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vector_calculus-2

# vector_calculus-2 - Vector Calculus 2 There's more to the...

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Vector Calculus 2 There’s more to the subject of vector calculus than the material in chapter nine. There are a couple of types of line integrals and there are some basic theorems that relate the integrals to the derivatives, sort of like the fundamental theorem of calculus that relates the integral to the anti-derivative in one dimension. 13.1 Integrals Recall the definition of the Riemann integral from section 1.6. Z b a dx f ( x ) = lim Δ x k 0 N X k =1 f ( ξ k ) Δ x k (13 . 1) This refers to a function of a single variable, integrated along that one dimension. The basic idea is that you divide a complicated thing into little pieces to get an approximate answer. Then you refine the pieces into still smaller ones to improve the answer and finally take the limit as the approximation becomes perfect. What is the length of a curve in the plane? Divide the curve into a lot of small pieces, then if the pieces are small enough you can use the Pythagorean Theorem to estimate the length of each piece. Δ k = p x k ) 2 + (Δ y k ) 2 Δ x k Δ y k The whole curve then has a length that you estimate to be the sum of all these intervals. Finally take the limit to get the exact answer. X k Δ k = X p x k ) 2 + (Δ y k ) 2 -→ Z d‘ = Z p dx 2 + dy 2 (13 . 2) How do you actually do this? That will depend on the way that you use to describe the curve itself. Start with the simplest method and assume that you have a parametric representation of the curve: x = f ( t ) and y = g ( t ) Then dx = ˙ f ( t ) dt and dy = ˙ g ( t ) dt , so d‘ = q ( ˙ f ( t ) dt ) 2 + ( ˙ g ( t ) dt ) 2 = q ˙ f ( t ) 2 + ˙ g ( t ) 2 dt (13 . 3) and the integral for the length is Z d‘ = Z b a dt q ˙ f ( t ) 2 + ˙ g ( t ) 2 where a and b are the limits on the parameter t . Think of this as R d‘ = R v dt , where v is the speed. James Nearing, University of Miami 1

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13—Vector Calculus 2 2 Do the simplest example first. What is the circumference of a circle? Use the parametrization x = R cos φ, y = R sin φ then d‘ = p ( - R sin φ ) 2 + ( R cos φ ) 2 = R dφ (13 . 4) The circumference is then R d‘ = R 2 π 0 R dφ = 2 πR . An ellipse is a bit more of a challenge; see problem 13.3 . If the curve is expressed in polar coordinates you may find another formulation preferable, though in essence it is the same. The Pythagorean Theorem is still applicable, but you have to see what it says in these coordinates. Δ k = p r k ) 2 + ( r k Δ φ k ) 2 r k Δ φ k Δ r k If this picture doesn’t seem to show much of a right triangle, remember there’s a limit involved, as Δ r k and Δ φ k approach zero this becomes more of a triangle. The integral for the length of a curve is then Z d‘ = Z p dr 2 + r 2 2 To actually do this integral you will pick a parameter to represent the curve, and that parameter may even be φ itself. For an example, examine one loop of a logarithmic spiral: r = r 0 e .
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vector_calculus-2 - Vector Calculus 2 There's more to the...

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