1 1 The Normal and Binormal Vectors2 2 The Normal and Binormal Vectors At a given point on a smooth space curve r(t), there are many vectors that are orthogonal to the unit tangent vector T(t). We single out one by observing that, because | T(t) | = 1 for all t, we have T(t) ! T'(t) = 0, so T'(t) is orthogonal to T(t). Note that T'(t) is itself not a unit vector. But at any point where ≠ 0 we can define the principal unit normal vector N(t) (or simply unit normal) as3 3 The Normal and Binormal Vectors The vector B(t) = T(t) × N(t) is called the binormal vector. It is perpendicular to both T and N and is also a unit vector. (See Figure 6.) Figure 64 4 The Normal and Binormal Vectors We summarize here the formulas for unit tangent, unit normal and binormal vectors, and curvature.Copyright © Cengage Learning. All rights reserved. 16.2 Line Integrals6 6 Line Integrals In this section we define an integral that is similar to a single integral except that instead of integrating over an interval [a, b], we integrate over a curve C. Such integrals are called line integrals, although “curve integrals” would be better terminology. They were invented in the early 19th century to solve problems involving fluid flow, forces, electricity, and magnetism.7 7 Line Integrals We start with a plane curve C given by the parametric equations x = x(t) y = y(t) a ≤ t ≤ b or, equivalently, by the vector equation r(t) = x(t) i + y(t) j, and we assume that C is a smooth curve. [This means that rʹ′ is continuous and rʹ′(t) ≠ 0.]8 8 Line Integrals If we divide the parameter interval [a, b] into n subintervals [ti – 1, ti] of equal width and we let xi = x(ti), and yi = y(ti), then the corresponding points Pi(xi, yi) divide C into n subarcs with lengths Δs1, Δs2, . . . , Δsn. (See Figure 1.) Figure 19 9 Line Integrals We choose any point in the ith subarc. (This corresponds to a point in [ti – 1, ti].) Now if f is any function of two variables whose domain includes the curve C, we evaluate f at the point multiply by the length Δsi of the subarc, and form the sum which is a Riemann sum.10 10 Line Integrals Then we take the limit of these sums and make the following definition by analogy with a single integral. We have found that the length of C is11 11 Line Integrals A similar type of argument can be used to show that if f is a continuous function, then the limit in Definition 2 always exists and the following formula can be used to evaluate the line integral: The value of the line integral does not depend on the parametrization of the curve, provided that the curve is traversed exactly once as t increases from a to b.12 12 Line Integrals If s(t) is the length of C between r(a) and r(t), then So the way to remember Formula 3 is to express everything in terms of the parameter t: Use the parametric equations to express x and y in terms of t and write ds as13 13 Line Integrals In the special case where C is the line segment that joins (a, 0) to (b, 0), using x as the parameter, we can write the parametric equations of C as follows: x = x, y = 0, a ≤ x ≤ b. Formula 3 then becomes and so the line integral reduces to an ordinary single integral in this case.14 14 Line Integrals Just as for an ordinary single integral, we can interpret the line integral of a positive function as an area. In fact, if f (x, y) ≥ 0, ∫C f (x, y) ds represents the area of one side of the “fence” or “curtain” in Figure 2, whose base is C and whose height above the point (x, y) is f (x, y). Figure 215 15 Example 1 Evaluate ∫C (2 + x2y) ds, where C is the upper half of the unit circle x2 + y2 = 1. Solution: In order to use Formula 3, we first need parametric equations to represent C. Recall that the unit circle can be parametrized by means of the equations x = cos t y = sin t and the upper half of the circle is described by the parameter interval 0 ≤ t ≤ π. (See Figure 3.) Figure 316 16 Example 1 – Solution Therefore Formula 3 gives cont’d17 17 Line Integrals Suppose now that C is a piecewise-smooth curve; that is, C is a union of a finite number of smooth curves C1, C2, …., Cn, where, as illustrated in Figure 4, the initial point of Ci + 1 is the terminal point of Ci. Figure 4 A piecewise-smooth curve18 18 Line Integrals Then we define the integral of f along C as the sum of the integrals of f along each of the smooth pieces of C:19 19 Line Integrals Any physical interpretation of a line integral ∫C f (x, y) ds depends on the physical interpretation of the function f. Suppose that ρ(x, y) represents the linear density at a point (x, y) of a thin wire shaped like a curve C. Then the mass of the part of the wire from Pi – 1 to Pi in Figure1 is approximately and so the total mass of the wire is approximately Figure 120 20 Line Integrals By taking more and more points on the curve, we obtain the mass m of the wire as the limiting value of these approximations: [For example, if f (x, y) = 2 + x2y represents the density of a semicircular wire, then the integral in Example 1 would represent the mass of the wire.]21 21 Line Integrals The center of mass of the wire with density function ρ is located at the point , where22 22 Line Integrals Two other line integrals are obtained by replacing Δsi by either Δxi = xi – xi –1 or Δyi = yi – yi – 1 in Definition 2. They are called the line integrals of f along C with respect to x and y:23 23 Line Integrals When we want to distinguish the original line integral ∫C f (x, y) ds from those in Equations 5 and 6, we call it the line integral with respect to arc length. The following formulas say that line integrals with respect to x and y can also be evaluated by expressing everything in terms of t : x = x(t), y = y(t), dx = x ʹ′(t) dt, dy = y ʹ′(t) dt.24 24 Line Integrals 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 When we are setting up a line integral, sometimes the most difficult thing is to think of a parametric representation …
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