ch14_02 - P1: OSO/OVY GTBL001-14-nal P2: OSO/OVY QC:...

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1130 CHAPTER 14 . . Vector Calculus 14-16 easier vector field, ± x , 2 x y ² . First, note that if x ³ ( t ) = x and y ³ ( t ) = 2 x y , then dy dx = 2 x y x = 2 y x . The flow lines will be the graphs of functions y ( x ) such that y ³ ( x ) = 2 y x ,or y ³ + 1 x y = 2. The left-hand side of the equation should look a little like a product rule. Our main goal is to multiply by a term called an integrating factor, to make the left-hand side exactly a product rule derivative. It turns out that for the equation y ³ + f ( x ) y = g ( x ), an in- tegrating factor is e ± f ( x ) .In the present case, for x > 0, we have e ± 1 / xdx = e ln x = x .(We have chosen the integration con- stant to be 0 to keep the integrating factor simple.) Multiply both sides of the equation by x and show that xy ³ + y = 2 x . Show that ³ + y = ( ) ³ . From ( ) ³ = 2 x , integrate to get = x 2 + c y = x + c x .To find a flow line passing through the point (1, 2), show that c = 1 and thus, y = x + 1 x .To find a flow line passing through the point (1, 1), show that c = 0 and thus, y = x .Sketch the vector field and highlight the curves y = x + 1 x and y = x . 14.2 LINE INTEGRALS In section 5.6, we used integration to find the mass of a thin rod with variable mass density. There, we had observed that if the rod extends from x = a to x = b and has mass density function ρ ( x ), then the mass of the rod is given by ± b a ρ ( x ) . This definition is fine for objects that are essentially one-dimensional, but what if we wanted to find the mass of a helical spring (see Figure 14.11)? Calculus is remarkable in that the same technique can solve a wide variety of problems. As you should expect by now, we will derive a solution by first approximating the curve with line segments and then taking a limit. FIGURE 14.11 A helical spring ( x n , y n , z n ) ( x 0 , y 0 , z 0 ) ( x 2 , y 2 , z 2 ) ( x 1 , y 1 , z 1 ) FIGURE 14.12 Partitioned curve In this three-dimensional setting, the density function has the form ρ ( x , y , z ) (where ρ is measured in units of mass per unit length). We assume that the object is in the shape of a curve C in three dimensions with endpoints ( a , b , c ) and ( d , e , f ). Further, we assume that the curve is oriented, which means that there is a direction to the curve. Fo rexample, the curve C might start at ( a , b , c ) and end at ( d , e , f ). We first partition the curve into n pieces with endpoints ( a , b , c ) = ( x 0 , y 0 , z 0 ) , ( x 1 , y 1 , z 1 ) , ( x 2 , y 2 , z 2 ) ,..., ( x n , y n , z n ) = ( d , e , f ), as indicated in Figure 14.12. We will use the shorthand P i to denote the point ( x i , y i , z i ), and C i for the section of the curve C extending from P i 1 to P i , for each i = 1 , 2 n . Our initial objective is to approximate the mass of the portion of the object along C i . Note that if the segment C i is small enough, we can consider the density to be constant on C i .In this case, the mass of this segment would simply be the product of the density and the length of C i .For some point ( x i , y i , z i )on C i ,we approximate the density on C i by ρ ( x i , y i , z i ). The mass of the section C i is then approximately ρ ( x i , y i , z i ) ± s i , where ± s i represents the arc length of C i . The mass m
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This note was uploaded on 10/04/2010 for the course CHE2C 929102 taught by Professor Carter during the Spring '10 term at UC Davis.

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ch14_02 - P1: OSO/OVY GTBL001-14-nal P2: OSO/OVY QC:...

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