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Unformatted text preview: 14-31SECTION 14.3..Independence of Path and Conservative Vector Fields114563.Prove Theorem 2.1 in the case of a curve in three dimensions.64.Prove Theorem 2.2.65.Prove Theorem 2.3.66.Prove Theorem 2.4.67.Prove Theorem 2.5.68.IfChas parametric equationsx=x(t),y=y(t),z=z(t),a≤t≤b,for differentiable functionsx,yandz,show thatCF·Tds=ba[F1(x,y,z)x(t)+F2(x,y,z)y(t)+F3(x,y,z)z(t)]dt, which is the work line integralCF·dr.69.If the two-dimensional vectornis normal (perpendicularto the tangent) to the curveCat each point andF(x,y)=F1(x,y),F2(x,y) ,showthatCF·nds=CF1dy−F2dx.70.IfT(x,y) isthe temperature function, the line inte-gralC(−k∇T)·ndsgives the rate of heat loss acrossC. ForT(x,y)=60ey/50andCthe rectangle with sidesx= −20,x=20,y= −5 andy=5,compute the rate of heatloss. Explain in terms of the temperature function why the in-tegral is 0 along two sides ofC.EXPLORATORY EXERCISES1.Look carefully at the solutions to exercises 5–6, 9–10 and15–16. Compare the solutions to integrals of the formba2x dxanddc3y2dy. Formulate a rule for evaluating line integrals ofthe formCf(x)dxandCg(y)dy. If the curveCis a closedcurve (e.g., a square or a circle), evaluate the line integralsCf(x)dxandCg(y)dy.14.3INDEPENDENCE OF PATH AND CONSERVATIVEVECTOR FIELDSAs you’ve seen, there are a lot of steps needed to evaluate a line integral. First, you mustparameterize the curve, rewrite the line integral as a definite integral and then evaluate theresulting definite integral. While this process is unavoidable for many line integrals, we willnow consider a group of line integrals that are the same along every curve connecting thegiven endpoints. We’ll determine the circumstances under which this occurs and see thatwhen this does happen, there is a simple way to evaluate the integral.y1x12C1FIGURE 14.22aThe pathC1We begin with a simple observation. Consider the line integralC1F·dr, whereF(x,y)=2x,3y2andC1is the straight line segment joining the two points (0, 0) and(1, 2). (See Figure 14.22a.) To parameterize the curve, we takex=tandy=2t,for≤t≤1.We then haveC1F·dr=C12x,3y2·dx,dy=C12x dx+3y2dy=1[2t+12t2(2)]dt=9,where we have left the details of the final (routine) calculation to you. For the same vectorfieldF(x,y), consider nowC2F·dr, whereC2is made up of the horizontal line segmentfrom (0, 0) to (1, 0) followed by the vertical line segment from (1, 0) to (1, 2). (SeeFigure 14.22b.) In this case, we haveC2F·dr=C22x,3y2·dx,dy=12x dx+23y2dy=9,1146CHAPTER 14..Vector Calculus14-32where we have again left the final details to you. Look carefully at these two line integrals.Although the integrands are the same and the endpoints of the two curves are the same, thecurves followed are quite different. You should try computing this line integral over severaladditional curves from (0, 0) to (1, 2). You will find that each line integral has the samevalue: 9. This integral is an example of one that is the same along any curve from (0, 0)to (1, 2)....
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