Stokes and the Surveyor s Shoelaces
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1 Stokes and the Surveyor s Shoelaces Dr. LaLonde UT Tyler Math Club February 15, 2017
2 Finding Areas of Polygons Problem: Is there a way to quickly find the area of a polygon just by knowing where its vertices are? (2, 6) (1, 4) (4, 5) (5, 5) (5, 4) (2, 3) (3, 3) (1, 2) (2, 1)
3 Finding Areas of Polygons One approach: Triangulate the polygon. (2, 6) (1, 4) (1, 2) (2, 3) (4, 5) (3, 3) (5, 5) (5, 4) A = = 19 2 (2, 1)
4 Finding Areas of Polygons I want an easier way. List out the coordinates of the points in counterclockwise order. Then cross multiply Now add up both columns, subtract the left from the right, and divide by 2: 1 2 (108 89) = 19 2 That s the area!
5 Areas of Triangles How do we know this algorithm works in general? Let s check it first for triangles. First assume things look like this: (0, 0) (x 2, y 2 ) (x 1, 0) It works, since the area is A = 1 2 x 1y x x 2 y x 1 y x 1 y 2
6 Areas of Triangles Now suppose our triangle isn t situated so nicely. Then rotate it. (x 2, y 2 ) (x 2 cos θ + y 2 sin θ, y 2 cos θ x 2 sin θ) (x 1, y 1 ) (0, 0) θ (x 1 cos θ + y 1 sin θ, y 1 cos θ x 1 sin θ) 2A = (x 1 cos θ + y 1 sin θ)(y 2 cos θ x 2 sin θ) (x 2 cos θ + y 2 sin θ)(y 1 cos θ x 1 sin θ) = x 1y 2 cos 2 θ x 1x 2 sin θ cos θ + y 1y 2 sin θ cos θ x 2y 1 sin 2 θ x 2y 1 cos 2 θ + x 1x 2 sin θ cos θ y 1y 2 sin θ cos θ + x 1y 2 sin 2 θ = x 1y 2 x 2y 1
7 Areas of Triangles There s a shorthand notation for this it s an example of a determinant. A = 1 2 x 1 y 1 x 2 y 2 = 1 2 (x 1y 2 x 2 y 1 ). You may have seen this sort of area formula in one of your classes that uses matrices.
8 Areas of Triangles Finally, what if the triangle isn t based at the origin? Then shift it. (x 2, y 2 ) (x 2 x 0, y 2 y 0 ) (x 1, y 1 ) (x 0, y 0 ) (x 1 x 0, y 1 y 0 ) (0, 0)
9 Areas of Triangles Now use the determinant formula: A = 1 2 x 1 x 0 y 1 y 0 x 2 x 0 y 2 y 0 [ ] x1 y 2 x 1 y 0 x 0 y 2 + x 0 y 0 x 2 y 1 + x 2 y 0 + x 0 y 1 x 0 y 0 = 1 2 = 1 2[ x0 y 1 + x 1 y 2 + x 2 y 0 x 1 y 0 x 0 y 2 x 2 y 1 ]. This is exactly what we get from the Shoelace Formula: x 0 y 0 x 1 y x 1 1 y 0 x x 2 y 0 y 1 x 2 2 y 1 x x 0 y 1 y 2 x 0 0 y 2 x 2 y 0 x 1 y 0 + x 2 y 1 + x 0 y 2 x 0 y 1 + x 1 y 2 + x 2 y 0
10 Proof by Example Our first example illustrates how the general proof would work. (1, 4) (1, 2) (2, 6) (2, 3) (2, 1) (4, 5) (3, 3) (5, 5) (5, 4)
11 Proof by Example Our first example illustrates how the general proof would work. (1, 4) (1, 2) (2, 6) (2, 3) (2, 1) (4, 5) (3, 3) (5, 5) (5, 4)
12 Approximating Areas Can we use the Shoelace Formula to estimate the areas of more complicated shapes? Yes approximate the shape with a polygon! Let s put this idea to use: We should expect that using a polygon with more sides yields a better approximation to the area. Letting the number of sides should give some sort of integral.
13 Approximating Areas Rewrite the shoelace formula as follows: A = 1 [ x 0 y 0 2 x 1 y 1 + x 1 y 1 x 2 y x n 1 x n y n 1 y n + x n y n x 0 y 0 ] For each i, let x i = x i x i 1 and y i = y i y i 1. It s easy to check that x i 1 y i 1 x i y i = x i 1 y i 1 x i y i Now use the Shoelace Formula to build Riemann sums : 1 n lim n 2 x i 1 y i 1 x i y i = 1 x y 2 dx dy = 1 x dy y dx 2 i=1 This is a special case of Green s theorem. C C
14 Green s Theorem Idea: An integral over a region D can be done by computing a different integral along the boundary C. D ( Q x P ) da = P dx + Q dy y C D C
15 Stokes s Theorem Green s theorem (and hence the Shoelace Formula) is a special case of a more general theorem about integrals and boundaries. Theorem (Stokes) Suppose M is an n-dimensional smooth, oriented manifold with boundary and ω is an (n 1)-form on M, then dω = ω. Implications: M The Fundamental Theorem of Calculus follows from Stokes s theorem. The Divergence Theorem and Stokes s Theorem from vector calculus are special cases. M
16 The End You can read more about the Shoelace Formula in: Bart Braden, The Surveyor s Area Formula, The College Mathematics Journal 17(4), From Saturday Morning Breakfast Cereal.
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