1 1 Vector Functions and Space Curves In general, a function is a rule that assigns to each element in the domain an element in the range. A vector-valued function, or vector function, is simply a function whose domain is a set of real numbers and whose range is a set of vectors. We are most interested in 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 V3 denoted by r(t).2 2 Vector Functions and Space Curves If f (t), g(t), and h(t) are the components of the vector r(t), then f, g, and h are real-valued functions called the component functions of r and we can write r(t) = 〈f (t), g(t), h(t)〉 = f (t)i + g(t)j + h(t)k We use the letter t to denote the independent variable because it represents time in most applications of vector functions.3 3 Example 1 – Domain of a vector function If r(t) = 〈t 3, ln(3 – t), 〉 then the component functions are f (t) = t 3 g(t) = ln(3 – t) h(t) = By our usual convention, the domain of r consists of all values of t for which the expression for r(t) is defined. The expressions t 3, ln(3 – t), and are all defined when 3 – t > 0 and t ≥ 0. Therefore the domain of r is the interval [0, 3).4 4 Vector Functions and Space Curves The limit of a vector function r is defined by taking the limits of its component functions as follows. Limits of vector functions obey the same rules as limits of real-valued functions.5 5 Vector Functions and Space Curves A vector function r is continuous at a if In view of Definition 1, we see that r is continuous at a if and only if its component functions f, g, and h are continuous at a. There is a close connection between continuous vector functions and space curves.6 6 Vector Functions and Space Curves Suppose that f, g, and h are continuous real-valued functions on an interval I. Then the set C of all points (x, y, z) in space, where x = f (t) y = g(t) z = h(t) and t varies throughout the interval I, is called a space curve. The equations in are called parametric equations of C and t is called a parameter. We can think of C as being traced out by a moving particle whose position at time t is (f (t), g(t), h(t)).7 7 Vector Functions and Space Curves If we now consider the vector function r(t) = 〈f (t), g(t), h(t)〉, then r(t) is the position vector of the point P(f (t), g(t), h(t)) on C. Thus any continuous vector function r defines a space curve C that is traced out by the tip of the moving vector r(t), as shown in Figure 1. Figure 1 C is traced out by the tip of a moving position vector r(t).8 8 Example 4 – Sketching a helix Sketch the curve whose vector equation is r(t) = cos t i + sin t j + t k Solution: The parametric equations for this curve are x = cos t y = sin t z = t Since x2 + y2 = cos2t + sin2t = 1, the curve must lie on the circular cylinder x2 + y2 = 1. The point (x, y, z) lies directly above the point (x, y, 0), which moves counterclockwise around the circle x2 + y2 = 1 in the xy-plane.9 9 Example 4 – Solution (The projection of the curve onto the xy-plane has vector equation r(t) = 〈cos t, sin t, 0〉.) Since z = t, the curve spirals upward around the cylinder as t increases. The curve, shown in Figure 2, is called a helix. cont’d Figure 210 10 Derivatives11 11 Derivatives The derivative r ʹ′ of a vector function r is defined in much the same way as for real valued functions: if this limit exists. The geometric significance of this definition is shown in Figure 1. Figure 1 (b) The tangent vector rʹ′(t) (a) The secant vector12 12 Derivatives If the points P and Q have position vectors r(t) and r(t + h), then represents the vector r(t + h) – r(t), which can therefore be regarded as a secant vector. If h > 0, the scalar multiple (1/h)(r(t + h) – r(t)) has the same direction as r(t + h) – r(t). As h → 0, it appears that this vector approaches a vector that lies on the tangent line. For this reason, the vector r ʹ′(t) is called the tangent vector to the curve defined by r at the point P, provided that r ʹ′(t) exists and r ʹ′(t) ≠ 0.13 13 Derivatives The tangent line to C at P is defined to be the line through P parallel to the tangent vector r ʹ′(t). We will also have occasion to consider the unit tangent vector, which is14 14 Example 1 (a) Find the derivative of r(t) = (1 + t3)i + te–t j + sin 2t k. (b) Find the unit tangent vector at the point where t = 0. Solution: (a) According to Theorem 2, we differentiate each component of r: r ʹ′(t) = 3t2i + (1 – t)e–t j + 2 cos 2t k15 15 Example 1 – Solution (b) Since r(0) = i and r ʹ′(0) = j + 2k, the unit tangent vector at the point (1, 0, 0) is cont’d16 16 Derivatives Just as for real-valued functions, the second derivative of a vector function r is the derivative of r ʹ′, that is, r ʺ″ = (r ʹ′)ʹ′. For instance, the second derivative of the function, r(t) = 〈2 cos t, sin t, t〉, is r ʺ″t = 〈–2 cos t, –sin t, 0〉17 17 Differentiation Rules18 18 Differentiation Rules The next theorem shows that the differentiation formulas for real-valued functions have their counterparts for vector-valued functions.19 19 Example 4 Show that if | r(t) | = c (a constant), then r ʹ′(t) is orthogonal to r(t) for all t. Solution: Since r(t) ! r(t) = | r(t) |2 = c2 and c2 is a constant, Formula 4 of Theorem 3 gives 0 = [r(t) ! r(t)] = r ʹ′(t) ! r(t) + r(t) ! r ʹ′(t) = 2r ʹ′(t) ! r(t) Thus r ʹ′(t) ! r(t) = 0, which says that r ʹ′(t) is orthogonal to r(t).20 20 Example 4 – Solution Geometrically, this result says that if a curve lies on a sphere with center the origin, then the tangent vector r ʹ′(t) is always perpendicular to the position vector r(t). cont’d21 21 Integrals22 22 Integrals The definite integral of a continuous vector function r (t) can be defined in much the same way as for real-valued functions except that the integral is a vector. But then we can express the integral of r in terms of the integrals of its component functions f, g, and h as follows.23 23 Integrals and so This means that we can evaluate an integral of …
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