Topic 5.1: Line Elements and Scalar Line Integrals. Textbook: Section 16.2
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1 Topic 5.1: Line Elements and Scalar Line Integrals Textbook: Section 16.2
2 Warm-Up: Derivatives of Vector Functions Suppose r(t) = x(t) î + y(t) ĵ + z(t) ˆk parameterizes a curve C. The vector: is: r (t) = x (t) î + y (t) ĵ + z (t) ˆk A. Perpendicular to the curve C. B. Orthogonal to the curve C. C. Tangent to the curve C. D. Normal to the curve C. E. I don t remember how to answer this question.
3 Big Ideas The vector line element dr is the infinitesimal tangent vector. The scalar line element ds = dr is the magnitude of the vector line element the infinitesimal Pythagorean theorem. ˆ Scalar line integrals f ds are one-dimensional integrals, where the region of integration is a curve. C
4 What We Are Doing Today... In Calc I/II, we integrated functions of a single variable, where the domain of integration was a line segment an interval on a coordinate axis. Now, we want to be able to compute integrals where the domain of integration is a curve in space. To do this, we will need to: Parameterize the curve this gives us a way of writing our integral in terms of a single variable (the parameter). Find a way to measure distance along the curve this is the infinitesimal yardstick ds, called the scalar line element.
5 Review for this Topic: Vector functions/parameterized curves (sec 13.1). Vector derivatives and tangent vectors (sec 13.2).
6 Recall: Curves and Vector Functions A curve C can be parameterized by a vector function: r(t) = x(t), y(t), z(t) = x(t) î + y(t) ĵ + z(t) ˆk Or by coordinate functions (aka parametric equations): x = x(t), y = y(t), z = z(t) The vector derivative, or tangent vector: r (t) = x (t) î + y (t) ĵ + z (t) ˆk is tangent to the curve C.
7 Example: Curves and Vector Functions Find vector functions r(t) and their derivatives r (t) for the following curves: The circle of radius r = 3, centered at the origin. The parabola y = x 2 1, from the point ( 1, 0) to the point (2, 3).
8 The Differentials dr and ds The vector differential is the vector: dr = dx î + dy ĵ + dz ˆk. If dx, dy, dz represent infinitesimal changes in x, y, and z, then dr is an infinitesimal change in position from the point (x, y, z) to the point (x + dx, y + dy, z + dz). The magnitude of the vector differential, denoted ds, is the infinitesimal distance: ds = dr = dx 2 + dy 2 + dz 2. ds is the infinitesimal version of the Pythagorean theorem! We will use ds as our infinitesimal yardstick to measure distance along curves.
9 The Differentials dr and ds The Bridge Book A great explanation of the differentials dr and ds can be found at the Bridge Book Wiki:
10 The Vector Line Element Computing dr on a Parametrized Curve Suppose a curve C is parameterized by the vector function: r(t) = x(t) î + y(t) ĵ + z(t) ˆk. The vector line element is the vector differential dr along C: dr = d [ x(t) ] î + d [ y(t) ] ĵ + d [ z(t) ] ˆk = x (t) dt î + y (t) dt ĵ + z (t) dt ˆk = ( x (t) î + y (t) ĵ + z (t) ˆk ) dt = r (t) dt Recall from u-sub in Calc I/II: If u = f (t), then du = f (t)dt.
11 The Scalar Line Element Computing ds on a Parameterized Curve Suppose a curve C is parameterized by the vector function: r(t) = x(t) î + y(t) ĵ + z(t) ˆk The scalar line element ds is the magnitude of the vector line element dr: ds = r (t) dt = [x (t) ] 2 + [ y (t) ] 2 + [ z (t) ] 2 dt ds is the length of an infinitesimal segment of the curve C. Or: ds is the distance traveled along C over an infinitesimal time increment.
12 Scalar Line Integrals The scalar line integral of f (x, y, z) over a curve C is: ˆ f (x, y, z) ds = ˆ b C a Where: C is the curve. f ( x(t), y(t), z(t) ) r (t) dt r(t) = x(t) î + y(t) ĵ + z(t) ˆk, for a t b, is a parameterization of the curve C. r(a) and r(b) are the endpoints of the curve C. ds = r (t) dt
13 Setting Up the Scalar Line Integral f (x, y, z) ds, Using a Parameterization C Parameterize the curve C that is, find r(t). Limits of Integration: Find the interval a t b so that r(a) and r(b) are the endpoints of C. Then the lower limit of integration is t = a, and the upper limit is t = b. Integrand: Substitute the component functions x(t), y(t), and z(t) of the parameterization for x, y, and z in the function f (x, y, z). After you do this, you will have a function only in t. Line Element: Compute the line element ds = r (t) dt.
14 Example: Scalar Line Integrals Set up the scalar line integral f (x, y) ds for each of the C following: f (x, y) = xy 2, and C is the right half of the circle x 2 + y 2 = 1. (HW 25 # 4) f (x, y, z) = 4x + 18z, and C is described by the coordinate functions: x = t, y = t 2, z = t 3, 0 t 1. (HW 25 # 3)
15 Applications of Scalar Line Integrals: Arc Length The length of a curve C is: ˆ s = ds Chop the curve into small pieces of length ds. C Add up all the lengths (integrate).
16 Example: Computing Arc Length Find the length of the curve r(t) = t 2 î + t 3 ĵ between the points P = (1, 1) and Q = (4, 8). (related to HW 25 # 7)
17 Applications of Scalar Line Integrals: Integrating Density Functions Suppose a density along a curve C is given by δ(x, y, z) 0. The total mass m along C is: ˆ m = δ(x, y, z) ds C Chop the curve into small pieces of length ds. The mass of a piece of length ds is dm = δ(x, y, z) ds. Add up the mass of all the pieces (integrate).
18 Applications of Scalar Line Integrals: Integrating Density Functions Suppose a curve C is parametrized by the vector function: r(t) = x(t) î + y(t) ĵ + z(t) ˆk a t b and δ(x, y, z) 0 represents a density along C. The total mass m along C is: ˆ m = C δ ds = ˆ b a δ ( x(t), y(t), z(t) ) r (t) dt Chop the curve into small pieces of length ds. The mass of a piece of length ds is dm = δ(x, y, z) ds. Add up the mass of each little piece (integrate).
19 Example: Mass Find the total mass of a wire bent into a helical shape, described by: r(t) = cos t, sin t, t 0 t 4π (distance measured in centimeters) assuming a mass density of δ(x, y, z) = z g/cm.
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