Lecture 11: Arclength and Line Integrals

Transcription

1 Lecture 11: Arclength and Line Integrals Rafikul Alam Department of Mathematics IIT Guwahati

2 Parametric curves Definition: A continuous mapping γ : [a, b] R n is called a parametric curve or a parametrized path and [a, b] is called the parameter space. The set := γ([a, b]) is called a geometric curve or a geometric path in R n. Examples: The parametric path γ(t) := (cos t, sin t) for t [, 2π] is a circle in R 2. The parametric path γ(t) := (cos t, sin t, t) for t [, 2π] is helix in R 3.

3 Polygonal approximations of paths Let γ : [a, b] R n be a parametric path. For a partition P := {t,..., t m } of [a, b], define l(p, γ) := m γ(t j ) γ(t j 1 ) j=1 and µ(p) := max{t j t j 1 : j = 1 : m}. Note that l(p, γ) l(q, γ) if Q is a refinement of P. Hence lim µ(p) l(p, γ) = sup l(p, γ). P

4 Arclength of a curve Definition: Let l(γ) := sup l(p, γ). If l(γ) is finite then γ is P said to be rectifiable (finite length) and l(γ) is said to be the arclength of γ. Theorem: Let γ : [a.b] R n be a C 1 (or PC 1 ) path. Then γ is rectifiable and b l(γ) = lim l(p, γ) = γ (t) dt. µ(p) a Proof: Use γ(t j ) γ(t j 1 ) = γ (t j 1 ) t j + e( t j ) t j with e( t j ) as t j and the Riemann sum of γ (t).

5 Arclength If f : [a, b] R is C 1 then the length of the graph of f is given by b 1 + f (x) 2 dx. a If γ : [a, b] R 2 is C 1 and γ(t) = (x(t), y(t)) then l(γ) = b a x (t) 2 + y (t) 2 dt. If γ : [a, b] R 3 is C 1 and γ(t) = (x(t), y(t)), z(t)) then l(γ) = b a x (t) 2 + y (t) 2 + z (t) 2 dt.

6 Examples The arclength of the helix γ(t) := (cos t, sin t, t) for t [, 2π] is given by 2π γ (t) dt = 2π 2dt = 2 2 π. The arclength of γ(t) := (cos t, sin t, cos(2t), sin(2t)) for t [, π] is given by π γ (t) dt = 2π 5dt = 5 π.

7 Invariance of arclength Obviously the arclength l(γ) depends to the parametrization γ. However, l(γ) is invariant under equivalent parametrization. Definition: Two parametric curves γ : [a, b] R n and ρ : [c, d] R n are said to be equivalent if there exits a continuous function u : [a, b] [c, d] such that u is strictly increasing u([a, b]) = [c, d] and γ = ρ u. Theorem: Let γ : [a, b] R n be rectifiable. Then every parametric curve equivalent to γ is rectifiable and has the same arclength l(γ).

8 Arclength differential Let γ : [a, b] R n be a C 1 path. Define s : [a, b] [, l] by where l := l(γ). Then s(t) := t a γ (τ) τ, ds dt = γ (t) or equivalently ds = γ (t) dt. ds is called the arclength differential and is written as ds = dx 2 + dy 2 when γ(t) = (x(t), y(t)) ds = dx 2 + dy 2 + dz 2 when γ(t) = (x(t), y(t), z(t))

9 Smooth parametrization A parametric curve : [a, b] R n is said to be smooth if γ is C 1 on [a, b] and γ (t) for t (a, b), piecewise smooth if γ is smooth on [t j 1, t j ] for some partition t,..., t m of [a, b].

10 Smooth parametrization by arclength Let γ : [a, b] R n be a smooth (or piecewise smooth) curve. Then ds t dt = γ (t) > = s(t) = γ (t) dt a is strictly increasing and s([a, b]) = [, l]. Consequently ρ : [, l] R n given by ρ(u) = γ(s 1 (u)) is equivalent to γ. Moreover, ρ (u) = 1 which gives l(ρ) = l ρ (u) du = l = l(γ).

11 Examples Consider the helix γ(t) := (cos t, sin t, t) for t [, 2π]. Then t s(t) = 2 dt = 2t t = s/ 2. Hence ρ : [, 2 2π] R 3 given by ρ(s) := (cos(s/ 2), sin(s/ 2), s/ 2) is a smooth parametrization of the helix γ by arclength. Consider γ(t) := (cos t, sin t) for t [, 2π]. Then s = t dt = t ρ(s) := (cos(s), sin(s)), s [, 2π] is a smooth parametrization of γ by arclength.

12 Partition of curves Let be a curve in R n paramatrized by r : [a, b] R n. Then a partion P := (a = t <... < t m = b) of [a, b] induces a partition of into m subarcs with arclenths s 1,..., s m. Define µ(p) := max 1 j m s j.

13 Riemann sum of scalar field w.r.t. arclength Let f : R. Then for any p j in the j-th subarc, consider the Riemann sum of f w.r.t. to the arclength S(P, f ) := m f (p j ) s j. j=1

14 Line integrals of scalar fields w.r.t. arclength Definition: Suppose that r is PC 1 and f : R. Then the line integral of f along w.r.t. the arclength is given by m f (x)ds := f (p j ) s j if the limit exists. lim S(P, f ) = µ(p) lim µ(p) j=1 Fact: If f is continuous and r(t) is PC 1 then we have b f (x)ds = f (r(t)) r (t) dt. a Proof: Since s r (t) t, i.e., ds = r (t) dt and t f (r(t)) r (t) is piecewise continuous, the result follows.

15 Line integrals of scalar fields For the plane curve : r(t) = (x(t), y(t)), t [a, b] we have b f (x, y)ds = f (x(t), y(t)) x (t) 2 + y (t) 2 dt. a Example: Evaluate (2 + x 2 y)ds, where is the upper half of the circle x 2 + y 2 = 1. Considering x(t) = cos t, y(t) = sin t, t π, we have π (2 + x 2 y)ds = (2 + cos 2 t sin t)dt = 2π + 2/3.

16 Application of line integrals Suppose a thin wire in the shape of a curve parametrized by a C 1 path r : [a, b] R 2 has density ρ(x, y). Then the total mass of the wire is given by m = f (x, y)ds. The center of mass ( x, ȳ) is given by x = 1 xρ(x, y)ds, ȳ = 1 m m yρ(x, y)ds. The moment of inertia about a line L is given by I L = δ(x, y) 2 ρ(x, y)ds, where δ(x, y) is shortest distance from (x, y) to L.

17 Example The wire W has the shape = 1 + 2, where 1 is parametrized by γ 1 (t) := (cos t, sin t), t [, π] and 2 is parametrized by γ 2 (t) := (t, ), t [ 1, 1]. The density is given by f (x, y) := x 2 + y 2. Then γ i (t) = 1, f (γ 1(t)) = 1 and f ( 2 (t)) = t. Thus π 1 m = fds = dt tdt + tdt = π + 1, 1 ( 1 π 1 ) x = cos tdt t 2 dt + t 2 dt =, π ( 1 π ) ȳ = sin tdt + = 2 π + 1 π + 1.

18 Example (contd.) For the moment of inertia I x about x axis, we have δ(x, y) = y and hence I x = π sin 2 tdt + = π 1 2 (1 cos(2t))dt = π 2. For the moment of inertia I y about y axis, we have δ(x, y) = x and hence I y = π cos 2 tdt 1 t 3 dt + 1 t 3 dt = π

19 Properties of line integrals of scalar fields Fact: Let be parametrized by a PC 1 curve r : [a, b] R n and f, g : R be continuous. Then the following hold: fds is invariant under equivalent parametrization of. (f + αg)ds = fds + α gds for α R. Let = m, where i is parametrized by C 1 curve r i : [a i, b i ] R n. Then fds = fds fds. m

20 Example Evaluate 2xds, where consists of the arc C 1 of the parabola y = x 2 from (, ) to (1, 1) followed by the line segment C 2 from (1, 1) to (1, 2). Then 2xds = 2xds + 2xds = 1 C 1 C 2 6 ( ). *** End ***