16.2 Line Integrals. location should be approximately 0 01 V(2 1) = 0 01 h4 3i = h i, so the particle should be approximately at the

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1 SECTION 6. LINE INTEGRALS ( ) = + ( ) =i +j. Thus, each vector ( ) has the same direction and twice the length of the position vector of the point ( ), so the vectors all point directly away from the origin and their lengths increase as we move away from the origin. Hence, is graph III. 3. ( ) =( + ) ( ) =( + ) i +( + ) j. The-and-components of each vector are equal, so all vectors are parallel to the line =. The vectors are along the line = and their length increases as the distance from this line increases. Thus, is graph II. 33. At =3the particle is at ( ) so its velocity is V( ) = h4 3i. After. units of time, the particle s change in location should be approximately V( ) = h4 3i = h4 3i, so the particle should be approximately at the point (4 3). 35. (a) We sketch the vector field F( ) = i j along with several approximate flow lines. The flow lines appear to be hyperbolas with shape similar to the graph of = ±, so we might guess that the flow lines have equations =. (b) If = () and = () are parametric equations of a flow line, then the velocity vector of the flow line at the point ( ) is () i + () j. Since the velocity vectors coincide with the vectors in the vector field, we have () i + () j = i j =, =. To solve these differential equations, we know = = ln = + = ± + = for some constant,and = = ln = + = ± + = for some constant. Therefore = = = constant. If the flow line passes through ( ) then () () = constant = = =,. 6. Line Integrals. = 3 and =,,sobyformula3 3 = 3 + = 3 (3 ) +() = = = 54 (453 ) or Parametric equations for are =4cos, =4sin,.Then 4 = (4 cos )(4 sin )4 ( 4sin) +(4cos) = 45 cos sin 4 6(sin +cos ) =4 5 (sin4 cos )(4) =(4) 6 5 sin5 = 46 = c Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

2 36 CHAPTER 6 VECTOR CALCULUS 5. If we choose as the parameter, parametric equations for are =, = for 4 and 3 = 4 ( ) 3 = = = = = + On : =, = =,. On : =, =3 =, 3. Then ( +) + = ( +) + + ( +) + = = 9. =sin, =, = cos,.thenbyformula9, (3 )+ ( ) = = = 5 = ( sin )()( cos) + + = 4 sin cos ( cos ) +() +(sin) = sin 4(cos +sin )+ = 5 sin = 5 cos + 4 sin = 5 = 5 integrate by parts with =, =sin. Parametric equations for are =, =, =3,. Then = ()(3) + +3 = 4 6 = 4 3. = ()( ) ( )( 3) = 4 5 = = 4 (6 ). = 5 ( )= ( ) 5 5. Parametric equations for are =+3, =, =,. Then + + = () 3 +(+3) + = = = 3 +3+= (a) Along the line = 3, the vectors of F have positive -components, so since the path goes upward, the integrand F T is always positive. Therefore F r = F T is positive. c Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

3 SECTION 6. LINE INTEGRALS 37 (b) All of the (nonzero) field vectors along the circle with radius 3 are pointed in the clockwise direction, that is, opposite the direction to the path. So F T is negative, and therefore F r = F T is negative. 9. r() = 4 i + 3 j,sof(r()) = ( 4 )( 3 ) i +3( 3 ) j = 7 i +3 6 j and r () =44 3 i +3 j.then. F r = F(r()) r () = ( ) = ( ) = =45. F r = sin 3 cos( ) 4 3 = (3 sin 3 cos + 4 ) = cos 3 sin + 5 = 6 cos sin F(r()) = ( ) i +sin j = i +sin j, r () = i j.then F r = = 5. =, = 3, = 4 so by Formula 9, F(r()) r () = sin +sin 9633 sin( + ) = 5 ( )sin( ) () +(3 ) +(4 3 ) = 5 sin( ) We graph F( ) =( ) i + j and the curve. We see that most of the vectors starting on point in roughly the same direction as, so for these portions of the tangential component F T is positive. Although some vectors in the third quadrant which start on point in roughly the opposite direction, and hence give negative tangential components, it seems reasonable that the effect of these portions of is outweighed by the positive tangential components. Thus, we would expect F r = F T to be positive. To verify, we evaluate F r. Thecurve can be represented by r() =cos i +sin j, 3, so F(r()) = ( cos sin) i +4cos sin j and r () = sin i +cos j. Then F r = 3 F(r()) r () = 3 [ sin( cos sin)+cos(4 cos sin )] =4 3 (sin sin cos +sincos ) =3 + [using a CAS] 3 9. (a) F r = 5 3 = +3 7 = = 8 c Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

4 38 CHAPTER 6 VECTOR CALCULUS (b) r() =, F(r()) = ; r =, F r = r() = h i, F(r()) = h i. 4 ; In order to generate the graph with Maple, we use the line command in the plottools package to define each of the vectors. For example, v:=line([,],[exp(-),]): generates the vector from the vector field at the point ( ) (but without an arrowhead) and gives it the name v. Toshow everything on the same screen, we use the display command. In Mathematica, we use ListPlot (with the PlotJoined - True option) to generate the vectors, and then Show to show everything on the same screen. 3. = cos 4, = sin 4, =,. Then = ( sin 4)(4) cos 4 = (4 sin 4 +cos4), = (cos 4)(4) sin 4 = ( 4cos4 +sin4),and =,so + + = ( ) [(4 sin 4 +cos4) +( 4cos4 +sin4) +] = 6(sin 4 +cos 4)+sin 4 +cos 4 +=3 Therefore 3 = ( cos 4) 3 ( sin 4) ( )(3 ) = 3 7 cos 3 4 sin 4= 7,74 5,63,75 ( 4 ) 33. We use the parametrization =cos, =sin,.then = + = ( sin) +(cos) =,so = = =(), = = ( cos ) = 4sin = 4, = = ( sin ) =. Hence ( ) = (a) = ( ), = ( ), = (b) = = 4sin +4cos +9 = 3 = 3, = = sin =, = 3 ( ) where = ( ). 3 cos =, 3 (3) = 3 =3. Hence ( ) =(3). c Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

5 SECTION 6. LINE INTEGRALS From Example 3, ( ) =( ), =cos, =sin,and =, = ( ) = sin [( sin )] = (sin sin 3 ) = ( cos ) ( cos )sin = + ( ) = 4 3 = ( ) = cos ( sin ) = = 3, using the same substitution as above. Let =cos, = sin in the second integral ( + cos ) cos sin 39. = F r = h sin 3 cos i h cos sin i = ( cos sin +sincos +3sin sin cos ) = ( cos +sin) = ( sin +cos) cos = integrate by parts in the second term 4. r() =h i,. = F r = ( ) () h i = ( ) = ( +8 ) = = (a) r() = i + 3 j v() =r () = i +3 j a() =v () = i +6 j, and force is mass times acceleration: F() = a() = i +6 j. (b) = F r = ( i +6 j) ( i +3 j) = ( ) = = Let F =85k. To parametrize the staircase, let =cos, =sin, = 9 = 5, 6 6 = F r = 6 h 85i sin cos 5 =(85) 5 6 = (85)(9) 67 4 ft-lb 47. (a) r() =hcos sin i,,andletf = h i. Then = F r = h i h sin cos i = ( sin + cos ) = cos + sin = + += (b) Yes. F ( ) = x = h i and = F r = h cos sin i h sin cos i = ( sin cos + sin cos ) = =. 49. Let r() =h()()()i and v = h 3 i.then v r = h3i h () () ()i = [ ()+ ()+ 3 ()] = ()+ ()+ 3 () =[()+()+3()] [ ()+()+3()] = [() ()] + [() ()] + 3 [() ()] = h 3 i h() ()() ()() ()i = h 3i [h()()()i h()()()i] =v [r() r()] c Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

6 3 CHAPTER 6 VECTOR CALCULUS 5. The work done in moving the object is F r = F T. We can approximate this integral by dividing into 7 segments of equal length =and approximating F T, that is, the tangential component of force, at a point ( ) on each segment. Since is composed of straight line segments, F T is the scalar projection of each force vector onto. If we choose ( ) to be the point on the segment closest to the origin, then the work done is F T 7 [F( ) T( )] =[++++++]()=. Thus, we estimate the work done to = be approximately J. 6.3 The Fundamental Theorem for Line Integrals. appears to be a smooth curve, and since is continuous, we know is differentiable. Then Theorem says that the value of r is simply the difference of the values of at the terminal and initial points of. From the graph, this is 5 = ( 3) = 3 =( ) and the domain of F is R which is open and simply-connected, so by Theorem 6 F is conservative. Thus, there exists a function such that = F, thatis, ( ) = 3 and ( ) = But ( ) = 3 implies ( ) = 3 + () and differentiating both sides of this equation with respect to gives ( ) = 3 + (). Thus = 3 + () so () =4 8 and () = 8 + where is a constant. Hence ( ) = is a potential function for F. 5. ( cos ) = sin, ( sin ) = sin. Since these are not equal, F is not conservative. 7. ( +sin) = +cos = ( + cos ) and the domain of F is R. Hence F is conservative so there exists a function such that = F. Then ( ) = +sin implies ( ) = + sin + () and ( ) = + cos + (). But ( ) = + cos so () = and ( ) = + sin + is a potential function for F. 9. (ln + 3 ) = +6 = (3 + ) and the domain of F is {( ) } whichisopenandsimply connected. Hence F is conservative so there exists a function such that = F. Then ( ) =ln + 3 implies ( ) = ln () and ( ) = +3 + (). But ( ) =3 + so () = () = and ( ) = ln is a potential function for F.. (a) F has continuous first-order partial derivatives and = = ( ) on R, which is open and simply-connected. Thus, F is conservative by Theorem 6. Then we know that the line integral of F is independent of path; in particular, the value of F r depends only on the endpoints of. Since all three curves have the same initial and terminal points, F r willhavethesamevalueforeachcurve. c Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.