講義請點此

14.1 Functions of Several Variables goo.gl/u9E7VM 1

Chapter 14

Partial Derivatives

14.1

Functions of Several Variables, page 888

GcrflrNwuq8 這一章開始正式進 入多變數函數的微 積分理論, 一開始 還是要先介紹二變 數 函 數 的 定 義 域、 對 應 域等 相 關 意 義。

Definition 1 (page 888). A function f of two variables (雙變數函數) is a rule that assigns to each ordered pair of real numbers (x, y) in a set D ⊂ R2 a unique real number denoted by f (x, y). The set D is the domain (定義域) of f and its range (值域) is the set of values that f takes on, that is, {f (x, y)|(x, y) ∈ D}.

We often write z = f (x, y) to make explicit the value taken on by f at the general point (x, y). The variables x and y are independent variables (獨立變數) and z is the dependent variable (依賴變數).

Example 2 (page 888–889).

(a) Function: f (x, y) = √x+y+1_{x−1 . Domain: D = {(x, y)|x + y + 1 ≥ 0, x 6= 1}.} (b) Function: g(x, y) = x ln(y2_{− x). Domain: D = {(x, y)|x < y}2_}.

x x y y 0 0 −1 −1 x= y2 x+ y + 1 = 0 x= 1 (a) (b)

Figure 1: (a) Domain of f (x, y) = √x+y+1_x−1 . (b) Domain of g(x, y) = x ln(y2_{− x).}

Exercise. Find and sketch the domain of the function f (x, y) = sin−1_(x2_{+ y}2_{− 2).}

Graphs, page 890

One way of visualizing the behavior of a function of two variables is to consider its graph. Definition 3 (page 890). If f is a function of two variables with domain D, then the graph of f (圖形_{) is the set of all points (x, y, z) ∈ R}3 _{such that z = f (x, y) and (x, y) is in D.}

The graph of a function f of two variables is a surface S with equation z = f (x, y). We can visualize the graph S of f as lying directly above or below its domain D in the xy-plane.

2 14.1 Functions of Several Variables goo.gl/u9E7VM 0 1 2 3 4 1 2 3 4 -1 0 1 0 1 2 3 4 0 1 2 3 4 x y z

Figure 2: The graph of f (x, y) = sin xy, 0 ≤ x ≤ 4, 0 ≤ y ≤ 4 and its level curves.

Level Curves, page 893

另外一種認識二變 數函數的方法是在 定義域上畫出它的 等高線, 就像平常 在地圖上看到的那 些線條, 也可以解 讀出地勢 (函數) 的地伏與變化。

Another method for visualizing functions, borrowed from mapmakers, is a contour map on which points of constant elevation are joined to form contour lines (等高線、 輪廓線), or level curves(等位線).

Definition 4 (page 893). The level curves of a function f of two variables are the curves with equations f (x, y) = k, where k is a constant (in the range of f ).

The level curves f (x, y) = k are just the traces of the graph of f in the horizontal plane z = k projected down to the xy-plane. The surface is steep where the level curves are close together. It is somewhat flatter where they are farther apart.



Figure 3: Level curves of the function f (x, y) = 5 − (x − 3)2_{− (y − 3)}2_.

Exercise(page 902). Match the function (a),(b),(c) with its graph (A),(B),(C) and its contour map (I), (II), (III). Give reasons for your choices.

(a) f (x, y) = sin x − sin y (b) g(x, y) = x − y

1 + x2_{+ y}2 (c) h(x, y) = e x_{cos y.}

14.1 Functions of Several Variables goo.gl/u9E7VM 3 -2 0 2 4 -4 -2 0 2 4 -200 20 40 60 -4 -2 0 2 4 -4 -2 0 2 4 -2 -1 01 2 -4 -2 0 2 4 -4 -2 0 2 4 -0.5 0 0.5 x x x y y y z z z (A) (B) (C) -4 -2 0 2 4 -4 -2 0 2 4 -4 -2 0 2 4 -4 -2 0 2 4 -4 -2 0 2 4 -4 -2 0 2 4

(I) (II) (III)

Figure 4: Match functions, graphs, and contour maps.

Functions of Three or More Variables, page 89

微積分課程中主要 以二變數函數與三 變數函數討論, 而 一些理論實際上與 維度無關, 故可推 廣至n-變數函數。

A function of three variables (三變數函數), f , is a rule that assigns to each ordered triple (x, y, z) in a domain D ⊂ R3 _{a unique real number denoted by f (x, y, z) ∈ R. For instance,}

the temperature T at a point on the surface of the earth depends on the longitude x and latitude y of the point and on the time, so we could write T = f (x, y, t).

In general, a function of n variables (n-變數函數) is a rule that assigns a number z = f (x1, x2, . . . , xn) to an n-tuple (x1, x2, . . . , xn) of real numbers. Sometimes are will use vector

notation to write such functions more compactly: If x = (x1, x2, . . . , xn), we often write f (x)

4 14.2 Limits and Continuity goo.gl/DJf3t8

14.2

Limits and Continuity, page 903

2IcaXKJPI58 二變數函數在一點 的極限, 給定誤差 ε >0後,必須要 找到一個範圍內所 有點的函數值與極 限值的差要在誤差 內。

Definition 1 (page 904). Let f be a function of two variables whose domain D includes points arbitrary close to (a, b). Then we say that the limit of f (x, y) as (x, y) approaches (a, b) is L (函數 f (x, y)靠近 (a, b)的極限值是 L) and we write

lim

(x,y)→(a,b)f (x, y) = L

if for every number ε > 0 there is a corresponding number δ > 0 such that if (x, y) ∈ D and 0 <p(x − a)2_{+ (y − b)}2 _{< δ, then |f (x, y) − L| < ε.}

Other notations for the limit are

lim x→a y→b f (x, y) = L and _{f (x, y) → L as (x, y) → (a, b).} 因為二變數函數的 定義 域 是 一 個 區 域, 會有各種可能 靠近 (a, b) 的方 法, 極限存在的意 思是不論用何種方 式靠近, 函數值都 要 接 近 同 一 個 數 字。

The definition refers only to the distance between (x, y) and (a, b). It does not refer to the direction of approach. Therefore, if the limit exists, then f (x, y) must approach the same limit no matter how (x, y) approaches (a, b). Therefore, we get

Property 2 _{(page 905). If f (x, y) → L}1 as (x, y) → (a, b) along a path C1 andf (x, y) → L2

as_{(x, y) → (a, b) along a path C}2, where L16= L2, then lim

(x,y)→(a,b)f (x, y) does not exist.



Example 3 (page 905). Show that lim

(x,y)→(0,0) x2

−y2

x2

+y2 does not exist.

Solution.

Example 4 (page 906). If f (x, y) = _x2xy

+y2, does lim

(x,y)→(0,0)f (x, y) exist?

14.2 Limits and Continuity goo.gl/DJf3t8 5

Example 5 (page 906). If f (x, y) = _xxy2_+y24, does lim

(x,y)→(0,0)f (x, y) exist? HLBkQmXy26Q 從各種方向直接切 入原點的極限值縱 使都一樣, 但仍然 不保證二變數函數 的極限值存在, 因 為選取路徑的方法 可以很任意, 像這 個例子, 考慮拋物 路徑的方式又會有 不同的極限值, 所 以二變數函數的極 限不存在。 Solution.



We can use polar coordinates to find the limit. Note that if (r, θ) are polar coordinates of the point (x, y) with r ≥ 0, then r → 0+ as (x, y) → (0, 0).

Example 6 (page 896). Find lim

(x,y)→(0,0) 3x2 y x2 +y2 if it exists. Solution.

Continuity, page 907

Definition 7 (page 908). A function f of two variables is called continuous at (a, b) if

lim

(x,y)→(a,b)f (x, y) = f (a, b) = f

 lim

(x,y)→(a,b)x,(x,y)→(a,b)lim y

 .

We say f is continuous on D if f is continuous at every point (a, b) in D.

A polynomial function of two variables (or polynomial 二變數多項式, for short) is a sum of terms of the form cxm_yn_{, where c is a constant and m and n are nonnegative integers.}

A rational function (有理函數) is a ratio of polynomials. All polynomials are continuous on R2. Any rational function is continuous on its domain because it is a quotient of continuous functions.

6 14.2 Limits and Continuity goo.gl/DJf3t8

Example 8 (page 908). Where is the function f (x, y) = x2−y2

x2_+y2 continuous?

Solution. The function f (x, y) is discontinuous at (0, 0) because it is not defined there. Since f (x, y) is a rational function, it is continuous on its domain D = {(x, y)|(x, y) 6= (0, 0)}. Example 9 (page 908). Let

f (x, y) = ( _x2

−y2

x2_+y2 if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0) .

Here f (x, y) is defined at (0, 0) but f (x, y) is still discontinuous there because lim

(x,y)→(0,0)f (x, y)

does not exist. (See Example 3.) Example 10. Let

f (x, y) =

( _3x2_y

x2_+y2 if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0) .

We know f (x, y) is continuous for (x, y) 6= (0, 0) since it is equal to a rational function there. From Example 6, we have

Therefore, , and so it is continuous on .

UkcGmlBL7oY 這裡列出合成函數 的連續性。 例題的 函數在多變數函數 理論中算是蠻重要 的函數。

Property 11(page 909). If f is a continuous function of two variables and g is a continuous function of a single variable that is defined on the range of f , then the composite function h = g ◦ f defined by h(x, y) = g(f (x, y)) is also a continuous function.

Example 12(page 909). Where is the function h(x, y) = tan−1 y

x
continuous?

Solution. The function f (x, y) = y_x is a rational function and therefore continuous except on . The function g(t) = tan−1_{t is continuous everywhere, so the composition function}

h(x, y) = g(f (x, y)) = tan−1 y

x
is continuous except where .

Functions of Three or More Variables, page 909

超過三個以上的變 數之函數理論也可 以同理類推。

Everything that we have done in this section can be extended to functions of three or more variables. The notation

lim

(x,y,z)→(a,b,c)f (x, y, z) = L

means that the values of f (x, y, z) approach the number L as the point (x, y, z) approaches the point (a, b, c) along any path in the domain of f . The function f is continuous at (a, b, c) if

lim

(x,y,z)→(a,b,c)f (x, y, z) = f (a, b, c) = f



lim

(x,y,z)→(a,b,c)x,(x,y,z)→(a,b,c)lim y,(x,y,z)→(a,b,c)lim z

 . For a function of n variables, we can write these definitions in a single compact form by vector notation. For instance, let x = (x1, x2, . . . , xn), a = (a1, a2, . . . , an), and f (x) is a

function of n variable, The function f is continuous at a if lim x_→af (x) = f (a) = f  lim x_→a x.

14.3 Partial Derivative goo.gl/xSBhh5 7

14.3

Partial Derivative, page 911

cyy0njb1qaY 若要研究二變數函 數, 當中的一個思 想是: 是否能夠利 用單變數函數所學 到的東西來理解多 變數。 雖然在極限 的討論似乎處處受 挫, 但是在微分的 層級下,在 「某些」 情況又可以如此處 理。 接下來的兩個 單元就是要研究這 個問題。 偏導數的意思就是 固 定 其 它 的 變 數, 讓函數變成單變數 函數, 去觀察函數 對於這個變數的變 化。

Definition 1 (page 913). If f is a function of two variables x and y, suppose we let only x vary while keeping y fixed, say y = y0, then g(x) = f (x, y0) is a function of a single variable

x. If g(x) has a derivative at x = x0, then we call it the partial derivative (偏導數) of f with

respect to x at (x0, y0) and denote it by fx(x0, y0). Thus

fx(x0, y0) = g′(x0) = lim h→0 g(x0+ h) − g(x0) h = limh→0 f (x0+ h, y0) − f (x0, y0) h .

Similarly, the partial derivative (偏導數) of f with respect to y at (x0, y0) and denote it by

fy(x0, y0), is obtained by keeping x fixed, say x = x0, and finding the ordinary derivative at

y = y0 of the function ˜g(y) = f (x0, y):

fy(x0, y0) = ˜g′(y0) = lim h→0 ˜ g(y0+ h) − ˜g(y0) h = limh→0 f (x0, y0+ h) − f (x0, y0) h .

Definition 2 (page 913). If f is a function of two variables, its partial derivatives (偏導函數) are the functions fx and fy defined by

fx(x, y) = lim h→0 f (x + h, y) − f (x, y) h , fy(x, y) = lim h→0 f (x, y + h) − f (x, y) h . 偏導數的記號有很 多種,也都很常見, 最後兩個記號與方 向導數有關, 可以 到 14.6的時候再 一起統整。

Notations for Partial Derivatives. If z = f (x, y), we write

fx(x, y) = fx= ∂f ∂x = ∂ ∂xf (x, y) = ∂z ∂x = f1 = Dxf = D1f, fy(x, y) = fy= ∂f ∂y = ∂ ∂yf (x, y) = ∂z ∂y = f2 = Dyf = D2f. Rule for Finding Partial Derivative of z = f (x, y).

(1) To find fx, regard y as a constant and differentiate f (x, y) with respect to x.

(2) To find fy, regard x as a constant and differentiate f (x, y) with respect to y.



Example 3 (page 914). If f (x, y) = x3+ x2y3_{− 2y}2, then (a) fx(x, y) =

fx(2, 1) =

(b) fy(x, y) =

8 14.3 Partial Derivative goo.gl/xSBhh5

Interpretations of Partial Derivatives, page 915

pkKfK7Pu-to 回想單變數函數在 一點求導, 幾何意 義就是在問函數圖 形在那一點的切線 斜率。 而偏導數的 幾何意義也可以類 似地去想它, 二變 數函數圖形與平面 x= x0或是y= y0相交,得到單變 數函數的圖形, 然 後去問在該點的切 線斜率。

The partial derivatives fx(x0, y0) and fy(x0, y0) can be interpreted geometrically as the slopes

of the tangent lines at P (x0, y0, f (x0, y0)) to the trace C1 and C2 of the surface S in the planes

y = y0 and x = x0. x y z C1 C₂ S (x0, y0,0) P

Figure 1: Geometric meaning of partial derivatives.

Example 4 (page 917). If f (x, y) = sin_1+yx , calculate ∂f_∂x and ∂f_∂y. Solution.

Example 5(page 917). Find ∂z_∂x and ∂z_∂y if z is defined implicitly as a function of x and y by the equation x3+ y3+ z3+ 6xyz = 1.

Solution. Exercise (page 927). If f (x, y) = x e sin(x2 y) (x2_{+ y}2₎32 , find fx(1, 0).

Functions of More Than Two Variables, page 917

mmkk1sfm23Y

認識多變數函數的 偏導函數之記號。

If z = f (x1, x2, . . . , xn) is a function of n variables, its partial derivative with respect to the

i-th variable xi is ∂z ∂xi = lim h→0 f (x1, . . . , xi−1, xi+ h, xi+1, . . . , xn) − f (x1, . . . , xi, . . . , xn) h . We also write ∂z ∂xi = ∂f ∂xi = fxi = fi = Dif.

14.3 Partial Derivative goo.gl/xSBhh5 9

Higher Derivatives, page 918

高次偏導函數的記 號要小心, 以函數 為主體, 由內而外 的順序操作, 所以 用下標註記的時候 順 序 是 由 左至 右, 用 ∂ 的符號註記 時是由右至左。 當 函數不是很好的時 候, 有可能偏微分 的順序不同而會有 不同的值。

If f is a function of two variables, then its partial derivatives fx and fy are also functions of

two variables, so we can consider their partial derivatives (fx)x, (fx)y, (fy)x, and (fy)y, which

are called the second partial derivatives (二次偏導數) of f . If z = f (x, y), we use the following notation: (fx)x= fxx = f11= ∂ ∂x  ∂f ∂x  = ∂ 2_f ∂x2 = ∂2_z ∂x2, (fx)y = fxy = f12= ∂ ∂y  ∂f ∂x  = ∂ 2_f ∂y∂x = ∂2z ∂y∂x, (fy)x= fyx= f21= ∂ ∂x  ∂f ∂y  = ∂ 2_f ∂x∂y = ∂2z ∂x∂y, (fy)y = fyy = f22= ∂ ∂y  ∂f ∂y  = ∂ 2_f ∂y2 = ∂2_z ∂y2.



Exercise. Let r(x, y) =px2_{+ y}2_{. For (x, y) 6= (0, 0), compute r}

x, ry, rxx, rxy, ryx, and ryy. 定理告知: 當函數 二次偏微分後的函 數都要是連續函數 時, 偏微分順序才 可以交換。 這一節 的最後會有一個例 子說明: 存在函數 其二次偏微分的順 序交換不同。

Clairaut’s Theorem (page 919). Suppose f is defined on a disk D that contains the point (x0, y0). If the functions fxy and fyx are both continuous on D, then

fxy(x0, y0) = fyx(x0, y0).



Exercise. Let f (x, y) = x

3_{− xy}2

x2_{+ y}2 .

(a) Determine the value f (0, 0) such that f (x, y) is continuous at (0, 0). (b) Find fx(x, y), fx(x, y), fx(0, 0) and fy(0, 0).

(c) Compute fxy(0, 0) and fyx(0, 0).

Partial Differential Equations, page 920

偏微分方程式的研 究是近代數學討論 的一個重點, 不論 方程式在工程科學 上 有重 要 的 應 用, 數學上方程式解的 存在、 唯一、 正則 性也是廣泛地被探 討。

Partial derivatives occur in partial differential equations (偏微分方程) that express certain physical laws. For instance,

(a) u = u(x, y), ∆u ≡ ∂∂x2u2 +

∂2_y

∂y2 = 0: Laplace’s equation (拉普拉斯方程).

(b) u = u(t, x),∂u_∂t = ∂_∂x2u2: heat equation (熱傳導方程).

(c) u = u(t, x),∂_∂t2u2 =

∂2

u

10 14.3 Partial Derivative goo.gl/xSBhh5

Example 6 (page 927). Let f (x, y) = ( _x3 y−xy3 x2_+y2 if (x, y) 6= (0, 0) 0 if (x, y) = (0, 0) . YvOYwcKytq8 這個例子將檢視偏 導數的定義, 還有 二次偏微分順序交 換的異同。 由於這 個函數的二次偏微 分 在 原點 不 連 續, 即Clairaut定理 的條件不成立, 而 這個例子告知順序 互換在函數不太好 的時候的確有差。

(a) Find fx(x, y) and fy(x, y) when (x, y) 6= (0, 0).

(b) Find fx(0, 0) and fy(0, 0).

(c) Find fxy(x, y) and fyx(x, y) when (x, y) 6= (0, 0).

(d) Find fxy(0, 0) and fyx(0, 0).

(e) Do the results of (c) and (d) contradict Clairaut’s Theorem? Solution.

(a) Direct computation gives

fx(x, y) = (x2+ y2)(3x2_{y − y}3_{) − (x}3_{y − xy}3)(2x) (x2_{+ y}2₎2 = x4y + 4x2y3_{− y}5 (x2_{+ y}2₎2 fy(x, y) = (b) By definition, we have fx(0, 0) = fy(0, 0) =

(c) Direct computation gives

fxy(x, y) = x6+ 9x4y2_{− 9x}2y4_{− y}6 (x2_{+ y}2₎3 fyx(x, y) = x6+ 9x4y2_{− 9x}2y4_{− y}6 (x2_{+ y}2₎3 . (d) By definition, we have fxy(0, 0) = fyx(0, 0) =

(e) Results of (c) and (d) don’t contradict to Clairaut’s Theorem because both fxy(x, y)

and fyx(x, y) are not continuous at (0, 0). We cant take path C1(x) = (x, 0), x 6= 0 and

C2(y) = (0, y), y 6= 0 to get fxy(x, y)|C1(x) = fyx(x, y)|C1(x) ≡ 1 and fxy(x, y)|C2(y) =

fyx(x, y)|C2(y)≡ −1. That is, lim

(x,y)→(0,0)fxy(x, y) and(x,y)→(0,0)lim fxy(x, y) do not exist.

14.4 Tangent Planes and Linear Approximations goo.gl/1nWupp 11

14.4

Tangent Planes and Linear Approximations,

page 927

Tangent Planes, page 928

vKAHbCqczHg

函數圖形在一點的 切平面是由坐標曲 線在那一點的切向 量所張出的平面。

Definition 1 (page 928). Suppose that a surface S has equation z = f (x, y), where f has continuous partial derivatives, and let P (x0, y0, z0) be a point on S. Let C1 and C2 be the

curves obtained by intersecting the vertical planes y = y0 and x = x0 with the surface S. Let

T1 and T2 be the tangent lines to the curves C1 and C2 at P . Then the tangent plane (切平

面) to the surface S at the point P is defined to be the plane that contains both tangent lines T1 and T2. x y z C1 C₂ S (x0, y0,0) P T₁ T2

Figure 1: The tangent plane contains the tangent lines T1 and T2.

函數圖形在一點的 切平面表示法可以 由 兩 種 觀 點 去 思 考, 一種是由平面 中的直線點斜式類 推, 另一種是由法 向量 的 觀 點 出 發, 用兩個切向量作外 積而得法向量。

An equation of the tangent plane to the surface z = f (x, y) at P (x0, y0, z0) is

z − z0 = fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0), or (點斜式)

fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0) − (z − z0) = 0 (用法向量看待)

Remark 2. Since tangent vectors to C1 and C2 at P are e1 = 1 i + 0 j + fx(x0, y0) k and

e2 = 0 i + 1 j + fy(x0, y0) k, a normal vector of the tangent plane is

n= e1× e2= −fx(x0, y0) i − fy(x0, y0) j + 1 k

// fx(x0, y0) i + fy(x0, y0) j − 1 k.



Example 3. Find the equation of the tangent plane of the surface z = ex−y _{at the point}

P (1, 1, 1).

12 14.4 Tangent Planes and Linear Approximations goo.gl/1nWupp

Linear Approximations, page 929

WMSKqix7Wp8 函數在一點的線性 估計就是用切平面 上 z 分量的取值 估計 原 函 數 的 值。 在 單變 數 函 數 理 論, 求導與微分同 義, 但是在多變數 函數理論, 偏導與 可微分將分成兩個 不同的概念。

Definition 4 (page 929). An equation of the tangent plane to the graph of the function z = f (x, y) at P (x0, y0, z0) is z − z0 = z − f (x0, y0) = fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0).

The linear function whose graph is this tangent plane, namely,

L(x, y) = f (x0, y0) + fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0)

is called linearization (線性化) of f at (x0, y0) and the approximation

f (x, y) ≈ f (x0, y0) + fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0) (1)

is called the linear approximation (線性估計) or tangent plane approximation of f at (x0, y0).

這 個 例 子 是 說 明: 雖然在x與y 方 向 有 好 的線 性 估 計, 但是兩者線性 組合出來的值可能 無法好好的近似函 數。

Example 5 (page 930). Consider the function f (x, y) =

( _xy

x2_+y2 if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0). (a) fx(0, 0) =

(b) fy(0, 0) =

(c) We take the path C1(t) = (t, t), t 6= 0, the function f (x, y)|C1(t) =

(d) A function of two variables can behave badly even through both of its partial derivatives exist. To rule out such behavior, we will define a differentiable function (可微分函數) of two variable. F3ViScKLk38 二變數函數可微分 的意思就是可以用 切平面對應的線性 函數做函數的線性 估計。

Definition 6 (page 931). If z = f (x, y), then f is differentiable (可微分的) at (x0, y0) if

∆x = x − x0, ∆y = y − y0, then f (x, y) satisfies

lim

(∆x,∆y)→(0,0)

f (x, y) − f (x0, y0) − fx(x0, y0)∆x − fy(x0, y0)∆y

p(∆x)2_{+ (∆y)}2 = 0.

Sometimes it is hard to use the definition to check the differentiability of a function, but the next theorem provides a convenient sufficient condition for differentiability.

當二變數函數的偏 微分是連續時才可 以線性估計。

Theorem 7 (page 932). If the partial derivatives fx and fy exist near (x0, y0) and are

con-tinuous at(x0, y0), then f is differentiable at (x0, y0).



14.4 Tangent Planes and Linear Approximations goo.gl/1nWupp 13

Differentials, page 932

For a differentiable function of two variables, z = f (x, y), we define the differentials (微分) dx and dy to be independent variables; that is, they can be given any values. Then the differential dz, also called the total differential (全微分), is defined by

dz = df = fx(x, y) dx + fy(x, y) dy = ∂f ∂xdx + ∂f ∂ydy = ∂z ∂xdx + ∂z ∂ydy. (2) If we take dx = ∆x = x − x0 and dy = ∆y = y − y0 in (2), then the differential of z is dz =

fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0), so in notation of differentials, the linear approximation

(1) can be written as f (x, y) ≈ f (x0, y0) + dz.

Figure 2 shows the geometric interpretation of the differential dx and the increment ∆z: dz represents the change in height of the tangent plane, whereas ∆z represents the change in height of the surface z = f (x, y) when (x, y) changes from (x0, y0) to (x0+ ∆x, y0+ ∆y).

x y z ∆x = dx ∆y = dy (x0, y0,0) (x0+ ∆x, y0+ ∆y, 0) (x0, y0, f(x0, y0)) (x0+ ∆x, y0+ ∆y, f (x0+ ∆x, y0+ ∆y)) dz ∆z z = f (x, y)

Figure 2: Geometric interpretation of the differential dz and the increment ∆z.

Example 8 (page 933). The base radius and height of a right circular cone are measured as 10 cm and 25 cm, respectively, with a possible error in measurement of as much as 0.1 cm in each. Use differentials to estimate the maximum error in the calculated volume of the cone. Solution.

Functions of Three or More Variables, page 932

Linear approximations, differentiability, and differentials can be defined in a similar manner for functions of more than two variables.

14 14.4 Tangent Planes and Linear Approximations goo.gl/1nWupp Example 9. Let f (x, y) = ( _x2 y x2_+y2 if (x, y) 6= (0, 0) 0 if (x, y) = (0, 0). SYgeaTX1B0k 這個例子也是重申 可微分與線性估計 之間的關係, 這個 函數在原點不是可 微分的, 所以函數 在原點做出來切平 面對應到的線性函 數在其它方向來看 無法確實呈現其線 性估計。

(a) f (x, y) is continuous at (0, 0) because

(b) fx(0, 0) =

(c) fy(0, 0) =

(d) For (x, y) 6= (0, 0), ∂f∂x =

(e) ∂f_∂x(x, y) is not continuous at (0, 0) because we take the path C1(t) = (t, t), t 6= 0, then

the function fx(x, y)|C1(t) =

(f) Compute for (x, y) 6= (0, 0)

f (x, y) − f (0, 0) − fx(0, 0)x − fy(0, 0)y =

and take the path C1(x) = (x, x), x 6= 0, we find

f (x, y) − f (0, 0) − fx(0, 0)x − fy(0, 0)y|C1(x)=

(g) From (e) and (f), we know that L(x, y) = f (0, 0) + fx(0, 0)x + fy(0, 0)y ≡ 0 is not a

14.5 The Chain Rule goo.gl/YA9B9n 15

14.5

The Chain Rule, page 937

pyUgdh2cf8A 鏈鎖律是二變數函 數微分的重點, 必 須清楚所有變量之 間的層級關係, 確 實討 論 其 變 化 率。 第一類型的鏈鎖律 若要用幾何的方式 去解釋, 則是探討 函數圖形上的曲線 之高度的變化率。

The Chain Rule, Case 1 (page 938). Suppose that z = z(x, y) is a differentiable function of x and y, where x = x(t) and y = y(t) are both differentiable function of t. Then z is a differentiable function of t and

dz dt = ∂z ∂x dx dt + ∂z ∂y dy dt.



Example 1 (page 938). If z = x2y + 3xy4, where x = sin 2t and y = cos t, find dz_dt when t = 0. Solution.

Example 2. Find the second derivative d_dt22z.

Solution. B1grFNAmbcI 第二類的鏈鎖律討 論層級是二變數與 二 變 數 之 間 的 變 化, 在幾何意義上 是在觀察函數在坐 標變換下的轉換關 係。

The Chain Rule, Case 2 (page 939). Suppose that z = z(x, y) is a differentiable function of x and y, where x = x(s, t) and y = y(s, t) are differentiable functions of s and t. Then

∂z ∂s = ∂z ∂x ∂x ∂s + ∂z ∂y ∂y ∂s, ∂z ∂t = ∂z ∂x ∂x ∂t + ∂z ∂y ∂y ∂t.



Case 2 of the Chain Rule contains three types of variables: s and t are independent variables, x and y are intermediate variables, and z is the dependent variable.

Example 3. If z = ex_{sin y, where x = st}2 _{and y = s}2_{t. Find} ∂z ∂s and

∂z ∂t.

Solution.

Example 4. Let z = y + f (x2_{− y}2_{) and f be a differentiable function of single variable. Find}

y∂z_∂x+ x∂z_∂y. Solution.

The Chain Rule, General Version (page 940). Suppose that u is a differentiable function of the n variables x1, x2, . . . , xn, and each xi is a differentiable function of the m variables

t1, t2, . . . , tm. Then u is a function of t1, t2, . . . , tm and

∂u ∂ti = ∂u ∂x1 ∂x1 ∂ti + ∂u ∂x2 ∂x2 ∂ti + · · · + ∂u ∂xn ∂xn ∂ti = n X j=1 ∂u ∂xj ∂xj ∂ti .



16 14.5 The Chain Rule goo.gl/YA9B9n

Coordinates Changes

In R2, denote (x, y) as the Cartesian coordinates and (r, θ) as the polar coordinates. We know relations between these coordinates are

( x = r cos θ y = r sin θ. ( r2= x2+ y2 tan θ = y_x .

So we know x = x(r, θ), y = y(r, θ) and r = r(x, y), θ = θ(x, y), and hence

x = x(r(x, y), θ(x, y)) and y = y(r(x, y), θ(x, y)).

Since x and y are independent variables, we have                  ⇒

So partial derivatives of coordinates changes form inverse matrices. Now we check this relation by computing partial derivatives directly.



dx ·dxdy = 1;多變數的情形: 矩陣相乘為單位矩陣。

14.5 The Chain Rule goo.gl/YA9B9n 17

Example 5. Consider z = f (x, y), where all the second partial derivatives of f are continuous. Let x = r cos θ and y = r sin θ.

7xafpX9Dsbo 更進一步地, 這裡 也用直角坐標與極 坐標的轉換關係學 習函數的二次偏微 分的鏈鎖律。 養成 好習慣, 先把結構 寫出來, 再一個一 個重問鏈鎖率, 這 樣比較不會漏寫或 出錯。 (a) Express ∂2 r ∂x2 and ∂2 θ ∂x2 in terms of r and θ.

(b) Express fxx in terms of r, θ, fr, fθ, frr, frθ, and fθθ.

Solution.

Implicit Differentiation (

隱函數微分

₎

The Chain Rule can be used to give a more complete description of the process of implicit differentiation. Suppose that an equation of the form F (x, y) = 0 defines y implicitly as a differentiable function of x, that is, y = y(x), where F (x, y(x)) = 0 for all x in the domain of y. If F is differentiable, we can apply the Chain Rule to differentiate both side of the equation F (x, y(x)) = 0 with respect to x to get

⇒

Implicit Function Theorem (page 942). If F (x, y) is defined on a disk containing (x0, y0),

where F (x0, y0) = 0, Fy(x0, y0) 6= 0, and Fx and Fy are continuous on the disk, then the

equation F (x, y) = 0 defines y as a function of x near the point (x0, y0) and the derivative of

this function is _dxdy _{= −}Fx

18 14.5 The Chain Rule goo.gl/YA9B9n

Exercise (page 946). If f (x, y) = 0 define y as a function of x, show that d2y dx2 = − fxxfy2− 2fxyfxfy+ fyyfx2 f3 y .

Now we suppose that z is given implicitly as a function z = z(x, y) by an equation of the form F (x, y, z) = 0. This means that F (x, y, z(x, y)) = 0 for all (x, y) in the domain of z. If F and z are differentiable, then we can use the Chain Rule to differentiable the equation F (x, y, z(x, y)) = 0 as follows: If ∂F ∂z 6= 0, we solve ∂z ∂x and obtain ∂z ∂x = ∂z ∂y = VqBMCTRPhPA 隱函數定理在分析 學中是一個非常重 要的定理, 在隱函 數的式子當中, 一 些變數之間的關係 隱諱不明, 這個定 理的重點是告知在 什麼情況下, 某個 變數可以完全用其 它變數的函數形式 表達, 也就是可以 清楚地知道哪些可 以設定成獨立的變 數, 哪些是依賴變 數。

Implicit Function Theorem (page 943). If F (x, y) is defined within a sphere containing (x0, y0, z0), where F (x0, y0) = 0, Fz(x0, y0, z0) 6= 0, and Fx, Fy and Fz are continuous inside

the sphere, then the equationF (x, y, z) = 0 defines z as a function of x and y near the point (x0, y0, z0) and the partial derivatives of this function are ∂z_∂x = −_FFx_z and ∂z_∂y = −F_Fy_z.

Example 6 (page 943). Find ∂z_∂x and ∂z_∂y if F (x, y, z) = x3_{+ y}3_{+ z}3_{+ 6xyz − 1 = 0.}

14.6 Directional Derivatives goo.gl/pj799z 19

14.6

Directional Derivatives and the Gradient

Vec-tor, page 946

Directional Derivatives, page 946

ZFOzRYkaego 方向導數顧名思義 就是去研究函數沿 著某個方向的變化 率, 利用定義域上 的單位向量指定方 向, 按照分量的比 例分配增加的程度 再算 差 商 的 極 限。 對應到的幾何意義 如左圖所示: 函數 的圖形限制在通過 P且包含指定向量 的平面, 研究這個 函數圖形在 P 點 的切線斜率。

Definition 1 (page 947). The directional derivative (方向導數) of f (x, y) at (x0, y0) in the

direction of a unit vector u = (a, b) is

Duf (x0, y0) = lim h→0

f (x0+ ha, y0+ hb) − f (x0, y0)

h if this limit exists.

x y z ha hb S (x0, y0,0) P u Figure 1: Directional derivative.



Theorem 2 (page 948). If f is a differentiable function of x and y, then f has a directional derivative in the direction of any unit vector u= (a, b) and

Duf (x, y) = fx(x, y)a + fy(x, y)b = (fx(x, y), fy(x, y)) · (a, b).

Proof. Define g(h) = f (x(h), y(h)) = f (x0+ha, y0+hb), then by the definition of a directional

derivative and the Chain Rule, we have

Duf (x0, y0) = g′(0) =  ∂f ∂x dx dh+ ∂f ∂y dy dh     h=0 = fx(x0, y0)a + fy(x0, y0)b = (fx(x0, y0), fy(x0, y0)) · (a, b).

If the unit vector u makes an angle θ with the positive x-axis, then we can write u = (cos θ, sin θ) and the directional derivative becomes

20 14.6 Directional Derivatives goo.gl/pj799z

The Gradient Vector, page 949

FbOc8tfJOl4 函數對每個變數偏 微分所組成的向量 非常重要, 我們把 它取名為梯度。 所 以前一頁的定理是 說: 函數的方向導 數會是梯度與該方 向的內積。

Definition 3(page 950). If f is a function of two variables x and y, then the gradient (梯度) of f is the vector function ∇f or grad f defined by

∇f (x, y) = grad f (x, y)def.= (fx(x, y), fy(x, y)) =

∂f ∂xi+

∂f ∂yj.

Theorem 4 (page 950). If f is a differentiable function of x and y, then f has a directional derivative in the direction of any unit vector u= (a, b) and

Duf (x, y) = ∇f (x, y) · u.





Maximizing the Directional Derivative, page 952

梯度向量的一個重 要意義是它會指向 函數增加最快的方 向。 這個性質由兩 向量內積的幾何意 義可以得知。

Theorem 5(page 952). Suppose f is a differentiable function of two variables. The maximum value of the directional derivative Duf (x, y) is |∇f (x, y)| and it occurs when u has the same

direction as the gradient vector _{∇f (x, y).} Proof. _{Since |u| = 1, we have}

Duf = ∇f · u = |∇f ||u| cos θ = |∇f | cos θ,

where θ is the angle between ∇f and u. The maximum value of cos θ is 1 and this occurs when θ = 0. Therefore the maximum value of Duf is ∇f when u is the same direction as

∇f .



Example 6. Let f (x, y) = 2x2_{− xy + y}2_{− 2x + y.}

(a) Find the directional derivative Duf (p), where p = (0, 0) and u =

√

3 2 ,12

 . (b) Find the unit vector v that the directional derivative Dvf (p) is maximal.

Solution.

Functions of Three Variables, page 950

0-toVDDT2fU

Using vector notation, we can write the directional derivative in the compact form:

Duf (x0) = lim h→0

f (x0+ hu) − f (x0)

h = ∇f (x0) · u. where x0 = (x0, y0) if n = 2 and x0= (x0, y0, z0) if n = 3.

14.6 Directional Derivatives goo.gl/pj799z 21

Tangent Planes to Level Surfaces, page 954

梯度向量的另一個 重 要 的 幾 何 意 義 是: 梯度向量與等 位面 (或等高線)

垂直。

Suppose S is a level surface with equation F (x, y, z) = k, and let P (x0, y0, z0) be a point on

S. Let C be any curve that lies on the surface S and passes through the point P , that is, C is parameterized by r(t) = (x(t), y(t), z(t)) and r(t0) = (x(t0), y(t0), z(t0)) = (x0, y0, z0). Since

C lies on S, we know

F (x(t), y(t), z(t)) = k. (3)

If x, y, and z are differentiable functions of t and F is also differentiable, then we can use the Chain Rule to differentiate both sides of equation (3) as follows:

∂F ∂x dx dt + ∂F ∂y dy dt + ∂F ∂z dz dt = 0 ⇒ ∇F · r ′_{(t) = 0.}

In particular, when t = t0, we have ∇F (x0.y0, z0) · r′(t0) = 0

The gradient vector at P , ∇F (x0, y0, z0), is perpendicular to the tangent vector

r′_{(0) to any curve C on S that passes through P .}

Definition 7 _{(page 954). If ∇F (x}0, y0, z0) 6= 0, it is therefore natural to define the tangent

plane to the level surfaceF (x, y, z) = k at P (等位面的切平面) as the plane that passes through P and has normal vector ∇F (x0, y0, z0). The equation is

Fx(x0, y0, z0)(x − x0) + Fy(x0, y0, z0)(y − y0) + Fz(x0, y0, z0)(z − z0) = 0. (4)



Definition 8(page 954). The normal line (法線) to S at P is the line passing through P and perpendicular to the tangent plane. The direction of the normal line is the gradient vector ∇F (x0, y0, z0), and so its symmetric equations are

x − x0 Fx(x0, y0, z0) = y − y0 Fy(x0, y0, z0) = z − z0 Fz(x0, y0, z0) . (5)



Example 9_{(page 941). Find the equations of the tangent plane and normal line at P (−2, 1, 3)} to the ellipsoid x₄2 + y2+z₉2 = 3.

22 14.6 Directional Derivatives goo.gl/pj799z

Significance of the Gradient Vector, page 955

Vs_Rhm-QaKs 這個地方就再次統 整 梯 度 的 幾 何 意 義, 對二變數函數 來說, 它的等高線 (畫 在 定 義 域 上) 是一條曲線, 而梯 度向量 (在定義域 的平面上) 與曲線 處處垂直。 對三變 數函數來說, 定義 域是三度空間中的 一個區域, 函數的 等 位 面 是 一 個 曲 面, 而梯度向量與 曲面處處垂直。

Consider a function of two variables f (x, y).

(1) The gradient vector ∇f is orthogonal to the level curve f (x, y) = k. (2) The gradient vector ∇f gives the direction of fastest increases of f .

x y

f(x, y) = k

curve of steepest ascent

p ∇f(p)

Figure 2: The gradient vector is orthogonal to the level curve.

Consider a function of three variables F (x, y, z).

(1) The gradient vector ∇F is orthogonal to the level surface F (x, y, z) = k. (2) The gradient vector ∇F gives the direction of fastest increases of F .

x y z F(x, y, z) = k p ∇F (p)

14.6 Directional Derivatives goo.gl/pj799z 23

Intersection of Two Surfaces

kUFrgcN_mTA 回想高中所學兩個 空間中的平面的交 線之方向向量, 計 算方式就是直接把 兩個平面的法向量 外積而得; 這裡也 是一樣, 兩個曲面 的交集若交出一條 空間曲線, 這條曲 線的切線找法, 也 可以直接把曲面各 自的梯度向量外積 而得。

Suppose S1and S2are two surfaces determined by two equations F (x, y, z) = 0 and G(x, y, z) =

0, respectively. The intersection of two surfaces is a space curve called C. Suppose that r(t) is a parametric equation of the space curve C and r(t0) = P , then r′(t0) is parallel to

∇F (p) × ∇G(p). p C ∇F (p) ∇G(p) r′_(t 0) S1: F = 0 S2 : G = 0

Figure 4: Intersection of two surfaces.

Example 10. Find the parametric equation of the tangent line to the curve of intersection of the surfaces x2_{+ 2y}2_{+ z}2 _{= 4 and x}2_{+ y}2_{− z}2 _{= 1 at the point (1, 1, 1).}

24 14.7 Maximum and Minimum Values goo.gl/8kiWHQ

14.7

Maximum and Minimum Values, page 959

uS8c92OCeQg 和單變數函數的情 形類似, 極值的意 思 也 分 成 兩 種 情 況: 局部極值與絕 對極值。 局部極值 是 存 在 一 個 鄰 域, 在鄰域內的函數值 與討論的函數值有 大小關係; 而絕對 極值是先指定一個 範圍, 然後在那個 範圍內的所有函數 值與討論的函數值 有大小關係。

Definition 1(page 960). A function of two variables has a local maximum (局部極大值處) at (x0, y0) if f (x, y) ≤ f (x0, y0) when (x, y) is near (x0, y0). (This means that f (x, y) ≤ f (x0, y0)

for all points (x, y) in some disk with center (x0, y0).) The number f (x0, y0) is called a local

maximum value (局部極大值_{). If f (x, y) ≥ f (x0}, y0) when (x, y) is near (x0, y0), then f has a

local minimum(局部極小值處) at (x0, y0) and f (x0, y0) is a local minimum value (局部極小值).

Definition 2 (page 960). If the inequalities in Definition 1 hold for all points (x, y) in the domain of f , then f has an absolute maximum (最大值) or absolute minimum (最小值) at (x0, y0). (a) x (b) x y y z z

Figure 1: (a) Local and absolute minimum. (b) Local and absolute maximum.

定理說明局部極值 的必要條件是函數 在該點的所有偏導 數都是零。 這個定 理雖然不是充份條 件, 但它提供了找 尋極值點以進行極 值分類的資訊。

Theorem 3(page 960). If f has a local maximum or minimum at (x0, y0) and the first-order

partial derivatives off exist there, then fx(x0, y0) = 0 and fy(x0, y0) = 0.

Proof. Let g(x) = f (x, y0). If f has a local maximum (or minimum) at (x0, y0), then g(x)

has a local maximum (or minimum) at x0, so by Fermat’s Theorem, we get g′(x0) = 0 =

fx(x0, y0). Similarly, by applying Fermat’s Theorem to the function ˜g(y) = g(x0, y0), we

obtain g′_(y 0) = 0 = fy(x0, y0).



Definition 4(page 960). A point (x0, y0) is called a critical point (臨界點) or stationary point

(平穩點、 駐點) of f if fx(x0, y0) = 0 and fy(x0, y0) = 0, or if one of these partial derivatives

does not exist.



14.7 Maximum and Minimum Values goo.gl/8kiWHQ 25

Example 5 (page 960). Consider f (x, y) = x2+ y2_{− 2x − 6y + 14, then}

kBxxO7emlqU 先從二次函數介紹 兩種 臨 界 點 的 長 相, 其中一個會產 生局部極小值 (若 函數全部加上負號 則變 成 局 部 極 大 值), 另一個會形 成鞍點。 fx(x, y) = 2x − 2 fy(x, y) = 2y − 6.

These partial derivative are equal to 0 when x = 1 and y = 3, so the only critical point is (1, 3). Since f (x, y) = 4 + (x − 1)2 + (y − 3)3 ≥ 4 for all x and y, f (1, 3) = 4 is a local minimum, and in fact it is the absolute minimum of f .

The graph of f is the with vertex (1, 3, 4). Example 6(page 960). Consider the function f (x, y) = y2_−x2_{. Since f}

x= −2x and fy = 2y,

the only critical point is _{. For points on the x-axis (x 6= 0) we have f (x, 0) = −x}2_{< 0}

and for points on the y-axis (y 6= 0) we have f (x, 0) = y2 _{> 0. Thus every disk with center}

(0, 0) contains points where f takes positive values and negative values. Therefore, f has no extreme value.

The graph of f is the .

Definition 7 (page 961). The graph of z = y2_{− x}2 has a horizontal tangent plane z = 0 at the origin. f (0, 0) = 0 is a maximum in the direction of x-axis but a minimum in the direction of the y-axis. Near the origin the graph has the shape of a saddle and so (0, 0) is called a saddle point (鞍點) of f .

x

y z

Figure 2: (0, 0) is a saddle point of f (x, y) = y2_{− x}2_.

For a function of one variable f (x), we use second derivative of f (x) to detect the critical points are local maximum or local minimum. Here we will introduce the Second Derivative Test for functions of two variables to investigate the properties of critical points.

回想單變數函數的 極值分類, 必須從 函數二次微分的符 號判斷之; 二變數 函數的二次微分依 變數個數及微分的 先後順序不同而有 四種可能, 將它寫 成矩陣的樣子稱為 赫氏矩陣。 以下介 紹的判別法會根據 赫氏矩陣的行列式 符號而下結論。

Definition 8. The Hessian matrix or Hessian (赫氏矩陣) of f (x, y) at (x0, y0) is

Hess (f )(x0, y0) = " fxx(x0, y0) fxy(x0, y0) fyx(x0, y0) fyy(x0, y0) # .

26 14.7 Maximum and Minimum Values goo.gl/8kiWHQ

Second Derivative Test(page 961). Suppose the second partial derivatives of f are contin-uous on a disk with center (x0, y0), and suppose that fx(x0, y0) = 0 and fy(x0, y0) = 0 (that

plS9h90CWUo 基本上赫氏矩陣的 行列式符號代表函 數凹口最大與最小 的乘積 (正號代表 凹口朝上、 負號代 表凹口朝下),所以 定理的結論可用推 理的方式得到, 比 方說(a)的條件知 道凹口最大的數與 凹口最小的數相乘 為正, 所以同正或 同負, 而 fxx > 0告知至少一數為 正, 因此另一數也 是正的, 於是凹口 朝上的臨界點會產 生局部極小值。 其 它情況也試著自行 推理。

is,(x0, y0) is a critical point of f ). Let

D(x0, y0) = det (Hess(f )(x0, y0)) = fxx(x0, y0)fyy(x0, y0) − (fxy(x0, y0))2.

(a) If D(x0, y0) > 0 and fxx(x0, y0) > 0, then f (x0, y0) is a local minimum.

(b) If D(x0, y0) > 0 and fxx(x0, y0) < 0, then f (x0, y0) is a local maximum.

(c) If D(x0, y0) < 0, then f (x0, y0) is not a local maximum or minimum.





Remark 9.

(1) 對稱矩陣可以正交對角化,所以 Hess(f ) = P DP−1_{,其中D是對角化矩陣; P 是坐標變換矩}

陣。

(2) 因為det(AB) = det(BA),所以行列式在坐標變換下不變;即det(Hess(f )) = det(P DP−1_{) =}

det(P P−1_{D) = det(ID) = det(D)。}

(3) fxx> 0 沿x 方向凹口向上,則對角矩陣其中一個值(特徵值、 固有值) 為正; fxx< 0 沿x 方 向凹口向下, 則對角矩陣其中一個值 (特徵值、 固有值) 為負。 2w0-IdJKnGA 例題示範如何將函 數的臨界點確實分 類。 用二次微分判 別法確定臨界點的 屬性。

Example 10. Find the extreme value (local maximum and minimum values and saddle points) of the function f (x, y) = 2x3_{− 4xy + 3y}2.

14.7 Maximum and Minimum Values goo.gl/8kiWHQ 27

Absolute Maximum and Minimum Values, page 965

aXDPv2LQqEQ 另一種極值問題是 要找絕對極值。 在 介紹二變數函數的 絕 對 極 值 定 理 前, 必須先了解什麼是 有界的閉集合。

Recall that for one variable function f (x), the Extreme Value Theorem says that if f is continuous on a closed interval [a, b], then f has an absolute maximum value and an absolute minimum value. Absolute maximum and absolute minimum points are happened at the critical points or endpoints.

We will introduce the Extreme Value Theorem of two variables in this section.

Definition 11 (page 965).

(a) A boundary point (邊界點_{) of a set D ⊂ R}2 _{is a point (x}

0, y0) such that every disk with

center (x0, y0) contains points in D and also points not in D.

(b) A closed set (閉集_{) D in R}2 _{is one that contains all its boundary points.}

(c) A bounded set (有界集_{) D in R}2 _{is one that is contained within some disk.}

(a) (b)

Figure 3: (a) Closed sets. (b) Sets that are not closed.

一個連續函數, 當 定義域是封閉且有 界時, 函數會有最 大值與最小值。 這 個定理會在高等微 積 分 課 程 詳 細 證 明, 而且到後期學 到 抽 象 化 的 概 念 下, 封閉且有界的 條 件 會 轉 變 成 緊 緻集 (compact), 而且歐氏空間的概 念也可以抽象化變 成 討 論 測 距 空 間 (metric space)。

Extreme Value Theorem for Functions of Two Variables(page 965). If f is continuous on a closed, bounded set D in R2, then f attains an absolute maximum value f (x1, y1) and

an absolute minimum value f (x2, y2) at some points (x1, y1) and (x2, y2) in D.

To find the absolute maximum and minimum values of a continuous function f on a closed, bounded set D:

(a) Find the values of f at the critical points of f in D. (b) Find the extreme values of f on the boundary of D.

(c) The largest of the values from step 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.

28 14.7 Maximum and Minimum Values goo.gl/8kiWHQ

Example 12. Find the extreme values of f (x, y) = x2 + xy + y2_{− 4x + 3y in the region} bounded by x = 0, y = 0, and x + y = 4. loeZzQiHXNQ 這 個 例 題 討 論 的 區 域 是 封 閉而 且 有界, 而且函數連 續, 所以絕對極值 必存在。 在區域內 部, 臨界點會是所 有產生極值的候選 點。 至於邊界, 這 個例子是三個線段 組成, 所以分別將 線段改寫, 則函數 限制在線段上變成 單變數函數, 再尋 求極值。 最後再把 所有的候選點與邊 界的極值點一起比 大小。 Solution. 5EmrV5_jyH0 這個例題也是在問 函 數 的絕 對 極 值, 因為區域的邊界是 圓形, 所以也可以 很容易地將邊界參 數化, 所以函數限 制在邊界上也形成 單變數函數, 這樣 就可以了解邊界上 函數的極值。 若區 域的邊界更複雜的 話, 下一個單元會 介紹拉格朗日乘子 法, 可用那個方法 處理。

Example 13. _{Find the absolute maximum and minimum values of f (x, y) = 4x+ 6y −x}2_−y2 in the region x2+ y2 _{≤ 1.}

14.7 Maximum and Minimum Values goo.gl/8kiWHQ 29

Appendix, page 967

Proof of the Second Derivative Test. We compute the second-order directional derivative of f in the direction of u = (h, k). The first-order derivative is

Duf = fxh + fyk.

Apply this theorem a second time, we have

D2uf = Du(Duf ) = ∇(Duf ) · u =  ∂ ∂x(fxh + fyk), ∂ ∂y(fxh + fyk)  · (h, k) = (fxxh + fyxk)h + (fxyh + fyy)k = fxxh2+ 2fxyhk + fyyk2 = fxx  fxxh + fxyk fxx 2 + k 2 fxx (fxxfyy− fxy2 ).

Remark that D(x0, y0) = det(Hess(f )(x0, y0)) = fxxfyy− fxy2

 

(x0,y0).

(a) If D(x0, y0) > 0 and fxx(x0, y0) > 0, then Du2f (x0, y0) ≥ 0 for any direction u. If

D2uf (x0, y0) = 0 for some direction u, then k = 0 and fxxh + fxyk = 0. However, it

implies (h, k) = (0, 0) and it contradicts to h2 + k2 = 1. Hence f (x0, y0) is a local

minimum.

(b) If D(x0, y0) > 0 and fxx(x0, y0) < 0, then Du2f (x0, y0) ≤ 0 for any direction u. If

D2

uf (x0, y0) = 0 for some direction u, then k = 0 and fxxh + fxyk = 0. However, it

implies (h, k) = (0, 0) and it contradicts to h2 _{+ k}2 _{= 1. Hence f (x}

0, y0) is a local

maximum.

(c) If D(x0, y0) < 0, we will find two different directions such that the signs of the second

derivatives of f are different.

⋆ If fxx > 0 and fyy < 0, then choosing u = (h, k) = (1, 0) implies Du2f = fxx > 0.

If we choose (h, k) = (0, 1), then D2uf = fyy < 0.

⋆ If fxx < 0 and fyy > 0, then choosing u = (h, k) = (1, 0) implies Du2f = fxx < 0.

If we choose (h, k) = (0, 1), then D2uf = fyy > 0.

⋆ If fxx > 0 and fyy > 0, then choosing u = (h, k) = (1, 0) implies Du2f = fxx > 0.

If we choose (h, k)//(fxy, −fxx) such that k 6= 0, then D2uf =

_f

xxfyy−fxy2

fxx



k2 _{< 0.}

⋆ If fxx < 0 and fyy < 0, then choosing u = (h, k) = (1, 0) implies Du2f = fxx < 0.

If we choose (h, k)//(fxy, −fxx) such that k 6= 0, then D2uf =

_f xxfyy−f 2 xy fxx  k2 _{> 0.}

30 14.8 Lagrange Multipliers goo.gl/cf32ER

14.8

Lagrange Multipliers, page 971

Q_ZJQm6btWo 拉格朗日乘子法是 在處理限制條件下 的 極 值 問 題, 前 一節討論的極值問 題, 因為變數之間 彼此獨立, 所以直 接對各個變數偏微 找出臨界點而尋求 答案。 現在因為變 數之間必須滿足一 個關係式, 所以若 用之前的方法不加 以修改, 臨界點不 見得會滿足這個限 制條件。 由此可以 再重新想一想拉格 朗日乘子法是如何 造出新的目標函數 以滿足所有條件。

Method of Lagrange Multipliers (page 972). To find the maximum and minimum values of f (x, y, z) subject to the constraint g(x, y, z) = k (assuming that these extreme values exist and_{∇g 6= 0 on the surface g(x, y, z) = k):}

(a) Find all values of x, y, z, and λ such that

∇f (x, y, z) = λ∇g(x, y, z) and g(x, y, z) = k.

(b) Evaluate f at all the points (x, y, z) that result from step (a). The largest (smallest) of these values is the maximum (minimum) value of f .

Proof. _{For λ ∈ R, consider F (x, y, z, λ) = f (x, y, z) − λ(g(x, y, z) − k), then on the level set} g(x, y, z) = k, the function F (x, y, z, λ) and f (x, y, z) take the same value. So to find the extreme value of f (x, y, z) on the surface g(x, y, z) = k is equivalent to find the extreme value of F (x, y, z, λ), and it implies ∇F (x, y, z, λ) = 0 ⇔ ∇f = λ∇g and g(x, y, z) = k. ul1c1LbqXtU 這裡先用各位在高 中時就會用空間幾 何的方法找出平面 外一點到平面的最 短距離的例子, 現 在試著用拉格朗日 乘子 的 方 法 求解, 目標與操作比較明 確; 高中的解法某 種程度來說比較流 於形式, 也只適用 於 點 到 平 面 的 情 況。 拉格朗日乘子 法適用於點到曲面 的情形。

Example 1 _{(page 964). Find the shortest distance from the point (1, 0, −2) to the plane} x + 2y + z = 4.

Solution.



14.8 Lagrange Multipliers goo.gl/cf32ER 31

Example 2 (page 964, 973). A rectangular box without a lid is to be made from 12 m2 of a cardboard. Find the maximum volume of such a box.

quJJXhw9rp0 製造無蓋紙盒有最 大的體積問題也可 以用拉格朗日乘子 法處理, 因為紙盒 的 表 面 積 有 限 制。 各位也可以想一想 這個問題如何用算 幾不等式處理。 Solution. zlbJk1dE54E 按照連續函數的極 值定理: 有界封閉 區域上的連續函數 必 有 最 大 最 小 值。 實作層面, 對於區 域內部直接找臨界 點, 而邊界的極值 就可以用拉格朗日 乘子法處理。 最後 再把所有的臨界點 與邊界的極值點一 起比大小。

Example 3. Find the maximum and minimum values of the function f (x, y) = x3+ 3x2y in the region x2+ 4xy + 5y2 _{≤ 5.}

32 14.8 Lagrange Multipliers goo.gl/cf32ER

Two Constraints, page 976

nxtcsb0gCPw 拉格朗日乘子法可 用於多個限制條件 下的極值問題, 若 有三個變數的函數 在兩個限制條件下 的極值問題, 每一 個限制條件就再創 一個變數, 得到新 的目標函數為五變 數函數, 因為在限 制條件下新的目標 函數與原目標函數 取值一樣, 所以改 研究新的目標函數 之極值, 而新的目 標函數對於新創的 兩個變數偏微令成 零之下即為限制條 件。

We want to find the maximum and minimum values of a function f (x, y, z) subject to two constraints of the form g(x, y, z) = k and h(x, y, z) = c. Let

F (x, y, z, λ, µ) = f (x, y, z) − λ(g(x, y, z) − k) − µ(h(x, y, z) − c).

For λ, µ ∈ R, under these two constraints, F (x, y, z, λ, µ) and f (x, y, z) take the same value. So to find the extreme values of f (x, y, z) with g(x, y, z) = k and h(x, y, z) = c is equivalent to find the extreme values of F (x, y, z, λ, µ). That is,

∇F = 0 ⇔ ∇f = λ∇g + µ∇h and g(x, y, z) = k and h(x, y, z) = c.

Example 4. The plane x + y + 2z = 2 intersects the paraboloid z = x2 + y2 in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.

14.8 Lagrange Multipliers goo.gl/cf32ER 33

Example 5 (page 977). Find the maximum value of the function f (x, y, z) = x + 2y + 3z on the curve of intersection of the plane x − y + z = 1 and the cylinder x2+ y2 = 1.

x9SmS_h5aS8 拉格朗日乘子法的 概念容易, 但是實 作層面上的難點是 聯立方程式的求解 困難, 如何適當地 分情況討論方程式 的 解 是 一 門 學 問。 若遇到更複雜的式 子求解那麼必須借 助電腦的幫忙。 Solution.