16.1 Vector Fields goo.gl/qERfs6 1
Chapter 16
Vector Calculus
16.1
Vector Fields, page 1068
Definition 1 (page 1069). qp7YA5mKyoo 第十六章是向量微 積分。 該主題起源 於物理, 許多力學 可以用場論描述像 是電場、 磁場、 重 力場等, 這之間蘊 含著 某 些 守 恆 關 係, 利用向量微積 分 的語 言 得 以 詮 釋。 向量 場 是 一 個 映 射, 從一個點對應 到一 個 向 量 的 關 係, 通常會用左圖 那樣的方式示意其 概念。
(1) Let D be a set in R2. A vector field on R2 (向量場) is a map F that assigns to each point (x, y) in D a two-dimensional vector F(x, y).
(2) Let E be a subset of R3. A vector field on R3 is a map F that assigns to each point (x, y, z) in E a three-dimensional vector F(x, y, z).
The best way to picture a vector field is to draw the arrow representing the vector F(x, y) starting at the point (x, y) for a few representative points in D.
x x y y z (x, y, z) F(x, y, z) (x, y) F(x, y)
Figure 1: Vector fields on R2 and on R3.
這裡的學習應時時 刻刻注意每個記號 代表的本質是向量 還是函數。
Since F(x, y) is a two-dimensional vector, we can write it in terms of its component func-tions (分量函數) P and Q as follows:
F(x, y) = P (x, y) i + Q(x, y) j = (P (x, y), Q(x, y)).
Functions P (x, y) and Q(x, y) are called scalar function (純量函數) or scalar fields.
Example 2 (page 1070). A vector field on R2 is defined by F(x, y) = −y i + x j. Denote
x= x i + y j by the position vector.
Figure 2: Vector fields F = −y i + x j = (−y, x) and x = x i + y j = (x, y).
Remark that x · F(x) = (x i + y j) · (−y i + x j) = −xy + xy = 0, so two vector fields are orthogonal (正交).
Example 3 (page 1071). Newton’s Law of Gravitation states that the magnitude of the gravitational force between two objects with masses m and M is
兩物體之間的吸引 力與 質 量 成 正 比, 與 距 離 平 方 成 反 比, 將牛頓的重力 定律用向量場的方 式描述, 可再將每 個位置重力的方向 標示出來。 重力場 的模型很重要, 之 後會繼續引申出位 勢的觀念, 進而證 明能量守恆定律。 |F| = GM mr2 ,
where r is the distance between the objects and G is the gravitational constant.
Let the position vector of the object with mass m be x = (x, y, z), then r2 = |x|2. Therefore the gravitational force acting on the object at x is
F(x) = −GM m |x|2 x |x| = − GM m |x|3 x, (1)
and we say the equation (1) is gravitational field (重力場).
電場的概念比起引
力場來說, 多了相
斥的可能。 所以描 述電場時必須再研 究電荷之正負號。
Example 4 (page 1072). Suppose an electric charge Q is located at the origin. According to Coulomb’s Law, the electric force F(x) (or electric field 電場) exerted by this charge on a charge q located at a point (x, y, z) with position vector x = (x, y, z) is
F(x) = εQq |x|2 x |x| = εQq |x|3 x,
where ε is a constant. For like charges, we have Qq > 0 and the force is repulsive; for unlike charges, we have Qq < 0 and the force is attractive.
Instead of considering the electric force F, physicists often consider the force per unit charge (電場強度):
E(x) = 1
qF(x) = εQ |x|3 x.
Gradient Fields (
梯度場
), page 1072
給 定 多 變 數 函 數, 每一個點都可以對 應到函數在該點的 梯度進而造出梯度 向量場, 由此延伸 出保守向量場的定 義: 若一個向量場 正好與某個函數的 梯度一致。 而相應 的函數稱為位勢函 數。
Recall that for a smooth function f (x, y), the gradient ∇f , or grad f , is defined by ∇f (x, y) = grad f = fx(x, y) i + fy(x, y) j.
Likewise, if f (x, y, z) is a scalar function of three variables, its gradient is a vector field on R3 given by ∇f (x, y, z) = fx(x, y, z) i + fy(x, y, z) j + fz(x, y, z) k.
Definition 5 (page 1072).
(a) For a scalar function f , we say ∇f is a gradient vector field (梯度向量場).
(b) A vector field F is called a conservative vector field (保守向量場) if it is the gradient of some scalar function, that is, if there exists a function f such that F = ∇f . In this situation f is called a potential function (位勢函數) for F.
16.1 Vector Fields goo.gl/qERfs6 3
Example 6 (page 1073). The gravitational field F is conservative because if we define a function qLgmuKvQmR0 重力場是一個保守 向量場, 這個例子 將確實找到位勢函 數並驗證它的梯度 與重力場一致。 f(x, y, z) = mM G px2+ y2+ z2, then 這裡再次複習 「函 數的梯度向量場與 函數的等高線互相 垂直」 之概念。
Example 7. Let f (x, y) be a smooth function, then the gradient vector field ∇f (x, y) is perpendicular to the level curves f (x, y) = k.
-4 -2 0 2 4 -4 -2 0 2 4
Figure 3: Level sets of f (x, y) = x2− y2 and the gradient field ∇f = 2x i − 2y j.
In general, all conservative vector vector field F is perpendicular to the level sets of its potential function f .
16.2
Line Integrals, page 1075
Line Integrals of Scalar Functions (
第一類曲線積分
), page 1075
HvZ-UovGbN8 線積分依照屬性區 分為兩類。 第一類 線 積 分 的 意 思 是: 給定一條曲線, 以 及定義在曲線上的 函數, 該如何計算 函數沿著曲線的積 分。 第一類線積分 的建構方式主要是 從曲線弧長的建構 當中, 選定樣本點 之後, 順勢將這個 樣 本 點 代 入 函 數 值, 最後再取和求 極限而得。
Suppose that a smooth plane curve C is given by the parametric equations r(t) = x(t) i + y(t) j, a ≤ t ≤ b. Recall that a curve is smooth means that r′(t) is continuous and r′(t) 6= 0. We will define an integral over a curve C.
a t1 t2· · · ti−1ti· · ·tn−1b t∗ i t P0 Pn P1 P2 Pi−1 Pi Pn−1 Pi∗(x∗ i, y∗i) x y
Figure 1: Line integrals of scalar functions.
(a) We divide the parameter interval [a, b] into n subintervals [ti−1, ti] of equal width. We
let xi = x(ti) and yi = y(ti), then the corresponding points Pi(xi, yi) divide C into n
subarcs with lengths ∆s1,∆s2, . . . ,∆sn.
(b) Choose any point Pi(x∗i, y∗i) (corresponding to t∗i ∈ [ti−1, ti]) in the i-th subarc.
(c) If f (x, y) is any function of two variables whose domain includes the curve C, we form the sum (similar to a Riemann sum)
n
P
i=1
f(x∗i, y∗i)∆si.
(d) Define the line integral of f (x, y) along C (函數 f(x, y) 沿曲線 C 的線積分) is
C f(x, y) ds = lim n→∞ n X i=1 f(x∗i, y∗i)∆si
if this limit exists.
Recall that the arc length function of C is s(t) =
t
ap(x′(u))2+ (y′(u))2du, and it implies
ds =p(x′(t))2+ (y′(t))2dt, so the line integral has the following expression:
C f(x, y) ds = b a f(x(t), y(t))p(x′(t))2+ (y′(t))2dt. 在將第一類線積分 轉變成可以計算的 積分公式時, 牽涉 到曲線的表達。(接 下頁)
Well-defined problem: (copy from Wikipedia) In mathematics, an expression is called well-defined(良好定義的) if its definition assigns it a unique interpretation or value. For example, a function is well-defined if it gives the same result when the representation of the input is changed without changing the value of the input. That is, if f (x) is a well-defined function defined on R, then f (0.5) must be equal to f (12), and f (1) must be equal to f (0.¯9).
16.2 Line Integrals goo.gl/cqZ7Z5 5
Suppose that r(v) = x(v) i + y(v) j, c ≤ v ≤ d is another parametrization of the plane curve C. To show the line integral is well-defined, we have to check that
因為曲線有各種可 能的表達式, 在數 學上還必須驗證的 是: 這樣的公式 與參數式的選取無 關。 C f(x, y) ds = d c f(x(v), y(v))p(x′(v))2+ (y′(v))2dv.
Check: This is because the arc length is s(v) =
v
a p(x′(u))2+ (y′(u))2du, and it implies
ds =p(x′(v))2+ (y′(v))2dv. qSVVaISYces 第一類線積分的應 用有很多, 比方說 考慮密度不均的線 圈, 將 密 度 函 數 線積分而得線圈之 質 量。 若 是 平 面 曲 線, 將 函 數 大 小呈現在 z 分 量下得到函數的圖 形, 其形狀像是一 個屏風, 則將函數 對於曲線進行第一 類線積分將得到屏 風的表面積; 跑步 的時候, 在不同階 段下熱量的消耗也 不同, 將熱量函數 對於路徑積分而得 消耗的總熱量。
The geometric meaning of the line integrals is to compute the area of one side of the “fence” or “curtain,” whose base is C and whose “height” at (x, y) is f (x, y).
ds是弧長參數,為正量,給定曲線參數式後,上、 下限代入終點與起點。 第一類曲線積分與 「路徑的方向無關」, 所以化為定積分時,下限總是小於上限。 x y z C (x, y) f(x, y)Figure 2: Line integrals: The area of the fence. The total calories of running.
例題示範第一類線 積分, 先將曲線參 數式, 把函數替換 成曲線參數式的符 號, 配上線元 ds, 積分時下限代入小 的數, 上限代入大 的數。
Example 1 (page 1076). Evaluate
C(2 + x2y) ds, where C is the upper half of the unit circle
x2+ y2 = 1. Solution. f2DwlqOVd0M 第一類曲線積分可 用在分段光滑曲線 的情況, 若曲線只 有有限個點無法定 義切向量, 這件事 並不影響積分值。
We can define the line integrals on piecewise smooth curve (分段光滑曲線) C, which is a union of a finite number of smooth curves C1, C2, . . . , Cn, and the initial point of Ci+1 is the
terminal point of Ci. The line integral of f (x, y) along C as the sum of the integrals of f
along each of the smooth pieces of C:
C f(x, y) ds = C1 f(x, y) ds + C2 f(x, y) ds + · · · + Cn f(x, y) ds.
Example 2 (page 1077). Evaluate
C2x ds, where C consists of the arc C1 of the parabola
y= x2 from (0, 0) to (1, 1) followed by the vertical line segment C2 from (1, 1) to (1, 2).
例題示範如何計算 對於分段光滑曲線 計 算 第 一 類 線 積 分, 其實就是分段 處理再相加即可。 Solution. B74qAVlmOjs 將密度函數對於曲 線積分得到曲線之 質量, 若將密度函 數再乘以 x 或 y 積分之後除掉質量 則為質心的概念。
Any physical interpretation of
Cf(x, y) ds depends on the physical interpretation of the
function f . For example, we define the mass (質量) m of the wire:
m= lim n→∞ n X i=1 ρ(x∗i, yi∗)∆si= C ρ(x, y) ds.
The center of mass (質心) of the wire with density function ρ(x, y) is (¯x,y) =¯ 1 m C xρ(x, y) ds, 1 m C yρ(x, y) ds . 例題示範密度不均 之線圈的質心。
Example 3 (page 1077). A wire takes the shape of the semicircle x2+ y2 = 1, y ≥ 0, and
is thicker near its base than near the top. Find the center of mass of the wire if the linear density at any point is proportional to its distance from the line y = 1.
16.2 Line Integrals goo.gl/cqZ7Z5 7
Definition 4 (page 1066).
(a) Define line integrals of f along C with respect to x (f 沿C 對於 x的線積分): wI_pijMjFKQ
函數沿著曲線對於 x 或 y 的線積 分的定義是為了要 引進第二類曲線積 分而準備。 特別注 意 dx 或 dy 是 有方向性的, 和坐 標軸同向的時候為 正, 反 向 的 時 候 為負, 所以會有正 負抵消的效果。 而 ds是線元,所以不 論曲線所在位置為 何、 如何行走, 線 元都是正的。 C f(x, y) dx = lim n→∞ n X i=1 f(x∗i, yi∗)∆xi = b a f(x(t), y(t))x′(t) dt.
(b) Define line integrals of f along C with respect to y (f 沿 C 對於 y 的線積分):
C f(x, y) dy = lim n→∞ n X i=1 f(x∗i, yi∗)∆yi= b a f(x(t), y(t))y′(t) dt.
(c) The line integral of f along C with respect to arc length is:
C f(x, y) ds = lim n→∞ n X i=1 f(x∗i, y∗i)∆si = b a f(x(t), y(t))p(x′(t))2+ (y′(t))2dt.
函數 f 沿 C 對於 x 的線積分, 因為dx可能正可能負, 因此積分與路徑的方向有關。 對 於 同一 條 曲 線, 有可能要同時計算 函數沿曲線對於x 或y的積分,這時 積分符號可以只留 最前面的記號, 並 不會引起混淆。It frequently happens that line integrals with respect to x and y occur together. When this happens, it’s customary to abbreviate by writing
C P(x, y) dx + C Q(x, y) dy = C P(x, y) dx + Q(x, y) dy. 例題示範如何計算 函數沿著曲線對於 x 或 y 的積分。 特別注意積分的下 限與下限, 它是遵 照起點與終點而設 定。 Example 5. Evaluate Cy2dx + x dy, where
(a) C = C1 is the line segment from (−3, −2) to (0, 1).
(b) C = C2 is the arc of the parabola x = 1 − y2 from (−3, −2) to (0, 1).
Line Integrals in Space, page 1080
-kIx2BM33Uo 上述所有概念都可 以推廣至三度空間 中的曲線, 基本上 就是增加z分量的 相關資訊於所有討 論中。Suppose that C is a smooth space curve given by the parametric equation x = x(t), y = y(t), z = z(t), a ≤ t ≤ b or vector equation r(t) = x(t) i + y(t) j + z(t) k. If f (x, y, z) is a smooth function that is continuous on some region containing C, then we define the line integral off along C as C f(x, y, z) ds = lim n→∞ n X i=1 f(x∗i, yi∗, z∗i)∆si = b a f(x(t), y(t), z(t))p(x′(t))2+ (y′(t))2+ (z′(t)) dt = b a f(r(t))|r′(t)| dt. Line integrals along C with respect to x, y, and z can also be defined. For example,
C f(x, y, z) dz = lim n→∞ n X i=1 f(x∗i, y∗i, zi∗)∆zi= b a f(x(t), y(t), z(t))z′(t) dt. Therefore, we evaluate integrals of the form
C
P(x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz by expressing everything (x, y, z, dx, dy, dz) in terms of the parameter t.
Line Integrals of Vector Fields (
第二類曲線積分
), page 1082
現在要正式介紹第 二類曲線積分, 它 是要研究分布在空 間中的向量場如何 作用在質點上, 造 成運行軌跡時所作 的 功。 同 樣 利 用 分割樣本點取和求 極限的過程設法把 積 分 式 表 達 出 來。 特別注意在加總的 階 段, 根 據 作 功 的 概 念, 是 在 計 算力對於質點運行 的方向之有效力乘 上運行的距離, 所 以在分割後的每一 小 段, 利 用 曲線 在樣本點上的力與 單位切向量兩者內 積, 再 乘 上 小 段 弧長, 加總取極限 而得第二類曲線積 分。
Suppose that F(x, y, z) = P (x, y, z) i + Q(x, y, z) j + R(x, y, z) k is a continuous force field on R3 such as gravitational field, electric force field, etc. We want to compute the work (功) done by this forcein moving a particle along a smooth curve C.
(a) We divide C into subarcs Pi−1Pi with lengths ∆si by dividing the parameter interval
[a, b] into subintervals of equal width.
(b) Choose Pi∗(x∗i, yi∗, zi∗) on the i-th subarc corresponding to t∗i.
(c) If ∆si is small, then as the particle moves from Pi−1 to Pi along the curve, it proceeds
approximately in the direction T(t∗
i), the unit tangent vector at Pi∗. Thus the total work
done by the force F in moving the particle along C is approximately
n
P
i=1
F(x∗i, y∗i, zi∗) · T(x∗i, y∗i, zi∗)∆si.
(d) When n tends to infinity, we define the work (功) W done by the force field F:
W = C F(x, y, z) · T(x, y, z) ds = C F· T ds. (2)
Hence, work is the line integral with respect to arc length of the tangential component of the force.
16.2 Line Integrals goo.gl/cqZ7Z5 9 P0 Pn P1 P2 Pi−1 Pi F(x∗ i, y∗i, zi∗) Pn−1 P∗ i(x∗i, y∗i, z∗i) T(t∗ i) x y z
Figure 3: Line integrals of vector fields.
這裡推得第二類曲 線積分的另一個表 達式, 由於單位切 向量必須除掉切向 量的長度, 另一方 面, 弧長轉換式中 又帶有切向量的長 度, 所以兩者相消 後公式變得簡單。
If the curve C is given by the vector equation r(t) = x(t) i + y(t) j + z(t) k, then we get the unit tangent vector is T = r′(t)
|r′(t)|, so we can rewrite (2) in the form
W = C F(r(t)) · r ′(t) |r′(t)| |r′(t)| dt = C F(r(t)) · r′(t) dt = C F(r(t)) · dr = C F(x(t), y(t), z(t)) · dr = C F· dr. DhMFx4OjtRk 這裡給予第二類曲 線 積 分 的 正 式 定 義。 注意到第二類 曲線積分與積分路 徑的方向有關係。
Definition 6 (page 1082). Let F be a continuous vector field on a smooth curve C given by a vector function r(t), a ≤ t ≤ b. Then the line integral of F along C is
C F· T ds = C F· dr = b a F(r(t)) · r′(t) dt.
第二類曲線積分與積分路徑的方向有關 (方向相反時,積分值變號)。 積分路徑的方向會用 C 與−C 表示,而 −CF· dr = − CF· dr。Connection between line integrals of vector fields and line
inte-gral of scalar functions, page 1084
第二類曲線積分若 改用分量的方式表 達, 其結 果 會 是 函數沿著曲線對於 x, y 或 z 的線積 分之加總。
Suppose that the vector field F on R3 is given in component form by the equation F(x, y, z) =
P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k. Then C F· dr = b a F(r(t)) · r′(t) dt = b a (P i + Q j + R k) · (x ′(t) i + y′(t) j + z′(t) k) dt = b a
(P (x(t), y(t), z(t))x′(t) + Q(x(t), y(t), z(t))y′(t) + R(x(t), y(t), z(t))z′(t)) dt =
C
Example 7. Find the work done by the force field F(x, y) = −y2i+ x2jin moving a particle along arc of the circle x2+ y2 = 2 traversed counterclockwise from (√2, 0) to (−√2, 0).
例題示範第二類曲 線積分的計算。 第 二類曲線積分具有 相當豐富的結構與 內容, 接下來的兩 單元要繼續針對第 二類曲線積分的性 質延伸。 Solution.
16.3 The Fundamental Theorem for Line Integrals goo.gl/C9i5sz 11
16.3
The Fundamental Theorem for Line Integrals,
page 1087
TVRKYVDwsg8 微積分基本定理是 告知單變數函數的 微分與積分是互逆 的 運 算。 而 線 積 分基本定理是在觀 察多變數函數的微 分與積分之間的關 係。 多變數函數牽 涉到的微分變成梯 度向量, 而積分的 範圍變成曲線, 所 以線積分基本定理 的結果是: 可微分 函數的梯度向量場 沿著曲線的第二類 線積分的結果, 會 是終點函數值與起 點函數值相減。 從證明的過程中也 可以很清楚地看到 線積分基本定理的 本質是微積分基本 定理, 也因此, 相 關的正負號效應也 都遵照當初微積分 基本定理的結果。Recall that Part 2 of the Fundamental Theorem of Calculus is
b a
F′(x) dx = F (b) − F (a), (3)
where F′(x) is continuous on [a, b]. We also called equation (3) the Net Change Theorem:
The integral of a rate of change is the net change.
Here we will introduce the Fundamental Theorem for line integrals, where we think of the gradient vector ∇f of a function f as a sort of derivative of f .
Theorem 1 (page 1087). Let C be a smooth curve given by the vector function r(t), a ≤ t ≤ b. Let f be a differentiable function of two or three variables whose gradient vector ∇f is continuous on C. Then C∇f · dr = f (r(b)) − f (r(a)).
若向量場來自於函數的梯度 (保守向量場), 則線積分的值為兩端點函數值的差。 第二類曲線積分的路徑有方向性, 即 C∇f · dr = − −C∇f · dr。 x x y y z A(x1, y1) B(x2, y2) A(x1, y1, z1) B(x2, y2, z2) C CFigure 1: The fundamental theorem for line integrals.
Proof. By the Chain Rule and the Fundamental Theorem of Calculus, we have
C∇f · dr = b a ∇f (r(t)) · r ′(t) dt = b a ∂f ∂x dx dt + ∂f ∂y dy dt + ∂f ∂z dz dt dt = b a d dtf(r(t)) dt = f (r(b)) − f (r(a)).
Theorem 1 is also true for piecewise smooth curves. This can be seen by subdividing C into a finite number of smooth curves and adding the resulting integrals.
Example 2 (page 1088). Find the work done by the gravitational field F(x) = −GM m|x|3 x in
moving a particle with mass m from the point (3, 4, 12) to the point (2, 2, 0) along a piecewise smooth curve C. 這 裡 以 重 力 場 為 例, 因為重力場是 某個函數的梯度向 量場, 於是線積分 基本定理告知, 第 二類曲線積分的結 果會是兩端函數值 相減。 這個例題中 並沒有明確告知曲 線的走法,換言之, 這個結果是與曲線 的路徑無關, 這件 事與物理上所述位 能差只與起點與終 點的位置有關。 Solution.
Independence of Path, page 1088
grK4CHZk6ek 由重力場的例子知 道有些場的第二類 曲線積分與路徑選 取無關, 只與起點 終點有關, 所以特 別把這個性質獨立 寫成一個定義, 之 後要研究的是: 除 了 保 守 向量 場 外, 到底有哪些向量場 也滿足積分與路徑 選取無關。
Suppose that C1 and C2 are two piecewise smooth curves, which are also called paths (路徑)
with have the same initial point A and terminal point B.
Definition 3 (page 1088). Suppose that F is a continuous vector field with domain D. We say the line integral
CF· dr is independent of path (積分和路徑選取無關) if C1 F· dr = C2 F· dr
for any two path C1 and C2 in D with the same initial and terminal points.
• In general vector field F,
C1F· dr 6=
C2F· dr. (See 16.2, Example 5).
• For conservative vector field F = ∇f , the Fundamental Theorem for line integrals tells us
C1∇f · dr =
C2∇f · dr = f (r(b)) − f (r(a)).
• The following discussion will say that the only vector fields that are independent of path are conservative vector fields.
保守向量場 ⇒ 第二類曲線積分與路徑選取無關。 因為曲線參數式是 將區間 t ∈ [a, b] 對應到平面或空間 上的點r(t),若要 呈現封閉曲線, 必 須要求區間上的兩 端點 a與b 對到 平面或空間中同樣 的位置。Definition 4 (page 1089). A curve is called closed (封閉曲線) if its terminal point coincides with its initial point, that is, r(b) = r(a).
16.3 The Fundamental Theorem for Line Integrals goo.gl/C9i5sz 13
Theorem 5 (page 1089). The line integral
CF· dr is independent of path in D if and only
if
CF· dr = 0 for every closed path C in D.
該定理表明第二類 線積分若與路徑選
取無關, 則它等價
於任何封閉曲線上 的線積分值為零。
Proof. (⇒) We choose any two points A and B on C and regard C as being composed of the path C1 from A to B followed by the path C2 from B to A. Then
C F· dr = C1∪C2 F· dr = C1 F· dr + C2 F· dr = C1 F· dr − −C2 F· dr = 0. (⇐) For any paths C1 and C2 from A to B in D, we define C to be the curve consisting of
C1 followed by −C2. Then we get
0 = C F· dr = C1 F· dr + −C2 F· dr = C1 F· dr − C2 F· dr, and hence C1F· dr = C2F· dr.
第二類曲線積分與路徑無關 ⇔ 封閉曲線上的第二類曲線積分為零。 在保守向量場 (例如重力場) 將一物沿封閉曲線做功為零。 Definition 6 (page 1089). KxFUI9mHagE 除了上述的等價敘 述外, 接下來要探 討第二類線積分與 路徑選取無關與位 勢函數的關係。 以 下的結果將牽涉到 區 域 的一 些 特 性, 所以必須講清楚區 域的屬性。 初學階 段可以先用幾個示 意圖去感受每個定 義最大的差別。(a) A domain D is open (開集合) if for every point P in D, there is a disk with center P that lies entirely in D. (D doesn’t contain any of its boundary points.)
(b) A domain D is path connected (路徑連通) if any two points in D can be joined by a path that lies in D.
Theorem 7 (page 1089). Suppose F is a vector field that is continuous on an open, path connected regionD. If
CF· dr is independent of path in D, then F is a conservative vector
該定理必須注意一 些關鍵字, 向量場 的連續性是為了第 二 類 線 積 分 有 意 義; 設定區域是開 集合的用意是定理 證明的過程中要用 到的手法: 在終點 附近可以上下左右 推出一個小範圍以 直線相連; 區域是 路徑連通是要確定 討論的兩點能夠用 一條曲線在範圍內 相連。 位勢函數的 找法就是先隨便取 一 條 路 徑 積 分 而 得, 因為積分與路 徑選取無關, 所以 函 數 值 可 以 確 定, 根據這個函數, 確 實驗證其梯度向量 與原向量場一致。 這 裡 必 須 強 調 的 是, 定理的敘述從 路徑選取無關推出 向量場是保守向量 場 是 單 向 的 敘 述, 而保守向量場是否 積分與路徑選取無 關這個反向的敘述 是否正確, 是後續 要討論的問題。
field onD; that is, there exists a function f such that ∇f = F.
Proof. Let A(a, b) be a fixed point in D. We construct the potential function f by f(x, y) = (x,y) (a,b) F· dr for (x, y) ∈ D. Since
CF· dr is independent of path, the function is well-defined.
Now we will show that ∇f = F:
C1 C2 C′ 1 C′ 2 (a, b) (a, b) (x, y) (x, y) (x1, y) (x, y1) D D x x y y
Figure 4: Choose suitable paths to prove ∇f = F.
Since D is open, there exists a disk contained in D with center (x, y). Choose any point (x1, y) in the disk with x1< xand let C consists of any path C1 from (a, b) to (x1, y) followed
by the horizontal line segment C2 from (x1, y) to (x, y). Then
f(x, y) = C1 F· dr + C2 F· dr = (x1,y) (a,b) F· dr + C2 F· dr. Notice that the first of these integrals does not depend on x, so
∂ ∂xf(x, y) = 0 + ∂ ∂x C2 F· dr. (4)
Consider C2 : r(t) = t i + y j, where t from x1 to x, then r′(t) = 1 i + 0 j and F(t) =
P(t, y) i + Q(t, y) j. Thus (4) gives ∂ ∂xf(x, y) = ∂ ∂x C2 F· dr = ∂ ∂x x x1 P(t, y) dt = P (x, y).
Similarly, using a vertical line segment, consider C = C1′ ∪ C2′, C2′ : r(t) = x i + t j, where t from y1 to y, then r′(t) = 0 i + 1 j and F(t) = P (x, t) i + Q(x, t) j. We have
∂ ∂yf(x, y) = ∂ ∂y C′ 2 F· dr = ∂ ∂y y y1 Q(x, t) dt = Q(x, y). Therefore, we know that
F(x, y) = P (x, y) i + Q(x, y) j = ∂f ∂xi+
∂f
∂yj= ∇f.
16.3 The Fundamental Theorem for Line Integrals goo.gl/C9i5sz 15
Next, we will determine whether or not a vector field F is conservative. Suppose that
vt7Oz-bDCr8 若將向量場的分量 寫 出, 而 且 分 量 函數是一次微分連 續 時, 由 函 數 二 次微分可交換這件 事, 可以得到向量 場是保守向量場的 必要條件。
F= P i + Q j is conservative, where P and Q have continuous first order partial derivatives. Then there is a function f such that F = ∇f , that is, P = ∂f∂x and Q =
∂f ∂y. By Clairaut’s Theorem, we know ∂P ∂y = ∂2f ∂y∂x = ∂2f ∂x∂y = ∂Q ∂x.
Hence Qx= Py is a necessary condition (必要條件) that F = P i + Q j is conservative.
Qx = Py 這個條 件, 必須加上區域 是 「單連通」 的時 候, 逆敘述才會成 立。 為此, 必須補 充說明單連通的意 義。
Theorem 8 (page 1090). If F(x, y) = P (x, y) i+ Q(x, y) j is a conservative vector field, where P and Q have continuous first order partial derivatives on a domain D, then throughout D we have
∂Q ∂x =
∂P ∂y.
The condition Qx = Py is a sufficient condition for a simply connected region.
Definition 9 (page 1090). 簡單曲線要描述的 是曲線不自交。 而 單連通的概念除了 要求是路徑連通之 外, 還要強調這個 區域的內部是沒有 「洞」。
(a) We say C is a simple curve (簡單曲線) if it doesn’t intersect itself anywhere between its endpoints. (r(t1) 6= r(t2) when a < t1 < t2< b).
Figure 5: (Left to right) Simple, not closed; simple closed; not simple, not closed; not simple, closed.
(b) D is a simply connected region (單連通區域) in a plane if it is path connected and every simple closed curve in D enclosed only points that are in D.
Theorem 10 (page 1091). Let F(x, y) = P (x, y) i + Q(x, y) j be a vector field on an open simply-connected region D. Suppose that P and Q have continuous first-order derivatives
b9KEObmI2uo 這個定理的重點是 若討論的區域是單 連通, 而且定義在 區域上的向量場一 次微分且連續的時 候, Qx = Py 的 條件才可以反推而 得向量場是保守向 量場。 and ∂Q ∂x = ∂P ∂y throughout D, then F is conservative.
We will prove Theorem 10 in the next section.
直觀上,單連通區域代表此區域沒有 「洞」 —虧格為零 (genus);而且無法分成兩塊。Finally, we will use “partial integration” to find the potential functions.
例題中的向量場處 處都是一次微分仍 連續, 所以只要檢 驗Qx= Py就可 確定它是否為保守 向量場。 若要得到 位勢函數, 則用積 分後再微分的方法 求得。 注意到若函 數對x積分後,所 有和y有關的函數 g(y)都會是 「積分 常數」。 Example 11(page 1091).
(a) If F(x, y) = (3 + 2xy) i + (x2− 3y2) j, find a function f such that F = ∇f .
(b) Evaluate the line integral
CF· dr, where C is the curve given by r(t) = e
tsin t i +
etcos t j, and t from 0 to π.
Solution. HDorsC5E4gg 對於空間中的向量 場, 同樣也是先利 用函數的二次微分 交換律而得到保守 向量 場 的 必 要 條 件, 它必須滿足三 個方程式。
Example 12 (page 1095). Show that if the vector field F = P i + Q j + R k is conservative and P, Q, R have continuous first order partial derivatives, then
∂P ∂y = ∂Q ∂x, ∂P ∂z = ∂R ∂x, and ∂Q ∂z = ∂R ∂y.
Proof. Since F = ∇f , we have P = fx, Q = fy, and R = fz. By Clairaut’s Theorem, we
know that Py = (fx)y = fxy = fyx = (fy)x = Qx, Pz = (fx)z = fxz = fzx = (fz)x = Rx, and
16.3 The Fundamental Theorem for Line Integrals goo.gl/C9i5sz 17
Example 13. If F(x, y, z) = y2i+ (2xy + e3z) j+ 3ye3zk, find a function f such that ∇f = F. 這個例題也是先練 習如何確實找出位 勢函數。 Solution. LE1g8HoZ7BQ 這個函數可以幫助 我們仔細思考這一 節所有理論的邏輯 關係,包括Qx = P y、 保守向量場、 積分與路徑選取無 關、 位勢函數的存 在性, 必須自己把 所有觀念重新整理 一次。
Example 14 (page 1095). Let F(x, y) = P (x, y) i + Q(x, y) j = x2−y+y2i+
x x2+y2j.
(a) Show that ∂Q∂x = ∂P∂y. (b) Show that
CF· dr is not independent of path.
(c) Compute ∇θ(x, y), where θ = θ(x, y) is the polar angle function. Solution.
Appendix: Conservation of Energy, page 1093
We will apply these ideas to a continuous force field F that moves an object along a path C given by r(t), a ≤ t ≤ b, where r(a) = A is the initial point and r(b) = B is the terminal point of C. 0-xRNBTrdqY 至此我們可以利用 現學到的微積分觀 念證明物理課所學 的能量守恆定律。
According to Newton’s Second Law of Motion, the force F(r(t)) at a point on C is related to the acceleration a(t) = r′′(t) by the equation F(r(t)) = m r′′(t), so the work done by the
force on the object is
W = C F· dr = b a F(r(t)) · r′(t) dt = b a m r′′(t) · r′(t) dt = m 2 b a d dt(r ′ (t) · r′(t)) dt = m 2 b a d dt|r ′ (t)|2dt = m 2 h |r′(t)|i b a = m 2(|r ′(b)|2− |r′(a)|2) = 1 2m|v(b)| 2−1 2m|v(a)| 2, (5)
where v(t) = r′(t) is the velocity.
The quantity 12m|v(t)|2, is called the kinetic energy (動能) of the object. Therefore we can rewrite Equation (5) as W = K(B) − K(A), which says that the work done by the force field along C is equal to the change in kinetic energy at the endpoints of C.
Now let’s further assume that F is a conservative force field; that is, we can write F = ∇f . In physics, the potential energy (位能) of an object at the point (x, y, z) is defined as P(x, y, z) = −f (x, y, z), so we have F = ∇f = −∇P . Then we have
W = C F· dr = − C∇P · dr = −(P (r(b)) − P (r(a))) = P (A) − P (B).
Comparing this equation with W = K(B) − K(A), we see that P(A) + K(A) = P (B) + K(B),
which says that if an object moves from one point A to another point B under the influence of a conservative force field, then the sum of its potential energy and its kinetic energy remains constant. This is called the Law of Conservation of Energy (能量守恆定律) and it is the reason the vector field is called conservative.
16.4 Green’s Theorem goo.gl/4MQ4go 19
16.4
Green’s Theorem, page 1096
XBngC_gleX8 在介紹格林定理之 前, 要先定義區域 邊界的方向性, 我 們會說曲線以逆時 針方向行走時為正 的定向, 順時針方 向行走時為負的定 向。
Green’s Theorem gives the relationship between a line integral around a simple closed curve C and a double integral over the plane region D bounded by C.
Definition 1 (page 1096). We say a simple closed curve C is positive orientation (正的定向) if the curve is traverses counterclockwise.
If C is given by the vector function r(t), a ≤ t ≤ b, then the region D is always on the left at the point r(t) traverses C.
C C D D x x y y
Figure 1: Positive orientation (left) and negative orientation (right).
關於格林定理, 若 與 前 一 節 觀 念 連 結, 則可想成封閉 路徑積分是否為零 與向量場是否為保 守 向量 場 的 差 有 關。
Green’s Theorem (page 1096). Let C be a positive oriented, piecewise smooth, simple closed curve in the plane and letD be the region bounded by C. If P(x, y) and Q(x, y) have continuous partial derivatives on an open region that contains D, then
C Pdx + Q dy = D ∂Q ∂x − ∂P ∂y dA. Remark 2. Sometimes we use the following notations
C Pdx + Q dy, C Pdx + Q dy, or ∂D Pdx + Q dy to indicate that the line integral is calculated in the positive orientation.
JhdcsgJbJuA
利用格林定理, 可
以把第二類線積分 的問題轉換為區域 內的二重積分。
Example 3 (page 1098). Evaluate
Cx4dx+xy dy, where C is the triangular curve consisting
of the line segments from (0, 0) to (1, 0), form (1, 0) to (0, 1), and from (0, 1) to (0, 0). Solution.
Example 4 (page 1098). Evaluate
C(3y − e
sin x) dx + (7x +py4+ 1) dy, where C is the
circle x2+ y2 = 9. 這個向量場若要直 接計算線積分會很 有 難 度, 因 為 向 量場的分量帶有積 不出來的函數。 取 而 代 之 的 是, 利 用格林定理, 那些 可怕的項微分後消 失, 所以可以順利 進行重積分。 Solution. 縱使在區域內部的 向量場很複雜, 但 是在區域在邊界上 的向量場都是零向 量時, 格林定理告 知: 區域內部對於 Qx− Py 重積分 的結果也是零。
Example 5(page 1098). If P (x, y) = Q(x, y) = 0 on a simple closed curve C, and P (x, y), Q(x, y) satisfy the hypotheses of Green’s Theorem, then
no matter what values P and Q assume in the region D.
LD8z1qffEGM 適當的設定 P, Q 讓 Qx − Py = 1, 則格林定理的 右式會對到區域面 積, 而左式可以轉 變為線積分。
Example 6(page 1099). If we take (P, Q) = (0, x), (P, Q) = (−y, 0), and (P, Q) = (−12y,12x), then Green’s Theorem gives
Example 7 (page 1102). 特別地, 格林定理 可以證明你以前可 能學過的多邊形面 積公式, 只要把多 邊形的頂點逆時針 排列, 把坐標依序 標下 (第一個點重 新標註) 之後, 就 有類似交叉相乘互 減除以二的公式。
(a) If C is the line segment connecting the point (x1, y1) to the point (x2, y2), then
C− 1 2ydx + 1 2xdy = 1 2(x1y2− x2y1).
16.4 Green’s Theorem goo.gl/4MQ4go 21
(b) If the vertices of a polygon, in counterclockwise order, are (x1, y1), (x2, y2), . . ., (xn, yn),
then the area of the polygon is A= 1
2((x1y2− x2y1) + (x2y3− x3y2) + · · · + (xn−1yn− yn−1xn) + (xny1− ynx1)).
Extended Versions of Green’s Theorem, page 1099
kU5eA_o5XY0 區 域 的 內部 有 洞 的時候, 在區域上 Qx− Py 的重積 分與邊界上的線積 分 必須 重 新 討 論, 對外部邊界而言是 取逆時針, 內部邊 界 而 言 要 取 順 時 針, 其結果才會一 致。
Green’s Theorem can be extended to apply to regions with holes (genus), that is, regions that are not simply connected.
C1 C2 D x x y y D′ D′′ (a) (b)
Figure 2: Region D is not simply connected.
See Figure 2 (a). Observe that the boundary C of the region D consists of two simple closed curves C1 and C2. We assume that these boundary curves are oriented so that the
region D is always on the left as the curve C is traversed. Thus the positive direction is counterclockwise for the outer curve C1 but clockwise for the inner curve C2.
If we divide D into two region D′ and D′′ by means of the lines shown in Figure 2 (b), then we applying Green’s Theorem to each of D′ and D′′ to get
D ∂Q ∂x − ∂P ∂y dA = D′ ∂Q ∂x − ∂P ∂y dA + D′′ ∂Q ∂x − ∂P ∂y dA = ∂D′ Pdx + Q dy + ∂D′′ Pdx + Q dy.
Since the line integrals along the common boundary lines are in opposite directions, they cancel and we get
D ∂Q ∂x − ∂P ∂y dA = C1 Pdx + Q dy + C2 Pdx + Q dy = C Pdx + Q dy.
Example 8 (page 1100). Let F(x, y) = x2−y+y2i+ x x2+y2j. j_ZkqHvIEPY 這 個 向 量 場 在 (0, 0) 處 沒 有 定 義, 也就是說, 向 量 場 除 了 坐 標 原 點 以 外 是 一 次 微 分 仍連 續 的 向 量 場, 向量場除了坐 標 原 點 外 都 滿 足 Qx = Py, 所以 當一個封閉曲線內 部不包含坐標原點 時, 線積分可用保 守向量場的觀念處 理, 其值為零。 當 卦閉曲線內部包含 坐標原點時, 積分 值不為零, 用推廣 版本的格定定理驗 證包含坐標原點的 封 閉 曲 線 的 積 分 (逆時針), 與一個 包含坐標原點的圓 (逆時針) 的積分 一樣, 而圓形就可 以確實地參數化並 求值。
(a) Show that
CF· dr = 0 for every simple closed curve that does not encloses the origin.
(b) Show that
CF· dr = 2π for every positively oriented simple closed path that encloses
the origin. Solution.
16.4 Green’s Theorem goo.gl/4MQ4go 23
Appendix, page 1097
BwISG5qgT6E 附錄是關於格林定 理的證明。 對於一 個區域來說, 進行 切割之後讓每一小 塊是滿足 type I 或是type II的區 域, 所以整個問題 就化簡成分別討論 type I或type II 區域之下格林定理 是否成立。 更進一 步地, 格林定理的 線積分與重積分的 關係, 可以看到和 P有關的是一個等 式,和Q有關的是 另一個等式, 格林 定理是將這兩個結 果合併。Proof of Green’s Theorem in which D is a simple region. It suffices to show that
C P(x, y) dx = − D ∂P ∂y dA and C Q(x, y) dy = D ∂Q ∂x dA. y = y2(x) y= y1(x) D x y C1 C2 C3 C4
Figure 3: Simple Region D.
We express D as a type I region D = {(x, y)| a ≤ x ≤ b, y1(x) ≤ y ≤ y2(x)}, where y1(x)
and y2(x) are continuous functions. By the Fundamental Theorem of Calculus, we have
− D ∂P ∂y dA = − b a y2(x) y1(x) ∂P ∂y(x, y) dy dx = − b a (P (x, y2(x)) − P (x, y1(x))) dx.
On the other hand, we know C = C1∪ C2∪ C3∪ C4. On C1, we write the vector function
r1(t) = t i + y1(t) j, and t from a to b. So C1 P(x, y) dx = b a P(x, y1(x)) dx.
On C3, we use the vector function r3(t) = t i + y2(t) j, t from b to a. Therefore
C3 P(x, y) dx = a b P(t, y2(t)) dt = − b a P(x, y2(x)) dx.
On C2 or C4, x is constant, so dx = 0 and hence
C2 P(x, y) dx = 0 = C4 P(x, y) dx. Hence C P(x, y) dx = C1 P(x, y) dx + C2 P(x, y) dx + C3 P(x, y) dx + C4 P(x, y) dx = b a P(x, y1(x)) dx − b a P(x, y2(x)) dx = − D ∂P ∂y dA. Equality C Q(x, y) dy = D ∂Q
Proof of Theorem 10 in section 16.3. If C is any simple closed path in D and R is the region that encloses, then Green’s Theorem gives
hLomuqZIf84 這裡要補充證明前 一節的定理: 在單 連通區域下的一次 微分仍連續的向量 場若滿足 Qx = Py,則它是保守向 量場。 各位若之後有學到 微分幾何的話, 我 們會用類似像外積 的記號將面元賦予 符號, 透過這種符 號可以將格林定理 的敘述與微積分基 本定理統一起來。 C F· dr = C Pdx + Q dy = R ∂Q ∂x − ∂P ∂y dA = R 0 dA = 0.
A curve that is not simple crossed itself at one or more points and can be broken up into a number of simple curve. We have shown that the line integral of F around these simple curves are all 0 and, adding these integrals, we see that
CF· dr = 0 for any closed curve C.
Therefore
CF· dr is independent of path in D, and F is a conservative vector field.
Remark 9. In differential geometry, we define the “wedge product” or “exterior operator” on vectors or differential forms. Given two differential forms dx, dy, their wedge product dx ∧ dy means the positive oriented area element, so we have
dA = dx ∧ dy = −dy ∧ dx, and dx ∧ dx = 0, and d(dx) = 0.
Green’s Theorem can be regarded as the relationship between the integral, differential forms, and wedge product:
C Pdx + Q dy = D d(P dx + Q dy) = D ∂P ∂x dx + ∂P ∂y dy ∧ dx + P d(dx) + ∂Q∂x dx + ∂Q ∂y dy ∧ dy + Q d(dy) = D ∂P ∂y dy ∧ dx + D ∂Q ∂x dx ∧ dy = D ∂Q ∂x − ∂P ∂y dA.
