13.1 Vector Functions and Space Curves goo.gl/qmVBuU 1
Chapter 13
Vector Functions
13.1
Vector Functions and Space Curves (page 848)
Wwwa5xVCCrE 這一章本身有一個 主體性, 是在了解 空間曲線理論; 另 一方面, 前兩節的 向量函數概念也是 之後雙變數函數理 論的前置作業。 簡 單說來, 之前都是 在 探討 單 變 數 函 數, 函數圖形也是 放在xy平面上觀 察, 從參數式的觀 點下, 現在要對於 「維度」 開始推廣, 變成現在所要討論 的向量函數。
We now study functions whose values are vectors because such functions are needed to describe curves and surfaces in space.
Definition 1(page 848). A vector-valued function, or vector function (向量函數), is a function whose domain is a set of real numbers and whose range is a set of vectors.
Here we will focus on vector functions r whose values are three-dimensional vectors. This means that for every number t in the domain of r there is a unique vector in R3 denoted by
r(t). We can write
r(t) = (f (t), g(t), h(t)) = f (t) i + g(t) j + h(t) k,
where f, g, h are real-valued functions of t called the component functions (分量函數) of r.
定義中 i= (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) 為R3 中的直角坐標向量。 課本用尖括號 r(t) = hf (t), g(t), h(t)i表示向量函數,但大部分文獻仍使用小括號。Example 2 (page 848). If r(t) = (t3,ln(3 − t),√t), then the component functions are
f(t) = t3, g(t) = ln(3 − t), and h(t) =√t.
By the usual convention, the domain (定義域) of r consists of all values of t for which the
expresstion for r(t) is defined. Therefore the domain of r is [0, 3).
Definition 3 (page 848). The limit of a vector function r is defined by taking the limits of its components functions as follows. If r(t) = (f (t), g(t), h(t)), then
lim
t→ar(t) =
lim
t→af(t), limt→ag(t), limt→ah(t)
= lim
t→af(t) i + limt→ag(t) j + limt→ah(t) k
provided the limits of the component functions exist.
Definition 4 (page 849). A vector function r is continuous at a if lim
t→ar(t) = r(a).
「連續」 可以看成 「函數」 與 「極限」 可以互換。 向量函數 r (在 t= a) 連續若且唯若所有分量函數f(t), g(t), h(t) (在 t= a) 連續。There is a closed connection between vector-valued functions and space curves.
gnA8LWTfFFY 研究空間曲線的方 式是用參數方程理 解它, 這樣就可以 用微積分的理論了 解 曲 線 的 幾 何 性 質。
Definition 5 (page 849). The set C of all points (x, y, z) in space, where
x= f (t), y= g(t), z= h(t), (1)
and t varies throughout the interval I, is called a space curve (空間曲線). The equations in
(1) are called parametric equations of C (參數方程) and t is called a parameter (參數).
有時候 r(t)也稱為位置向量 (position vector)。2 13.1 Vector Functions and Space Curves goo.gl/qmVBuU
Example 6 (page 849). The curve r(t) = cos t i + sin t j + t k is called a helix (螺旋線).
螺旋線算是空間曲 線論中最標準的一 個模型, 我們會用 螺旋線認識各式幾 何量, 再深入探討 其理論。 x y z
Figure 1: A helix r(t) = cos t i + sin t j + t k, 0 ≤ t ≤ 6π.
Example 7. 現在要學習如何把 曲線的參數方程表 達; 特別是有些曲 線是兩個曲面的交 集, 該如何寫出其 參數方程是需要一 些經驗的累積。
(a) Find a vector equation and parametric equations for the line that join the point A(a1, a2, a3)
to the point B(b1, b2, b3).
(b) Find a vector equation and parametric equations that represents the curve of intersec-tion of the cylinder x2+ y2 = 1 and the plane y + z = 2.
Solution.
Using computers to draw space curves, page 851
現在有非常多的數 學繪圖軟體, 只要 輸入參數方程, 它 就可以在電腦上呈 現曲線的長相, 可 幫助我們更快認識 空間曲線。
(a) Toroidal spiral: x = (4 + sin 7t) cos t, y = (4 + sin 7t) sin t, z = cos 7t, 0 ≤ t ≤ 2π. (b) Trefoil knot: x = (2 + cos 1.5t) cos t, y = (2 + cos 1.5t) sin t, z = sin 1.5t, 0 ≤ t ≤ 4π.
(c) Twisted cubic: x = t, y= t2, z= t3. x y z x y z x y z
Figure 2: (a) Toroidal spiral. (b) Trefoil knot. (c) Twisted cubic.
Exercise (page 855). Find a vector function that represents the curve of intersection of the two surfaces.
(a) The cone z =px2+ y2 and the plane z = 1 + y.
13.2 Derivatives and Integrals of Vector Functions goo.gl/nCXC9e 3
13.2
Derivatives and Integrals of Vector Functions
(page 855)
Derivatives, page 855
qVOC-rk4i6I 這一節要認識一些 向量函數的微積分 理論, 向量函數的 導函數就是每個分 量 都 各 自 算 導 函 數, 放在一起就形 成一個向量, 它會 是 曲 線 的 切 向 量; 只要切向量不是零 向量, 那麼就可以 定義曲線的切線以 及單位切向量。Definition 1 (page 855). The derivative (導函數) r′
(t) of a vector function r(t) is dr dt = r ′ (t) = lim h→0 r(t + h) − r(t) h . Definition 2 (page 856).
(a) The vector r′
(t0) is called the tangent vector (切向量) to the curve C defined by r(t) at
the point P = r(t0), provided that r′(t0) exists and r′(t0) 6= 0.
(b) The tangent line (切線) to the curve C at P = r(t0) is defined to be the line through P
parallel to the tangent vector r′
(t0).
(c) The unit tangent vector (單位切向量) is
T(t) = r
′
(t) kr′
(t)k.
Theorem 3 (page 856). If r(t) = (f (t), g(t), h(t)) = f (t) i + g(t) j + h(t) k, where f, g, and h are differentiable functions, then
r′ (t) = (f′ (t), g′ (t), h′ (t)) = f′ (t) i + g′ (t) j + h′ (t) k. Ps6fqT_Cxsg 向量函數的四則運 算 與 微 分 的 規 則, 因為向量之間又還 有內積與外積的計 算, 所以又會多了 一些關係式; 在看 定理前, 必須先了 解 每 個 符 號 的 本 質: 在運算之前與 之 後 到 底 是 數 字、 函數、 還是向量必 須確認清楚。
Theorem 4 (page 858). Suppose u(t) and v(t) are differentiable vector functions, c is a
scalar, and f(t) is a real-valued function. Then
(1) d dt(u(t) + v(t)) = u ′ (t) + v′ (t). (2) d dt(c u(t)) = c u ′ (t). (3) d dt(f (t) u(t)) = f ′ (t) u(t) + f (t) u′ (t). (4) dtd(u(t) · v(t)) = u′ (t) · v(t) + u(t) · v′ (t). (5) dtd(u(t) × v(t)) = u′ (t) × v(t) + u(t) × v′ (t). (6) dtd(u(f (t))) = u′ (f (t))f′ (t).
Proof. Let u(t) = u1(t) i + u2(t) j + u3(t) k and v(t) = v1(t) i + v2(t) j + v3(t) k.
(1) d dt(u(t) + v(t)) = limh→0 u(t + h) + v(t + h) − (u(t) + v(t)) h = lim h→0 u(t + h) − u(t) h + limh→0 v(t + h) − v(t) h = u ′ (t) + v′ (t).
4 13.2 Derivatives and Integrals of Vector Functions goo.gl/nCXC9e (2) d dt(c u(t)) = limh→0 c u(t + h) − c u(t) h = c limh→0 u(t + h) − u(t) h = c u ′ (t). (4) d dt(u(t) · v(t)) = d dt(u1(t)v1(t) + u2(t)v2(t) + u3(t)v3(t)) = u′ 1(t)v1(t) + u ′ 2(t)v2(t) + u ′ 3(t)v3(t) + u1(t)v ′ 1(t) + u2(t)v ′ 2(t) + u3(t)v ′ 3(t) = u′ (t) · v(t) + u(t) · v′ (t). (6) d dt(u(f (t))) = d dt(u1(f (t)) i + u2(f (t)) j + u3(f (t)) k) = u′ 1(f (t))f ′ (t) i + u′ 2(f (t))f ′ (t) j + u′ 3(f (t))f ′ (t) k = (u′ 1(f (t)) i + u ′ 2(f (t)) j + u ′ 3(f (t)) k)f ′ (t) = u′ (f (t))f′ (t).
Exercise. Show that
(3) d dt(f (t) u(t)) = f ′ (t) u(t) + f (t) u′ (t). (5) dtd(u(t) × v(t)) = u′ (t) × v(t) + u(t) × v′ (t). xUFKBF-aRUk 這個式子算式看起 來簡單, 但很容易 被人忽略。 我們可 以把這件事情賦予 幾何意義: 若曲線 落在半徑為 c 的 球上, 球心與坐標 中心一致, 則曲線 的位置向量與切向 量處處垂直。 向量 函 數 的 積 分, 基本上就是每個分 量函數各自積分的 意思。
Example 5 (page 858). If kr(t)k = c (a constant), then r(t) · r(t) = c2 and d
dt(r(t) · r(t)) = Thus r′
(t) · r(t) = 0, which says that .
Exercise (page 861). If r(t) 6= 0, show that d
dtkr(t)k =
r(t) · r′
(t)
kr(t)k .
Integrals, page 859
The definite integral (定積分) of a continuous vector function r(t) is Z b a r(t) dt = lim n→∞ n X i=1 r(t∗ i)∆t = lim n→∞ n X i=1 f(t∗ i)∆t ! i+ n X i=1 g(t∗ i)∆t ! j+ n X i=1 h(t∗ i)∆t ! k ! = Z b a f(t) dt i+ Z b a g(t) dt j+ Z b a h(t) dt k.
We can extend the Fundamental Theorem of Calculus to continuous vector functions as fol-lows: Z b a r(t) dt = hR(t)i t=b t=a= R(b) − R(a),
where R is an antiderivative of r, that is, R′
(t) = r(t). We use the notation R r(t) dt for indefinite integrals (不定積分).
