y, z). Then w becomes a function of t by composition: w = g(α(t)) = g(x(t), y(t), z(t)). As a matter of notation, from here on, we use w to denote the value of the composition, reserving g to stand for the original function of x, y, z. The composition is a real-valued function of one variable, so the Little Chain Rule applies. It says that the derivative of the compositon is “gradient dot velocity”: dw dt = ∇g(α(t)) · α′(t) = ∂g ∂x, ∂g ∂y, ∂g ∂z · dx dt , dy dt , dz dt = ∂g ∂x dx dt + ∂g ∂y dy dt + ∂g ∂z dz dt , (6.5) where the partial derivatives ∂g ∂x, ∂g ∂y, ∂g ∂z are evaluated at the point α(t) = (x(t), y(t), z(t)). This calculation can be visualized using a “dependence diagram,” as shown in Figure 6.2. It illustrates w = g x y z t Figure 6.2: A dependence diagram for the Little Chain Rule the dependence of the various variables involved: g is a function of x, y, z, and in turn x, y, z are functions of t. The Little Chain Rule (6.5) says that, to find the derivative of the top with respect to the bottom, multiply the derivatives along each possible path from top to bottom and add these products. Tweak 1. Suppose that x, y, z are functions of two variables s and t. In other words, replace α: R →R3 with a smooth function f : R2 →R3, where f(s, t) = (x(s, t), y(s, t), z(s, t)). Then the composition g◦f : R2 →R3 →R is a function of s and t: w = (g◦f)(s, t) = g(x(s, t), y(s, t), z(s, t)). The corresponding dependence diagram is shown in Figure 6.3. To compute ∂w ∂s (or ∂w ∂t ), only one variable actually varies. Because the composition is again real-valued, the Little Chain Rule applies with respect to that variable, just as in the previous case. Notationally, because w, x, y, and z are no longer functions of just one variable, we need to replace | Multivariable_Calculus_Shimamoto_Page_175_Chunk3390 |
164 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE w = g x y z s t Figure 6.3: Dependence diagram for tweak 1 the ordinary derivative terms d ds (or d dt) in equation (6.5) with partial derivatives ∂ ∂s (or ∂ ∂t). For instance, for ∂w ∂s , the Little Chain Rule says: ∂w ∂s = ∂g ∂x ∂x ∂s + ∂g ∂y ∂y ∂s + ∂g ∂z ∂z ∂s. (6.6) The relevant paths in the dependence diagram are shown in bold in the figure. Similarly: ∂w ∂t = ∂g ∂x ∂x ∂t + ∂g ∂y ∂y ∂t + ∂g ∂z ∂z ∂t . (6.7) In fact, equations (6.6) and (6.7) can be rewritten as a single matrix equation: ∂w ∂s ∂w ∂t = h ∂g ∂x ∂g ∂y ∂g ∂z i ∂x ∂s ∂x ∂t ∂y ∂s ∂y ∂t ∂z ∂s ∂z ∂t . (6.8) The entries corresponding to equation (6.6) are shown in red. The point is that the matrix on the left of equation (6.8) is the derivative D(g◦f), while those on the right are Dg and Df, respectively. Tweak 2. We alter g so that it is vector-valued, say g: R3 →R2, where g(x, y, z) = (g1(x, y, z), g2(x, y, z)), and leave f : R2 →R3 as in tweak 1. The composition g ◦f : R2 →R2 is vector-valued: w = g | Multivariable_Calculus_Shimamoto_Page_176_Chunk3391 |
6.4. THE CHAIN RULE: A COMPUTATIONAL APPROACH 165 as highlighted in bold in the figure. Again, the collection of the four individual partial derivatives of the composition computed in this way can be organized into a single matrix equation: " ∂w1 ∂s ∂w1 ∂t ∂w2 ∂s ∂w2 ∂t # = " ∂g1 ∂x ∂g1 ∂y ∂g1 ∂z ∂g2 ∂x ∂g2 ∂y ∂g2 ∂z # ∂x ∂s ∂x ∂t ∂y ∂s ∂y ∂t ∂z ∂s ∂z ∂t . Once again, D(g ◦f) appears on the left, while the product of Dg and Df is on the right. The same pattern persists in general for any composition. The individual partial derivatives of the composition can be calculated using the Little Chain Rule and then combined as the matrix equation D(g ◦f)(a) = Dg(f(a)) · Df(a). In a way, the matrix version of the chain rule is a bookkeeping device that keeps track of many, many Little Chain Rule calculations! Indeed, since we presented a rigorous proof of the Little Chain Rule in Exercise 5.6 of Chapter 4, we can think of this approach as supplying a proof of the general chain rule as well. Example 6.12. The substitutions x = r cos θ and y = r sin θ convert a smooth real-valued function f(x, y) of x and y into a function of r and θ: w = f(r cos θ, r sin θ). For instance, if f(x, y) = x2y3, then w = (r cos θ)2(r sin θ)3 = r5 cos2 θ sin3 θ. We find general formulas for ∂w ∂r and ∂w ∂θ in terms of ∂f ∂x and ∂f ∂y . As just discussed, we can compute these individual partials using the Little Chain Rule and a dependence diagram (Figure 6.5). w = f x y r θ Figure 6.5: Finding partial derivatives with respect to polar coordinates Hence: ∂w ∂r = ∂f ∂x ∂x ∂r + ∂f ∂y ∂y ∂r = ∂f ∂x cos θ + ∂f ∂y sin θ and ∂w ∂θ = ∂f ∂x ∂x ∂θ + ∂f ∂y ∂y ∂θ = −∂f ∂x r sin θ + ∂f ∂y r cos θ. Example 6.13 (Implicit differentiation). Let f : R2 →R be a smooth real-valued function of two variables, and let S be the level set corresponding to f = c, where c is a constant. That is, S = {(x, y) ∈R2 : f(x, y) = c}. Assume that the condition f(x, y) = c defines y implicitly as a smooth function of x, y = y(x), for some interval of x values. See Figure 6.6. Find the derivative dy dx in terms of f and its partial derivatives. We start with the condition f(x, y(x)) = c that defines y(x). The expression on the left is the composition x 7→(x, y(x)) 7→f(x, y(x)). Call this last value w. The dependence diagram is in Figure 6.7. The derivative dw dx can be computed using the chain rule: dw dx = ∂f ∂x dx dx + ∂f ∂y dy dx = ∂f ∂x + ∂f ∂y dy dx. | Multivariable_Calculus_Shimamoto_Page_177_Chunk3392 |
166 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE Figure 6.6: The level set f(x, y) = c defines y implicitly as a function of x on an interval of x values. w = f x y x Figure 6.7: A dependence diagram for implicit differentiation On the other hand, w has a constant value of c, so dw dx = 0. Hence: ∂f ∂x + ∂f ∂y dy dx = 0. (6.9) Solving for dy dx gives dy dx = −∂f/∂x ∂f/∂y. For example, consider the ellipse x2 9 + y2 4 = 1. It’s the level set of f(x, y) = x2 9 + y2 4 corresponding to c = 1. Hence, by the previous example, along the ellipse, we have: dy dx = −∂f/∂x ∂f/∂y = − 2 9x 1 2y = −4x 9y . Of course, implicit differentiation problems of this type appear in first-year calculus, and the multivariable chain rule is never mentioned. There, to find the derivative along the ellipse, one simply differentiates both sides of the equation of the ellipse and uses the one-variable chain rule. In other words, one computes d dx | Multivariable_Calculus_Shimamoto_Page_178_Chunk3393 |
6.5. EXERCISES FOR CHAPTER 6 167 (b) Let U be an open set in Rn, and let f : U →Rm be a function, where f(x) = (f1(x), f2(x), . . . , fm(x)). Show that, if f is continuous at a point a of U, then so are f1, f2, . . . , fm. 1.2. Let U be an open set in Rn, a a point of U, and f : U −{a} →Rm a function defined on U, except possibly at a. Write out in terms of ϵ’s and δ’s what it means to say that limx→a f(x) = L. Section 2 Differentiability revisited 2.1. Let f : R2 →R2 be given by f(x, y) = (x2 −y2, 2xy). (a) Find Df(x, y). (b) Find Df(1, 2). 2.2. Let f : R3 →R2 be given by f(x, y, z) = | Multivariable_Calculus_Shimamoto_Page_179_Chunk3394 |
168 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE (a) Find Df(x, y) and Dg(x, y). (b) Show that f and g are differentiable at every point (x, y) of R6. Section 3 The chain rule: a conceptual approach 3.1. Let f : R2 →R3 and g: R3 →R2 be given by: f(s, t) = | Multivariable_Calculus_Shimamoto_Page_180_Chunk3395 |
6.5. EXERCISES FOR CHAPTER 6 169 4.3. The spherical substitutions x = ρ sin ϕ cos θ, y = ρ sin ϕ sin θ, and z = ρ cos ϕ convert a smooth real-valued function f(x, y, z) into a function of ρ, ϕ, and θ: w = f(ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ). (a) Find formulas for ∂w ∂ρ , ∂w ∂ϕ , and ∂w ∂θ in terms of ∂f ∂x, ∂f ∂y , and ∂f ∂z . (b) If f(x, y, z) = x2 + y2 + z2, use your answer to part (a) to find ∂w ∂ϕ , and verify that it is the same as the result obtained if you first write w in terms of ρ, ϕ, and θ directly, say by substituting for x, y, and z, and then differentiate that expression with respect to ϕ. 4.4. The substitutions x = s2 −t2 and y = 2st convert a smooth real-valued function f(x, y) into a function of s and t: w = f(s2 −t2, 2st). Find a formula for ∇w = | Multivariable_Calculus_Shimamoto_Page_181_Chunk3396 |
170 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE Assume that the condition f(x, y, z) = c defines x and y implicitly as smooth functions of z on some interval of z values in R, that is, x = x(z) and y = y(z) are functions that satisfy f(x(z), y(z), z) = c. Then the corresponding portion of the level set is parametrized by the path α(z) = (x(z), y(z), z). Show that x′(z) and y′(z) satisfy the matrix equation: " ∂f1 ∂x ∂f1 ∂y ∂f2 ∂x ∂f2 ∂y # x′(z) y′(z) = − " ∂f1 ∂z ∂f2 ∂z # , where the partial derivatives are evaluated at the point (x(z), y(z), z). (Hint: Let w = (w1, w2) = f(x(z), y(z), z).) (b) Let f(x, y, z) = (x −cos z, y −sin z), and let C be the level set of f corresponding to c = (0, 0). Find the functions x = x(z) and y = y(z) and the parametrization α of C described in part (a), and describe C geometrically. Then, use part (a) to help find Dα(z), and verify that your answer makes sense. 4.11. Let f(x, y) be a smooth real-valued function of x and y defined on an open set U in R2. The partial derivative ∂f ∂x is also a real-valued function of x and y (as is ∂f ∂y ). If α(t) = (x(t), y(t)) is a smooth path in U, then substituting for x and y in terms of t converts ∂f ∂x into a function of t, say w = ∂f ∂x(x(t), y(t)). (a) Show that: dw dt = ∂2f ∂x2 dx dt + ∂2f ∂y ∂x dy dt , where the partial derivatives are evaluated at α(t). (b) Now, consider the composition g(t) = (f ◦α)(t). Find a formula for the second derivative d2g dt2 in terms of the partial derivatives of f as a function of x and y and the derivatives of x and y as functions of t. (Hint: Differentiate the Little Chain Rule. Note that this involves differentiating some products.) (c) Let a be a point of U. If α: I →U, α(t) = a+tv = (a1 +tv1, a2 +tv2), is a parametriza- tion of a line segment containing a and if g(t) = (f ◦α)(t), use your formula from part (b) to show that: d2g dt2 = v2 1 ∂2f ∂x2 + 2v1v2 ∂2f ∂x ∂y + v2 2 ∂2f ∂y2 , where the partial derivatives are evaluated at α(t). (Compare this with equation (4.22) in Chapter 4 regarding the second-order approximation of f at a.) 4.12. Let f(x, y) be a smooth real-valued function defined on an open set in R2. The polar co- ordinate substitutions x = r cos θ and y = r sin θ convert f into a function of r and θ: w = f(r cos θ, r sin θ). (a) Show that: ∂2w ∂θ2 = −r cos θ ∂f ∂x −r sin θ ∂f ∂y + r2 sin2 θ ∂2f ∂x2 −2r2 sin θ cos θ ∂2f ∂x ∂y + r2 cos2 θ ∂2f ∂y2 . (Hint: Start with the results of Example 6.12.) | Multivariable_Calculus_Shimamoto_Page_182_Chunk3397 |
6.5. EXERCISES FOR CHAPTER 6 171 (b) Find an analogous expression for the mixed partial derivative ∂2w ∂r ∂θ. (c) Find an analogous expression for ∂2w ∂r2 . (d) If f(x, y) = x2 +y2 +xy, use your answer to part (b) to find ∂2w ∂r ∂θ. Then, verify that it is the same as the result obtained by writing w in terms of r and θ directly by substituting for x and y and then differentiating that expression with respect to r and θ. | Multivariable_Calculus_Shimamoto_Page_183_Chunk3398 |
172 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE | Multivariable_Calculus_Shimamoto_Page_184_Chunk3399 |
Chapter 7 Change of variables We now turn to the counterpart for integrals of the chain rule. It is called the change of variables theorem. For functions of one variable, the corresponding notion is the method of substitution. The one-variable and multivariable versions are expressed in equations that are similar formally, but the approaches one might take to understand where they come from are quite different. It’s exciting when a change in perspective still leads to a result of recognizable form. We make a few remarks comparing the two versions later, after the theorem has been stated. In many respects, the change of variables theorem and the chain rule hold together the foun- dation of the more advanced theory of multivariable calculus. We begin, however, with a concrete example where changing variables just seems like the right thing to do. Example 7.1. Let D be the disk x2 + y2 ≤4 in the xy-plane. Find RR D p x2 + y2 dx dy. For instance, if one wanted to know the average distance to the origin on D, one would evaluate this integral and divide by Area (D) = 4π. Figure 7.1: The disk x2 + y2 ≤4 Setting up the integral with the order of integration dy dx in the usual way (Figure 7.1) gives: ZZ D p x2 + y2 dx dy = Z 2 −2 Z √ 4−x2 − √ 4−x2 p x2 + y2 dy dx. While not impossible, this looks potentially messy. On the other hand, expressing the problem in terms of polar coordinates seems promising for two reasons. • The function to be integrated is simpler: p x2 + y2 = r. • The region of integration is simpler: in polar coordinates, D is described by 0 ≤r ≤2, 0 ≤θ ≤2π, in other words, by a rectangle in the rθ-plane. 173 | Multivariable_Calculus_Shimamoto_Page_185_Chunk3400 |
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