Vector Calculus. Chapter Vector Fields A vector field describes the motion of the air as a vector at each point in the room

hapter 5 Vector alculus 5. Vector Fields 5.. A vector field describes the motion of the air as a vector at each point in the room. 5.. y 4 4 4 4 5..3 At selected points a, b, plot the vector fa, b,ga, b. 5..4 The gradient of a function at a point is a vector describing the direction in which the value of the function is increasing most rapidly. The collection of these vectors over all points is a vector field. 5..5 The gradient field gives, at each point, the direction in which the temperature is increasing most rapidly and the amount of increase. 5..6 y 5..7 4 4 4 4 477

478 HAPTE 5. VETO ALULU 5..8 y 5..9 4 4 4 4 5.. 5.. y 4 3 4 4 5.. 5..3 y 4 4 4 4 opyright c 3 Pearson Education, Inc.

5.. VETO FIEL 479 5..4 5..5 y 4 4 4 4 5..6 a corresponds to, since has zero component, and the y component increases as y does. b corresponds to, since the component appears to be zero along the line y. c corresponds to B since the y component is zero on the -ais and the component is zero on the y ais. Finally, d corresponds to A since the component is zero on the -ais and the y component is zero on the y-ais. 5..7 Here is the circle of radius, so a vector tangent to the circle is y,. o F is normal to at, y since y,, y. 5..8 is the circle of radius, so a vector tangent at, y is y,. Then F is a scalar multiple of the tangent vector, so it is tangent to at all points. 5..9 is the vertical line at, so the tangent vector is a multiple of,y at all points. Then, y is never a multiple of, y for any point on since there, and, y,y y is zero for y, so that F is normal to at,. opyright c 3 Pearson Education, Inc.

48 HAPTE 5. VETO ALULU 5.. is the circle of radius, so a vector tangent at, y is y,. F is a multiple of the tangent vector for y or for, so that F is tangent to at ±, and at, ±. F y, y,so that F is normal to at ±, ±. 5.. 5.. 5..3 5..4 5..5 The gradient field is ϕ,ϕ y, y. opyright c 3 Pearson Education, Inc.

5.. VETO FIEL 48 5..6 The gradient field is ϕ,ϕ y, y. +y +y 5..7 The gradient field is ϕ,ϕ y,. y 3 3 3 3 5..8 The gradient field is ϕ,ϕ y y,. opyright c 3 Pearson Education, Inc.

48 HAPTE 5. VETO ALULU 5..9 ϕ ϕ,ϕ y y y, y. 5..3 ϕ ϕ,ϕ y y y, y y y,. 5..3 ϕ ϕ,ϕ y /y, /y. 5..3 ϕ ϕ,ϕ y +y y, r y,. 5..33 ϕ, y, z r. 5..34 ϕ + +y +z, y + +y +z, z + +y +z r r +. 5..35 ϕ + y + z 3/, y, z r r 3. 5..36 ϕ e z cos + y,e z cos + y, e z sin + y. 5..37 a. The gradient field is, 3. b. The equipotential curve at, is +3y 5, which is a line of slope 3 so has a tangent vector at, y parallel to, 3. But, 3, 3, so the gradient field is normal to the equipotential line through,. c. The equipotential curve at any point is a line of slope 3 and thus has a tangent vector at, y parallel to, 3 y. The same argument as in part b shows that this is normal to the gradient field. d. y 4 5 4 4 5 4 5..38 a. The gradient field is, y. b. At,, the tangent vector is parallel to ϕ y,,ϕ,,, which is normal to the gradient at, which is,. c. At, y, the tangent vector is parallel to y,, and, y y,, so the gradient is everywhere normal to the equipotential curves. opyright c 3 Pearson Education, Inc.

5.. VETO FIEL 483 d. 5..39 a. The gradient field is e y, e y. b. At,, the tangent vector is parallel to ϕ y,,ϕ,,, which is normal to the gradient, at,. c. At, y, the tangent vector is parallel to e y,e y, and e y, e y e y,e y, so the gradient is everywhere normal to the equipotential curves. d. 5..4 a. The gradient field is, 4y. b. At,, the tangent vector is parallel to ϕ y,,ϕ, 4,, which is normal to the gradient, 4 at,. c. At, y, the tangent vector is parallel to 4y,, which is normal to the gradient field. opyright c 3 Pearson Education, Inc.

484 HAPTE 5. VETO ALULU d. 5..4 a. True. ϕ ϕ 3, and ϕ y ϕ y. b. False. It is constant in magnitude magnitude but not direction. c. True. For eample, it points outwards along the line y but horizontally along the line. 5..4 a. The magnitude of the vector field is + y.thuson, the magnitude is on the boundary of and less than elsewhere. On, the magnitude is a maimum at the corners of, where it is. Finally, on, the magnitude is a maimum at the corners, where it is. b. The vector field is everywhere directed outwards. 5..43 a. This is a rotational field with magnitude + y at, y. Thus the answer to this question is the same as for the previous question: on, the magnitude is on the boundary of and less than elsewhere. On, the magnitude is a maimum at the corners of, where it is. Finally, on, the magnitude is a maimum at the corners, where it is. opyright c 3 Pearson Education, Inc.

5.. VETO FIEL 485 b. For and the field is directed out of the region on line segments between any verte and the midpoint of the boundary line when proceeding in a counterclockwise direction; on the vector field is tangent to the boundary curve everywhere. 5..44 For eample, F y,. 5..45 For eample, F y, or F,. 5..46 For eample, F y,. The magnitude is + y. 5..47 For eample, F, y r +y r, F,. 5..48 a. V, y k + y /,soe V V,V y k + y 3/, y. b. From the above formula, the field is a varying multiple of, y, which is a radial field pointing away from the origin. The radial component of E is thus E k + y 3/, y k + y 3/ + y / k r. k c. The equipotential curves are curves of the form + y so that + y k and the equipotential curves are circles. Thus the tangent vectors to the equipotential curves are proportional to y, and thus are normal to E, which is proportional to, y. 5..49 a. V c +y r r c r c r r r r. + y and similarly, V y cy +y, so that E V c +y, y b. From the above formula, the field is a varying multiple of, y, which is a radial field pointing away c from the origin. The radial component of E is thus E + y + y c + y. r c. The equipotential curves are curves of the form c ln K, so are solutions to +y 5..5 r +y e ck and thus of the form + y K for some constant K. Hence the equipotential curves are circles, so have tangent vectors proportional to y, ; these are clearly normal to E, which is proportional to, y. a. U GMm + y + z / GMm + y + z 3/ GMm + y + z 3/ and similarly for Uy and U z. Thus F U GMm + y + z 3/, y, z. opyright c 3 Pearson Education, Inc.

486 HAPTE 5. VETO ALULU b. From the above formula the field is a varying multiple of, y, z, which is a radial field pointing away from the origin. The radial component of F is thus F GMm + y + z 3/ + y + z GMm r. c. The equipotential surfaces are solutions to GMm + y + z K and so are spheres. The tangent plane at,y,z isu,y,z +U y,y,z y y +U z,y,z z z and so a normal to the plane is U,U y,u z, which is proportional to F. 5..5 The flow curve y of the vector field F at, y is defined to be a continuous curve through, y that is aligned with the vector field, i.e. whose tangent at, y is given by F, y f, y,g, y. The slope of the tangent line is then g,y f,y, so this is y. 5..5 The streamlines satisfy y, so that y +. 5..53 The streamlines satisfy y, so that y +. 5..54 The streamlines satisfy y d y. But also y d yy, so that y d y d y d thus y.thusy + and the streamlines are the hyperbolas y K. d and opyright c 3 Pearson Education, Inc.

5.. VETO FIEL 487 y yy, we have y d dy 5..55 The streamlines satisfy y y. Because d d y. Thusy + and the streamlines are the circles + y. d d y and thus y 5..56 u r is the unit vector forming an angle of θ with i, so it is of unit length and proportional to cos θ, sin θ. But cos θ+sin θ, so this is in fact the unit vector. Thus u r cos θ i+sin θ j. imilarly, u θ forms an angle of θ+ π with i, so it is the unit vector cos θ + π, sin θ + π sin θ, cos θ. The other two formulas can be found by solving these for i, j as linear equations. For eample, u r cos θ i + sin θ j, u θ sin θ i + cos θ j. Multiply the first equation by sin θ and the second by cos θ and add to obtain u r sin θ+u θ cos θ sin θ j + cos θ j j. 5..57 For θ,u r is coincident with i, and u θ with j. From the formula u r cos i + sin j i, u θ sin i + cos j j. Forθ π, the picture implies that we should have u r j, u θ i. From the formulas, u r cos π i+sin π j j, uθ sin π i+cos π j i. For θ π, ur is coincident with i, and u θ with j. From the formula u r cos π i+sin π j i, u θ sin π i+cos π j j. For θ 3π, the picture implies that we should have u r j, u θ i. From the formulas, u r cos 3π i+sin 3π j j, u θ sin 3π i + cos 3π j i. 5..58 F r, θ u r cos θ i + sin θ j + y i + y + y j i + yj. + y y 5..59 F r, θ u θ sin θ i + cos θ j + y i + + y j yi + j. + y opyright c 3 Pearson Education, Inc.

488 HAPTE 5. VETO ALULU 5..6 F r, θ r u θ + y sin θ i + cos θ j + y + y yi + j y,. 5..6 F, y y u r cos θ u θ sin θ + u r sin θ+u θ cos θ r sin θ cos θ u r sin θ u θ +r cos θ sin θ cos θ u r + cos θ u θ r sin θ + cos θ u θ ru θ. 5. Line Integrals 5.. A single-variable integral integrates along a segment while a line integral integrates along an arbitrary curve. 5.. The integral is evaluated by evaluating the integral off r t wherer t epresses the velocity of the parameterization with respect to arc length. 5..3 r t r + r y +4t 5..4 hoose a parameterization r t for ; then T r t r t and F T ds becomes b a F r t dt. 5..5 Because T t,y t,z t, b a f t+gy t+hz t dt is simply a rewriting of the dot product. 5..6 The circulation measures the degree to which the vector field is positively aligned with has a positive dot product with the curve as is traversed with a particular orientation. 5..7 Take the line integral of F T along the curve using arc length parameterization. 5..8 The flu measures the degree to which the vector field is outwards normal to the curve as is traversed with a particular orientation. 5..9 Take the line integral of F n along the curve using arc length parameterization. 5.. One parameterization for the curve is cos t, sin t, t π. opyright c 3 Pearson Education, Inc.

5.. LINE INTEGAL 489 5.. r s is an arc length parameterization, so we have y ds π cos s sin s ds. 5.. hoose the arc length parameterization r s cos s, sin s then we have + y ds π cos s + sin s ds sin s cos s π. 5..3 With r s s, s, r t, so that we have 4 s ds s3 6 4 3 3. 5..4 With r s s, s, r t, so that yds 4 3 3 6. 5..5 a. r t 4 cos t, 4 sin t, t π. b. r t 4 sin t + 4 cos t 4. c. 5..6 + y ds π 4 6 cos t + 6 sin t dt π 64 dt 8π. a. r t 5t, 5t, t. b. r t 5 + 5 5. c. 5..7 + y ds 5t 5 dt 5 t dt 5 3. a. r t t, t, t. b. r t +. c. 5..8 +y ds t t +t dt t dt ln. a. r t t, t, t. b. r t +t +4t. c. y/3 ds t 3 /3 +4t dt t +4t dt 5 5. opyright c 3 Pearson Education, Inc. y ds 4 s s ds s s s ds 3 6 4 s4 8

49 HAPTE 5. VETO ALULU 5..9 a. r t cos t, 4 sin t, t π/. b. r t 4 sin t + 6 cos t 4+cos t +3cos t. c. y ds π/ 6 sin t cos t +3cos tdt. Let u + 3 cos t dt so that du 6 cos t sin tdt. ubstituting gives 8 4 3 u/ du 6 9 u 3/ 4 6 9 8 9. 5.. a. r t t,t, r t t, t, t. b. r t, r t. c. 3y ds t 3t dt + t 3 t dt 4t 5 dt 3. 5.. Let r t t +, 4t +, t. Then r t 7 and +y ds t ++4t + 7 dt 7 9t +3dt 5 7. The length of the line segment is 7, so the average value is 5. 5.. Let r t 9 cos t, 9 sin t, t π. Then r t 9, and +4y π ds 8 cos t +4 8 sin t 9 dt 3645π. The circumference of the circle is π 98π, so the average value is 3645 8 45. 5..3 Let rt t, t 3/ for t 5. Then r t +9t/4 and 5 5 4+9y /3 ds 4+9t +9t/4 dt 4+9t dt 4t +9t / 5 65 4. The length of the curve is 5 +9/4 d 335 7 so the average value is 43 68. 5..4 Let r t cos t, sin t, t π. Then r t and ey ds π cos te sin t dt. Thus, the average value is. 5..5 r t 4 sin t ++4cos t,so + y + z ds π cos t+sint dt. 5..6 r t +9sin t + 9 cos t 3,so y +z ds 3 π 3 cos t+6sint dt 6π. 5..7 Let r t t, t, 3t, t. Then r t 4, so yz ds 4 6t3 dt 4 3t 4 4. 3 5..8 Let r t t +, t +4, t +, t. Then r t 3, so 3 t+t+4 t+ dt 3 t +4dt 3. 5..9 r t, so y z ds π 3 sin t t dt π. 5..3 r t, so eyz ds e te 8t dt 4 6e. 3 opyright c 3 Pearson Education, Inc. y z ds

5.. LINE INTEGAL 49 5..3 The length of the curve is the line integral of along the curve. Then r t simplifies to 5 cost/4, 5 sint/4,, and then r t 5 + 4 so that the arc length is dt. 5..3 The length of the curve is the line integral of along the curve. r t 9 cos t + 6 cos t + 5 sin t 5 so that ds 5 π dt π. 5..33 r t 4, t, and F r t 4t, t 4, t 6t +t 3, so that F T ds F r t dt 6t +t 3 dt 7. 5..34 r t 4 sin t, 4 cos t, sof r t 4 sin t, 4 cos t 4 sin t, 4 cos t 6. Then F T ds π F r t dt π 6 dt 6π. 5..35 Let r t 4t+, 9t+, t ; then r t 4, 9 and F r t 9t+, 4t+ 4, 9 7t+3. Then F T ds 7t + 3 dt 49. 5..36 Let r t 8t +, 8t +, t ; then r t 8, 8 and F r t + y 3/, y 8, 8 +y 8. Then +y 3/ F T ds 8 6t+4 dt 8 8t+ 3 dt 8t+ 5. 5..37 r t t, 6t ; then F r t t 4 3/ t, 3t t, 6t t3 t 3 dt 4 3. t 6 t 3 and F T ds 5..38 r t, 4, so that F r t 7t t, 4t, 4 t. Then F T ds t dt ln. 5..39 r t t, t, r t, 4t, t. Then r t,, r t, 4, so that F r t dt t, t, dt + 4t,, 4 dt dt. 5..4 r t t, 8t, r t t, 8, t. Then r t, 8, r t,, so that F r t dt t, 8t, 8 dt + t, 8, dt 69t dt 67. 5..4 r t t, 8t, t. Then r t, 6t, so F r t dt 8t, t, 6t dt 48t dt 6. 5..4 r t t +, 8 4t, t. Then r t, 4, so F r t dt 8 4t, t, 4 dt dt. 5..43 r t 4 sin t, 4 cos t, 4 sin t. Thus we have F r t dt π 4 cos t, 4 sin t, 4 cos t 4sin t, 4 cos t, 4 sin t dt π 6 sin t cos t dt. 5..44 r t sin t, cos t, π,so F r tdt π sin t, cos t, t π 4+ t 4π dt +8π. π sin t, cos t, π dt 5..45 Let r t t +,t +,t +, t 9, so that r t,,. Then F r t dt 9 t +,t+,t+,, dt 9 3t+ 3/ 3 t +dt 9 t+ 3 3 dt t+ 3 3. opyright c 3 Pearson Education, Inc.

49 HAPTE 5. VETO ALULU 5..46 Let r t 7t +, 3 t +,t +, t, so that r t 7, 3,. Then F r t dt 7t +, 3 t +,t+ 7, 3, dt 59t+ dt ln + 7t+ +3t+ +t+ 7t+ +3t+ +t+ ln 7 ln 7. 5..47 a. Looking at the vector field, it appears that the vector field points counterclockwise just as much as it points clockwise at the boundary of the region, so we would epect the circulation to be zero. b. r t sin t, cos t, so F r t dt π sin t cos t, cos t sin t, cos t dt 4 π cos t + sin t cos t sin t dt. 5..48 a. Looking at the vector field, it appears that the vector field points counterclockwise just as much as it points clockwise at the boundary of the region, so we would epect the circulation to be zero. b. Parameterize the boundary by four paths: r t, +4t, r t 4t,, r 3 t, 4t, r 4 t +4t,, all with t. Then r t, 4, r t 4,, r 3 t, 4, r 4 t 4,. Then F r t dt, +4t, 4 dt + 8 6t+6t 8 6t+6t 4t, 4, dt +, 4t, 4 dt + 8 6t+6t 8 6t+6t +4t, 4, dt 8+6t 8+6t 8+6t 8+6t dt 64t 3 8 6t+6t 8 6t+6t dt. 5..49 a. Looking at the vector field, the inward-pointing vectors in quadrants II and IV appear larger than the outward-pointing vectors in quadrants I and III. Thus we would epect the flu to be negative. b. r t sin t, cos t, so that F n ds π sin t cos t cos t cos t sin t dt 4 π cos t sin t cos t dt 4π. 5..5 a. The vector field points outwards everywhere, so we would epect the flu to be positive. b. Parameterize the boundary by four paths: r t, +4t, r t 4t,, r 3 t, 4t, r 4 t +4t,, all with t. Then r t, 4, r t 4,, r 3 t, 4, r 4 t 4,. Then F n ds 4 dt + 8 6t+6t 8 6t+6t + 4 dt + 4 dt + 8 6t+6t + 4 dt 3 8 6t+6t dt 8ln 8 6t+6t +3 4.. 5..5 a. True. This is the definition of an arc length parameterization. b. True. Let r t cos t, sin t. Then r t sin t, cos t, and F r t dt π sin t, cos t sin t, cos t dt π cos t sin t dt. F n ds π sin t cos t cos t sin t dt π sin t cos tdt. c. True. d. True. It is the line integral F nds. 5..5 The work done on either path is simply F r t dt. a. Here r t t, 5, for t, so r t, and F r t dt 5 dt 3 opyright c 3 Pearson Education, Inc.

5.. LINE INTEGAL 493 b. Here r t cos t, sin t and r t sin t, cos t, so F r t dt π 5 sin tdt 3. The same amount of work is done along each path. 5..53 a. For the first path, the work done is F r t dt 4 dt 8, and for the second path F r t dt π 4 sin t 5 cos t dt 8, so again they are equal. b. For the first path, the work done is F r t dt 4 dt 8, while for the second path, F r t dt π 4 sin t 5 cos t dt 8, so the amount of work is still equal along the two paths. 5..54 a. Let rt cos t, sin t for t π so that r t. Then fds π cos t + sin t dt. b. Let rt cos t, sin t for t π so that r t. Note that this parameterization traces out the unit circle, but clockwise. Then fds π cos t sin t dt. c. The two integrals are equal. 5..55 a. Let rt t, t for t so that r t 4t +, and fds t 4t +dt 5 5. b. rt t, t for t. Then r t fds t 4t 8t +5dt 5 5. c. The two integrals are equal. + t 4t 8t + 5 and 5..56 Letting rt cos t, sin t and r t sin t, cos t as usual, we have F r t dt b sin t + c cos t dt π c b, so that the circulation is zero only for b c. π 5..57 Let rt r cos t, r sin t for a circle of radius r, so that r t r sin t, r cos t. Then F r t dt π ar cos t + br sin t rsin t+cr cos t + dr sin trcos t dt r π a sin t cos t b sin t + c cos t + d sin t cos t dt π c b r, so that the circulation is zero provided b c. 5..58 Let rt cos t, sin t ; then the flu is F n ds π a cos t cos t d sin t sin t dt a + d π, so the flu is zero if a d. 5..59 Let rt r cos t, r sin t for a circle of radius r; then the flu is F n ds π ar cos t + br sin trcos t cr cos t + dr sin t r sin t dt r π a cos t + b sin t cos t + c sin t cos t + d sin t dt r a + d π, so the flu is zero provided that a d. 5..6 Parameterize by rt t, t for t. Then r t,, and F T ds t + t dt dt. Parameterize by rt cos t, sin t for t π. Then r t sin t, cos t, and F T ds π/ sin t sin t + cos tcos t dt π/ dt π. Finally, parameterize 3 by the two paths r t t, and r t,t for t, so that r t, and r t,. Then 3 F T ds + t dt + t + dt. None of the three path integrals are equal. opyright c 3 Pearson Education, Inc.

494 HAPTE 5. VETO ALULU 5..6 Using the same parameterizations as for the previous problem, we have F T ds t + t dt t dt, F T ds π/ sin t sin t + cos t cos t dt π/ cos t sin t dt, 3 F T ds + t dt + t + dt. All three are equal to zero. 5..6 r θ sin θ, cos θ, so that r θ and ρds π θ π + dθ π. 5..63 Parameterize by rt t, t for t 3. Then r t +6t and ρds 3 +t 3 +6t dt 49.5. 5..64 a. b. The gradient is 5i 5j. c. F 5i +5j. d. Parameterize the boundary by rt,t for t. Then r t, and F n ds 5 5 dt 5. e. Parameterize the boundary by rt t, for t note: we do not use t, because we need counterclockwise orientation. Then r t, and F n ds 5 5 dt 5. 5..65 r t,, and r t t 3, so the work is F T ds a a. For p, we have a t dt lna. b. The work is not finite. c. For p 4, we have a 3 t 3 dt a 6t 6 6a. d. As a, the work approaches 6. 3t t 3 dt p 3 p/ a t p dt e. For the general p>, the analysis above shows that the integral is for p 3 p/ p t p a +a p while for p the integral is from part a ln a. 3 p/ p e. This approaches a limit only for p> when p<. This limit is 3 p/ p. 5..66 a. r t sin t, cos t, so the flu along the quarter circle is F n ds π/ sin t cos t cos t sin t dt π/ 8 sin t cos tdt4. opyright c 3 Pearson Education, Inc.

5.3. ONEVATIVE VETO FIEL 495 b. With r t as above, the flu is F n ds π 8 sin t cos t dt 4. π/ c. Both the normal vectors and the vector field F in the third quadrant are the negatives of their values in the first quadrant, so their dot product is the same. Thus the flu is identical. d. Both the normal vectors and the vector field F in the fourth quadrant are the negatives of their values in the second quadrant, so their dot product is the same. Thus the flu is identical. e. The total flu is 4 4+4 4. 5..67 We use four line segment parameterizations for the rectangle, all for t : r t at,, so r t a,. r t a, bt, sor t,b. r 3 t a at, b, sor 3 t a,. r 4 t,b bt, so r 4 t, b. Then dy at +a b +a at + b dt ab dt ab. 5..68 Parameterize by rt a cos t, a sin t, so that r t a sin t, a cos t. Then yd π a sin t a sin t dt a π sin tdt πa. 5.3 onservative Vector Fields 5.3. A simple curve has no self-intersections; the initial and terminal points of a closed curve are identical. 5.3. A region is connected, roughly speaking, if it consists of one piece. A simply connected region has the property that every closed loop can be contracted to a point. 5.3.3 If F f,g is a vector field in and if f g, then F is conservative. 5.3.4 If F f,g,h is a vector field in 3 and if f conservative. g f and z h g and z h, then F is 5.3.5 Integrate f with respect to to get an answer where the constant is actually a function of y. Take the partial with respect to y and equate with g to compute the constant. 5.3.6 The integral is ϕ B ϕ A where ϕ F. 5.3.7 The integral is zero. 5.3.8 There eists a potential function ϕ such that F ϕ. F dr ϕb ϕa for all points A and B in and all smooth oriented curves from A to B path independence. F dr on all simple smooth closed oriented curves in. 5.3.9 Yes, because y. 5.3. Yes, because y y. 5.3. Yes, because y y. 5.3. No. y y, but + y. 5.3.3 Yes. e cos y e sin y e sin y. 5.3.4 Yes. 3 + y y, and y 3 + y y. 5.3.5 Yes. y. A potential function is +y. opyright c 3 Pearson Education, Inc.

496 HAPTE 5. VETO ALULU 5.3.6 Yes. y. A potential function is y. 5.3.7 No. 3 y, and + y. 5.3.8 Yes. +y y y +y potential function is ln + y. 5.3.9 Yes. +y a potential function of + y. y +y 3/ +y. Integrating +y y. Integrating +y with respect to, we see that a +y with respect to, we obtain 5.3. Yes. y, z y and z y. A potential function is y + z. To see this, integrate y with respect to to get y + f y, z; differentiating with respect to y gives + f y y, z, so that f y, z is actually a function of z, sayg z. But then differentiating y + g z with respect to z gives g z,sog z z. 5.3. Yes, because the mied partials are pairwise equal. A potential function is z + y. 5.3. Yes, because the mied partials are pairwise equal. A potential function is yz. 5.3.3 Yes, because the mied partials are pairwise equal. To find the potential function, integrate y + z with respect to to get y + z+f y, z; differentiating with respect to y gives + z + f y y, z so that f y y, z z andf y, z yz. Thus a potential function is y + yz + z. 5.3.4 Yes, because the mied partials are pairwise equal. As in problem 8, a potential function is ln + y + z. 5.3.5 Yes, because the mied partials are pairwise equal. As in problem 9, a potential function is + y + z. 5.3.6 Yes, because the mied partials are pairwise equal and zero. A potential function is 4 4 +y 4 z4. 5.3.7 a. ϕ y,, so ϕ r t dt π sin t, cos t sin t, cos t dt π cos t dt. b. ince ϕ is obviously conservative, the integral is simply ϕ cos π, sin π ϕ cos, sin. 5.3.8 a. ϕ, y, so ϕ r t dt π sin t, cos t cos t, sin t dt π dt. b. The integral is ϕ sin π, cos π ϕ sin, cos. 5.3.9 a. ϕ, 3, so ϕ r t dt, 3, dt dt 4. b. The integral is ϕ, ϕ, 6 4. Note thatϕ, is ϕ evaluated at the point where t. 5.3.3 a. ϕ,,, so ϕ r t dt π,, cos t, sin t, π dt π b. The integral is ϕ,, ϕ,,. cos t sin t + π dt. opyright c 3 Pearson Education, Inc.

5.3. ONEVATIVE VETO FIEL 497 5.3.3 a. ϕ, y, z, so ϕ r t dt π cos t, sin t, t π sin t, cos t, π dt π t π dt. b. The integral is ϕ,, ϕ,, 5. 5.3.3 a. ϕ y + z, + z, + y, so ϕ r t dt 4 5t, 4t, 3t,, 3 dt 4 tdt 76. b. The integral is ϕ 4, 8, ϕ,, 76 76. 5.3.33 Parameterize by rt 4 cos t, 4 sin t, t π. Then F dr π 4 cos t, 4sint 4 sin t, 4 cos t dt π dt. 5.3.34 Parameterize by rt 8 cos t, 8 sin t, t π. Then F dr π 8 sin t, 8 cos t 8 sin t, 8 cos t dt 8 π cos t sin t dt. 5.3.35 Parameterize by three paths, all for t : r t t, t, sor t,. r t t, t, so r t,. r 3 t, t, sor 3 t,. Then F dr t, t, dt + t, t, dt +, t, dt t +t +t + t t dt 8t 4 dt. 5.3.36 Parameterize by rt 3 cos t, 3 sin t, t π. Then F dr π 3 sin t, 3 cos t 3 sin t, 3 cos t dt 9 π dt 8π. This integral is not zero because the vector field y, is not conservative: y, while. 5.3.37 Using the given parameterization, π F dr cos t, sin t, sin t, cos t, dt π dt. 5.3.38 Using the given parameterization, π F dr sin t cos t,, cos t sin t sin t, cos t, sin t dt 5.3.39 π dt. a. False. Parametrize the curve by 4 cos t +, y 4 sin t, for t π. Then F dr π 4 sin t, 4 cos t + 4sin t, 4 cos t dt π 6 sin t + 6 cos t + 4 cos t dt 3π. b. True. This is because F is conservative. c. True. If the vector field is a, b, then a potential function is a + by. d. True. This is because f g y. 5.3.4 Write ϕ, y, z + yz. Then using the Fundamental Theorem, this integral is equal to ϕ cos 8π, sin 8π, 4π ϕ cos, sin, ϕ,, 4π ϕ,,. 5.3.4 Write ϕ, y e cos y. Then using the Fundamental Theorem, this integral is equal to ϕ ln, π ϕ, e ln cos π e cos. 5.3.4 If t, y y t, then e cos yd+ sin ydy F dr, where F e cos y,e sin y. This is a conservative vector field, because e cos y e sin y e sin y, so that the integral around the closed curve is zero. opyright c 3 Pearson Education, Inc.

498 HAPTE 5. VETO ALULU 5.3.43 F is a conservative vector field; a potential function can be found by integrating with respect to y to obtain y+f, z; differentiate with respect to z to get f z, z z, so thatf, z z +g. Thus the potential function is y +z +g ; differentiating with respect to gives y +z +g y +z, so that g and we may take g. oifϕ y + z, then ϕ F and thus the integral is zero because both sine and cosine, and thus ϕ, have the same values at the two endpoints of. 5.3.44 ds is the length of the curve, which is π. The other two integrals are zero, because they are the same as integrating the conservative vector fields,,, and,, respectively around a closed curve. 5.3.45 This is a conservative vector field with potential function ϕ, y +y, so the work is ϕ, 4 ϕ,. 5.3.46 This is a conservative vector field with potential function ϕ, y + y, so the work is ϕ 3, 6 ϕ, 45 43. 5.3.47 This is a conservative vector field with potential function ϕ, y, z + y + z, so the work is ϕ, 4, 6 ϕ,, 8 3 5. 5.3.48 This is not a conservative vector field, because for eample z e+y, while ze+y ze +y. Parameterize by t, t, 4t, t ; then F dr et,e t, 4te t,, 4 dt et +6te t dt e + 5. 5.3.49 For, the vector field points against the curve for most of its length, and with larger magnitude, so the integral is negative. For, the vector field points with the curve for its entire length, so the integral is positive. 5.3.5 For, the vector field points against the curve for its entire length; for, the vector field points with the curve, so the integral over is negative and the integral over is positive. 5.3.5 F a, b, c is a conservative force field with potential function ϕ, y, z a + by + cz, so the work done is ϕ B ϕ A F B F A F B A F AB. 5.3.5 a. The acceleration is the time derivative of velocity, so Newton s second law says that m dv dt ma F ϕ d dv b. By the product rule, dt v v dt v + v dv dt dv dt v, and the desired equation follows. c. Multiplying part a by v r and using b, we have letting be a path from A to B m dv dt v ϕ r m d dt v v. 5.3.53 Thus, ϕ r m B A d dt v t v t dt,soϕ A ϕ B m v B. The last equality follows A because the integrand is a conservative vector field. Thus m v B m v A ϕ A ϕ B, so m v B + ϕ B m v A + ϕ A. a. Away from the origin where the denominator of the force field equation is undefined, the force field is conservative because, for eample, GMm + y + z GMm y 3/ + y + z 5/ GMm y + y + z. 3/ b. A potential function for the force field is ϕ, y, z GMm + y + z / GMm r. opyright c 3 Pearson Education, Inc.

5.3. ONEVATIVE VETO FIEL 499 c. The work done in moving the point from A to B, because the force field is conservative, is ϕ B ϕ A GMm B A GMm r r. d. Because the field is conservative, the work done does not depend on the path. 5.3.54 This vector field is F, y, z + y + z p/, so away from the origin is conservative with potential function ϕ, y, z p + y + z p/ as long as p. When p, the potential function is ϕ ln + y + z. The field is conservative at the origin if it is defined and if its potential function is defined, i.e. if both p and p are nonnegative, which happens only if p. 5.3.55 a. This field is F y, + y p/, and we have y + y p/ + y p/ + + y p/ py +y p/ +y + y p/ + py + y p/ +p y. +y +p/ + y p/ p +y p/ +y + y p/ p + y p/ p +y +y +p/. For the force field to be conservative, these two would have to be equal. However, their difference is N M + y p/ p + y + y p/ + y p/ p + y p/ p + y p/ which is in general nonzero. b. From the above formula, if p, then the mied partials are equal, so that F is conservative. c. For p,f +y y,. Integrating the component of F with respect to gives ϕ tan y. 5.3.56 a. Because b. Because 5.3.57 a. a + by b and c + dy c, the field is conservative when b c. a by by and cy cy, the field is conservative when c b. b. Parameterize by two paths: r t t,, t, and r t, t, t y. Then F dr t, t, dt + y t, +t, dt tdt+ y t dt + y y. c. Parameterize by the paths r t,t, t y, and r t t, y, t. Then F dr y t, t, dt + t y, t +y, dt y tdt+ t y dt + y y. 5.3.58 Using problem 57, we have the same paths r, r, and F dr, t, dt + y t,, dt dt + y dt y. 5.3.59 Using problem 57 and the same paths r, r,wehave F dr t,, dt+ y, t, dt tdt+ y tdt + y. 5.3.6 Using problem 57 and the same paths r, r, note that F r r, y + y /, and F dr,, dt + y, t +t +t, dt dt + y t dt + + y + y. +t 5.3.6 Using problem 57 and the same paths r, r,wehave F dr t3,, dt + y 3 + t, t 3 + t, dt t3 dt + y t 3 + t dt 4 + y + y 4. opyright c 3 Pearson Education, Inc.

5 HAPTE 5. VETO ALULU 5.4 Green s Theorem 5.4. As with the Fundamental Theorem of alculus, it allows evaluation of the integral of a derivative by looking at the value of the underlying function on the boundary of a region or, in the case of the Fundamental Theorem, an interval. 5.4. The line integral for flu corresponds to the double integral of the divergence; the line integral for circulation to the double integral of the curl. 5.4.3 The curl is g f y +4 3 4 3 y. 5.4.4 The divergence is f + g y +y. 5.4.5 The area is dy yd where is the boundary of the region. 5.4.6 Because the curl being zero is an equivalent condition to the field being conservative. 5.4.7 Because the flu is the integrand in Green s theorem, so the integral vanishes. 5.4.8 A conservative vector field such as, y will have zero curl: 5.4.9 5.4. A conservative and a source-free field each have functions a potential function in the case of a conservative field; a stream function in the case of a source-free field that closely reflect the vector field. The properties of the partials of these functions are such that the curl or divergence, for a source-free field vanish. 5.4. a. The curl is g f. b. F dr π cos t, sin t sin t, cos t dt. opyright c 3 Pearson Education, Inc. g f da da.

5.4. GEEN THEOEM 5 c. The vector field is conservative because its curl is zero. 5.4. a. The curl is g f. b. F dr,t, dt + t,, dt +, t, dt + t,, dt. g f da da. c. The vector field is conservative because its curl is zero. 5.4.3 a. The curl is g f 4. b. F dr π, t, dt + π sin t, t, cos t dt sin t t cos t dt 8. π g f da 4 da 4 π sin d 8. c. It is not conservative because the curl is nonzero. 5.4.4 a. The curl is g f 3+36. b. F dr, 3t, dt + 6t, 3 3t, dt + 3 t,, dt 6t +6 6t dt 6. g f da 6 da 6 d 6. c. No, because the curl is nonzero. 5.4.5 a. The curl is g f b. F dr,t, dt + t t,t t t, t dt t t+t t t g f da da. c. Yes, because the curl is zero. 5.4.6 a. The curl is g f. dt. b. F dr π, sin t + cos t sin t, cos t dt π cos tdt. g f da da π r cos θ rdrdθ. c. No, because the curl is nonzero. 5.4.7 Parameterize the boundary by 5 cos t, y 5 sin t, t π. Then d 5 sin tdt, dy 5 cos tdt, and the area is dy yd π 5 cos t 5 cos t 5 sin t 5 sin t dt 5 π dt 5π. opyright c 3 Pearson Education, Inc.

5 HAPTE 5. VETO ALULU 5.4.8 Parameterize the boundary by 6 cos t, y 4 sin t, t π. Then d 6 sintdt, dy 4 cos tdt, and the area is dy yd π 6 cos t 4 cos t 4 sin t 6 sin t dt π dt 4π. 5.4.9 Parameterize the boundary by 4 cos t, y 4 sin t, t π. Then d 4 sintdt, dy 4 cos tdt, and the area is dy yd π 4 cos t 4 cos t 4 sin t 4 sin t dt 8 π dt 6π. 5.4. This is an elliptical region whose boundary is an ellipse with semimajor ais 5 and semiminor ais 3. Parameterize the ellipse by 5 cos t, y 3 sin t, then d 5 sin tdt and dy 3costdt. The area is then dy yd π 5 cos t 3 cos t 3 sin t 5 sin t dt 5 π dt 5π. 5.4. Traverse the first path from to, then the second path back from to. The area is then dy yd t 4 t t dt + t t t dt t dt + t + t dt 3. 5.4. y..8.6.4..4...4 We have d t + t t dt 3t dt and dy tdt. We parameterize the curve from t to t so that we traverse the region counterclockwise; then the area is dy yd t t t t 3t dt 8 5. 5.4.3 a. The divergence is f + g +. b. F n ds 4 π cos t cos t sin t sin t dt 8π. f + g da da 4π 8π. c. It is not source-free because its divergence is nonzero. 5.4.4 a. The divergence is f + g +. b. F n ds + t dt + t + dt + + t dt + t + dt t + t +t dt 4t dt. f + g da da. opyright c 3 Pearson Education, Inc.

5.4. GEEN THEOEM 53 c. Yes, because its divergence is zero. 5.4.5 a. The divergence is f + g. b. F n ds + 3t dt + 4 t t +3t dt. f + g da da. c. Yes, because its divergence is zero. 5.4.6 a. The divergence is +. b. Parameterize the triangle by the three paths: r t 3t,, r t 3 3t, t, and r 3 t, t, all for t. Then F n ds + 9t 3 dt + 3t + 9 9t 3 dt + 3t 3 dt 7t 3t 7+7t 3t +3dt 48t 4 dt. f + g da da. c. Yes, because its divergence is zero. 5.4.7 a. The divergence is f + g y y. b. Parameterize the region by r t t,, and r t t, t t for t ; traverse the second path from t to t to make a counterclockwise closed curve. Then F n ds t dt + t t t t t dt. f + g da da c. Yes, because its divergence is zero. 5.4.8 a. The divergence is f + g. sin t + cos t cos t + sin t dt π b. F n ds π f + g da c. No, because its divergence is nonzero. da π cos θ rdrdθ. cos tdt. 5.4.9 The line integral, using the flu form of Green s theorem, is equal to + e y + 4y + e da +8y da +8y d dy +8y dy 6. 5.4.3 Using the flu form of Green s theorem, the integral is equal to 6 da 6 area of 6π. 5.4.3 Using the flu form of Green s theorem, the integral is equal to 4 dyd 4 d 8 3. opyright c 3 Pearson Education, Inc. 3y+ 3 +4y da + y da

54 HAPTE 5. VETO ALULU 5.4.3 Using the flu form of Green s theorem, the integral is equal to note the leading minus sign to correct for the orientation + y da +4y dy d y +y y d y + d 8 3. 5.4.33 Using the circulation form of Green s theorem, the integral is equal to 4 + y 3 + y π sin da 4 y da 4 y dy d 8 π. 5.4.34 Using the flu form of Green s theorem, the integral is equal to e y + ey da e y + e y da e y + e y dy d e y + e y y y d e e d e + e. 5.4.35 a. Using Green s theorem, the circulation is y da da, b. Using Green s theorem, the flu is 3π. + y da da area of 4π π 5.4.36 a. Using Green s theorem, the circulation is y da da area of 9π π 6π. b. Using Green s theorem, the flu is y+ da. 5.4.37 a. Using Green s theorem, the circulation is b. Using Green s theorem, the flu is 4 6π π 5 π. 5.4.38 a. Using Green s theorem, the circulation is b. Using Green s theorem, the flu is 5.4.39 4y + y da + y+ 4y da a. True. This is the definition of work along a path. +y y da y+ +y da da. da area of da. 3 da 3 area of 6. b. False. ivergence corresponds to flu, so if the divergence is zero throughout a region, the flu is zero across the boundary. opyright c 3 Pearson Education, Inc.

5.4. GEEN THEOEM 55 c. True. This follows from Green s theorem. 5.4.4 a. The circulation is 3 y +y da 3 π b. The flu is π 3r cosθ r 5.4.4 tan y ln + y da r sinθ r ln + y + rdrdθ π rdrdθ 3 π tan y da y +y y +y da sin θ dr dθ 3 π sin θ dθ. 3 +y da 3 cos θ dr dθ π 3 cos θ dθ. a. Because F is conservative, the circulation on the boundary of is zero. b. F + y /, y, so the flu is + y +y da +y + y / π 3 da r rdrdθ π 3 dr dθ π. 5.4.4 a. The circulation is sin y cos da π/ π/ cos dy d π/ cos da π cos d π. b. The flu is y cos + sin da π/ π/ y sin dy d 8 π. y sin da 5.4.43 Note that the region is the area between 3y and 36 y, which intersect at y 3. a. The circulation is 3 3 b. The flu is y + y da 96 7y 7y +8y 3 8y 4 dy 85 5. + y + y da da. 5.4.44 By Green s theorem, d d +dy dy d +dy da da. 5.4.45 Because f g, the integral is zero because F is conservative. opyright c 3 Pearson Education, Inc. y da 3 36 y y d dy 3 3y da da. imilarly,

56 HAPTE 5. VETO ALULU 5.4.46 Let f, y,g, y y + y 4 ; then y +4y 3 f da + g da dy gd y + y 4 d by Green s theorem, where is the boundary of the triangle. To evaluate this line integral, parameterize by three paths, all for t : r t t,, sothatr t,. r t t, t, so that r t,. r 3 t, t, so that r 3 t,. Then y + y 4 d t 4 dt t t dt t + t 4 dt t 4 t 3 + t dt 7 6. 5.4.47 By Green s theorem, y d + y + dy y + y da y + y da da area of A. 5.4.48 Using the circulation form of Green s theorem, the integral is ay d + b dy b a da b a area of A. 5.4.49 a. The divergence is 4 +. b. ψ 4y. 5.4.5 a. The divergence is y + +. b. ψ 3 y 3 3. 5.4.5 a. The divergence is e sin y + e cos y e sin y e sin y. b. ψ e cos y. 5.4.5 a. The divergence is b. ψ y. 5.4.53 + y. a. The curl and divergence are curl F e sin y e cos y e sin y +e sin y. div F e cos y + e sin y e cos y e cos y. b. ϕ, y e cos y. ψ, y e sin y. c. ϕ +ϕ yy e cos y+ e cos y e cos y e cos y. ψ +ψ yy e sin y+ e sin y e sin y e sin y. opyright c 3 Pearson Education, Inc.

5.4. GEEN THEOEM 57 5.4.54 a. The curl and divergence are curl F y 3 3 y 3 3y 6y +6y. div F 3 3y + y 3 3 y 3 3y +3y 3 b. ϕ, y 4 4 + y 4 3 y. ψ, y 3 y y 3. c. ϕ +ϕ yy 4 4 + y 4 3 y + 4 4 + y 4 3 y 3 3y + y 3 3 y 3 3y +3y 3. ψ + ψ yy 3 y y 3 + 3 3y 6y 6y. 3 y y 3 + 3 y y 3 5.4.55 a. The curl and divergence are curl F ln + y tan y. div F tan y + ln + y. b. ϕ, y tan y y + y ln + y. ψ, y ln + y dy y tan y ln + y +. c. ϕ + ϕ yy tan y y + y ln + y + tan y y + y ln + y tan y y +y + y +y + +y + ln + y + y +y tan y + ln + y. ψ + ψ yy y tan y ln + y + + y tan y ln + y + y +y +y ln + y + + tan y + y +y y +y ln + y + tan y. 5.4.56 a. The curl and divergence are curl F div F +y + y +y. b. ϕ, y ln + y. ψ, y tan y. c. ϕ + ϕ yy ln + y + 5.4.57 y. ψ +y +ψ yy tan y y. +y y +y a. The velocity field is 4 cos sin y, 4 sin cos y. ln + y + tan y +y, and +y + y y +y + +y +y y + +y y +y opyright c 3 Pearson Education, Inc.

58 HAPTE 5. VETO ALULU b. The field is source-free if its divergence is zero. div F 4 cos sin y+ 4 sin cos y 4sinsin y 4 sin sin y, so the field is source-free. c. The field is irrotational if its curl is zero. curl F 4 sin cos y 4 cos sin y 4 cos cos y + 4 cos cos y 8 cos cos y, so the field is not irrotational. d. ince the field is source-free, it has zero flu across the boundary. e. The circulation around the boundary of the rectangle is by Green s theorem given by π/ π/ π/ 8 cos cos yda 8 cos cos ydyd 6 cos d3. π/ π/ 5.4.58 If f is continuous, then the circulation form of Green s theorem says that f c dy df c d da. The right side of this equation evaluates to df c d da b c df c a d dy d b df a d d. To evaluate the left side, parameterize the boundary of with four paths, each for t : r t a +b a t,, so r t b a,. r t b, c t, so r t,c. r 3 t b +a b t, c, so r 3 t a b,. r 4 t a, c ct, so r 4 t, c. Then we evaluate F r i for each i and add: c f dy f a +b a t +f b c + f b +a b t +f a c dt f b f a. c 5.4.59 If f is continuous, then the flu form of Green s theorem says that f c d df c d da. The right side of this equation evaluates to df c d da b c df c a d dy d b df a d d. To evaluate the left side, parameterize the boundary of with four paths, each for t : r t a +b a t,, so r t b a,. r t b, c t, sor t,c. r 3 t b +a b t, c, sor 3 t a b,. r 4 t a, c ct, so r f 4 t, c. Then we evaluate F r i for each i and add: c dy c + f b c ++f a c dt f b f a so that b a 5.4.6 a. The curl is +y y +y. π/ df d d f b f a. b. Take a line integral around the unit circle, parameterized as cos t, sin t. The circulation is then y +y d + +y dy π sin t sin t + cos t cos t dt π dt π. c. The vector field is not defined everywhere in ; specifically, it is undefined at the origin. 5.4.6 a. The divergence is +y + y +y. b. Take a line integral around the unit circle, parameterized as cos t, sin t. The flu is then y +y d π cos t cos t sin t sin t dt π dt π. c. The vector field is not defined everywhere in ; specifically, it is undefined at the origin. +y dy + opyright c 3 Pearson Education, Inc.

5.5. IVEGENE AN UL 59 5.4.6 a. Green s theorem does not apply to a region including the origin because F is not defined at the origin. b. + y +y da + y / π da +y r rdrdθπ. c. dy y d π cos t cos t sin t sin t dt π dt π. +y +y d. They do agree. Because Green s theorem does not apply, there is no particular reason why they should. 5.4.63 Because ψ is a stream function, dψ ψ d + ψ y dy, so the flu integral is F n ds fdy gd ψ y dy ψ d dψ ψ B ψ A, so that the integral is independent of the path. 5.4.64 howing that F is tangent to the level curves of the stream function is the same as showing that F is normal to the gradient of the stream function. But that gradient is ψ,ψ y, and F ψ, ψ y ψ y,ψ ψ,ψ y. 5.4.65 howing that the level curves of ϕ and ψ are orthogonal is equivalent to showing that the gradients of ϕ andψ are orthogonal. But ϕ ψ f,g g, f. 5.4.66 a. The stream function is found by taking d 3 3. A plot together with some streamlines is b. The curl of F is, so the curl on is ; on 4 it is ;on,itis, and on,itis. c. The circulation is by Green s theorem da d dy dy. d. The curl is positive for negative and negative for positive. These cancel, giving a net circulation of zero. This can easily be seen from the picture - any circulation resulting from the top boundary y is cancelled by the circulation in the opposite direction resulting from the bottom boundary. 5.5 ivergence and url 5.5. The divergence is f + g + h z. 5.5. The divergence measures the epansion or contraction of the vector field at each point. 5.5.3 It means that the field has no sources or sinks. opyright c 3 Pearson Education, Inc.

5 HAPTE 5. VETO ALULU 5.5.4 The curl is F h g z i + f z h j + g f k. 5.5.5 The curl indicates the ais and speed of rotation of a vector field at each point. 5.5.6 It means that the vector field is irrotational. 5.5.7 F ; see Theorem 5.. 5.5.8 Here u is a potential function, so u is the curl of a conservative vector field, which is. 5.5.9 + 4y+ z 3z 3. 5.5. y+ 3+ z z. 5.5. + 6y+ z 6z. 5.5. yz + y z + z yz yz yz yz yz. 5.5.3 y + y z + z z +y +z. 5.5.4 ey + ez y + z e z e y + e z y + e z. 5.5.5 + +y + y + +y + z z + +y +y +3 + +y. 5.5.6 yz sin + z cos y + z y cos z yz cos z sin y y sin z. 5.5.7 +y +z + y +y +z + z +y +z +y +z z + y + + z y + + y z +y +z + y + z +y +z 3/ + 5.5.8 y + +y +z 3/ z +y +z z + y + + z y + + y z. 5/ 5.5.9 +y +z + y +y +z + z z z +y +z 3/ z +y +z +y +z z + y 3 + + z 3y + + y 3z 3 +y +z + y + z 3 r 4 5.5. + y + z + y + y + z + z z + y + z 3 + y + z + +3y + z + + y +3z 5 r r. 5.5. a. At both P and Q, the arrows going away from the point are larger in both number and magnitude than those going in, so we would epect the divergence to be positive at both points. b. The divergence is + + y +, so is positive everywhere. c. The arrows all point roughly away from the origin, so we the flu is outward everywhere. d. The net flu across should be positive. 5.5. a. At P, the divergence should be positive, while at Q, the larger arrows point in towards Q, so the divergence should be negative. b. The divergence is + y +y; atp,, this is 3, while at Q,, it is. opyright c 3 Pearson Education, Inc.

5.5. IVEGENE AN UL 5 c. The flu is outward above the line y approimately; below this line, the flu is inward across. d. The size of the arrows pointing outward at the top of the circle seems to roughly equal those pointing inward at the bottom, so the remaining outward-pointing arrows result in a net positive flu across. 5.5.3 i j k a. The ais of rotation is,,, the -ais. F zj + yk + i + y z j + k i. It is in the same direction as the ais of rotation. b. The magnitude of the curl is i. 5.5.4 i j k a. The ais of rotation is,,. F zi zj+ + y k + i + y z j+ k,, and the curl is in the same direction as the ais of rotation. b. The magnitude of the curl is,, 5.5.5 a. The ais of rotation is,,. i j k F y + z i + z j + + y k + i+ j++k y z,,, and the curl is in the same direction as the ais of rotation. b. The magnitude of the curl is,, 3. 5.5.6 a. The ais of rotation is,, 3. i j k F 3 3y z i + 3 z j + + y k + i + j + y z 3 3 k,, 3, and the curl is in the same direction as the ais of rotation. b. The magnitude of the curl is,, 3 4. 5.5.7 y,y,z i + j +y +y k 3yk. 5.5.8,z y, yz z z i + j + k 3zi. 5.5.9 z,, z i + z z j + k 4zj. 5.5.3, y, z i + j + k. 5.5.3 +y +z 3/, y, z +y +z 5/ 3yz +3yz i + 3z +3z j + 3y +3y k. opyright c 3 Pearson Education, Inc.

5 HAPTE 5. VETO ALULU 5.5.3 +y +z /, y, z +y +z 3/ yz + yz i + z + z j + y + y k. 5.5.33 z sin y,z cos y, z sin y z cos y z cos y i +z sin y z sin y j + z cos y z cos y k. 5.5.34 3z 3 e y, z 3 e y, 3z e y 6yz e y 6z e y i + 9z e y 3z e y j + z 3 e y 6yz 3 e y k z e y 6y 6 i +9 3 j +z 6yz k. 5.5.35 imply compute it:, +y +z 3/, +y +z 3/ z +y +z 3/ 3 3y 3z,, +y +z 5/ +y +z 5/ +y +z 3r. 5/ r 5 5.5.36 +y +z, +y +z, z +y +z y z,, +y +z +y +z +y +z r. r 4 5.5.37 r r, from Problem 36; applying Theorem 5.8 we have r r 4 r r 4 3 4. r 4 r 4 5.5.38 ln r ln + y + z ln + y + z r +y +z, y, z. r 5.5.39 a. False. For eample, F y, z, has zero divergence yet is not constant. b. False. For eample, F, y, z is a countereample. c. False. For eample, consider the vector field, from problem 66 in the previous section. d. False. For eample, F,, has divergence. e. False. For eample, the curl of z, z,y is,,. 5.5.4 a. F u f,g,h,, z u f + g + h z u f u + g u + h u z. b. Because F,,, F y z 3 y z 3 + y z 3 + z y z 3 y z 3 +yz 3 +3y z. 5.5.4 a. No; divergence is a concept that applies to vector fields. b. No; the gradient applies to functions. c. Yes; this is the divergence of the gradient and is thus a scalar function. d. No, since ϕ does not make sense part a. e. No; curl applies to vector fields, ϕ does not make sense. f. No, since F is a function, so that applying to it does not make sense. g. Yes, this is the curl of a vector field and is thus a vector field. h. No, since F is a function, not a vector field. i. Yes; this is the curl of the curl of a vector field and is thus a vector field. 5.5.4 Let a a,a,a 3 ; then F a r a z a 3 y i +a 3 a z j +a y a k, so that F a z a 3 y+ a 3 a z+ z a y a. opyright c 3 Pearson Education, Inc.

5.5. IVEGENE AN UL 53 5.5.43 Let a a,a,a 3 ; then F a r a z a 3 y i +a 3 a z j +a y a k, so that F a + a i + a + a j + z a 3 + a 3 k a. 5.5.44 The field switches from inward-pointing to outward-pointing at points where it is tangent to the circle + y, i.e. where it is orthogonal to the normal to the circle. The normal to the circle at, y is a multiple of, y, so we want to find, y so that, y,y 3 + y with + y. Thus 3 +. The solutions are and y ±. 5.5.45 div F +yz + yz + ; this function clearly achieves its maimum magnitude at,,,,,,,,, and,,, where its magnitude is 6. 5.5.46 For F z,, y, curl F,,. a. The component of curl F in the direction,, is. b. The component of curl F in the direction,, is,,,,. c. The component of curl F in the direction a, b, c is b a a +b +c ; this has its maimum in the direction,,. 5.5.47 curl F +, +,,,. Ifn a, b, c, then curl F n when a + b c so that c a + b; thus all such vectors are of the form a, b, a + b, where a, b are real numbers. 5.5.48 F z,,, orf,,, so it is not unique. 5.5.49 F y + z,, or F, y, z, so it is not unique. 5.5.5 a. Looking at the picture, it is clear that the distance from P to a is r. b. The velocity field is a r, so the speed, which is the magnitude of velocity, is a r. Now a r a r sin θ n a n, where n is a vector normal to the plane determined by a and r. Thus the motion of the particle is always perpendicular to this plane, so it rotates about the ais a. It is moving at a speed a around a circle of radius, so its angular speed is a a. c. Because v a, it follows from part b that ω a v. 5.5.5 The curl of this vector field is,,. The component of the curl along some unit vector n is F n. a.,,,,, so the wheel does not spin. b.,,,,, so the wheel spins clockwise looking towards positive y. c.,,,,, so the wheel does not spin. 5.5.5 The curl of the vector field is v,,. opyright c 3 Pearson Education, Inc.

54 HAPTE 5. VETO ALULU a. The wheel is placed with its ais in the direction,,, so the component of velocity in that direction is,,,,, and ω. b. The wheel is placed with its ais in the direction,,, so the component of velocity in that direction is,,,,, and the wheel does not turn. c. The wheel is placed with its ais in the direction,,, so the component of velocity in that direction is,,,,, and ω. 5.5.53 The curl of the vector field is v,,. Because the wheel is placed with its ais normal to the plane + y + z, its ais must point in the direction,, with unit vector 3,,. Thus, the component of velocity along that direction is 3,,,, 3 and then ω is the absolute value of one half of that amount, or ω 5 3 or π.989 revolutions per time unit. 3 5.5.54 F k e +y +z ke +y +z, y, z. Looking at the component, its contribution to the divergence is k +y +z [ ] e +y +z k +y +z y z e +y +z and similarly for the +y +z +y +z 3/ y and z components. Thus the divergence is the sum of these three terms, which is k e +y +z + y + z 3/ +y +z + y + z k e +y +z + y + z. 3/ +y +z 5.5.55 F k e +y +z ke +y +z, y, z, so the divergence is k e +y +z + ye +y +z + ze +y +z z k e +y +z + e +y +z + e +y +z +y e +y +z + e +y +z +z e +y +z k e +y +z + + y + z e +y +z k + y + z /, y, z, and thus the diver- k +y +z + e +y +z. 5.5.56 F k + + + y + z gence is F k +y +z 5.5.57 a. F ϕ GMm [ ] +y +z, [ ] +y +z, z [ ] +y +z GMm + y + z 3/, y, z GMm + y + z 3/, y, z GMm r r 3. b. + y + z 3/ 3y + y + z 5/. Applying this pattern in computing the curl gives F GMm + y + z 5/ 3yz +3yz i + 3z +3z j + 3y +3y k, so the field is irrotational. 5.5.58 Note: this is identical to the previous problem ecept for the constant. a. F ϕ q 4πɛ [ ] +y +z, q 4πɛ + y + z 3/, y, z [ ] +y +z, z [ ] +y +z q 4πɛ + y + z 3/, y, z b. + y + z 3/ 3y + y + z 5/. Applying this pattern in computing the curl gives F q 4πɛ + y + z 5/ 3yz +3yz i + 3z +3z j + 3y +3y k, so the field is irrotational. opyright c 3 Pearson Education, Inc. q 4πɛ r r 3.

5.5. IVEGENE AN UL 55 5.5.59 Using Eercise 4, we have ρ u t, v t, w t + so that 5.5.6 ρ u t + v t + w t u ρ t + uu + v u + w u z v ρ t + u v + v v + w v z w t + uw + v w + w w z. u, v, w p t, p t, p t + μ t + t + t u, v, w, p p + μ u + u + u z + μ v + v + v z p z + μ w + w + w z a., 3y, 5z and y, z, y so they are both irrotational. b. If ψ is defined as stated, then ψ ψ + ψ + z ψ v k u, v, k v u zi+ z j+ v u k c. u ψ + u. u v while ζ v u so that ψ ζ as desired. ψ sin cos y and v cos sin y. The velocity field looks like d. The vorticity function is ζ ψ v u sin sin y + sin sin y sin sin y The diagram shows level curves for ζ at 4,, 3 4, and 3 from outer to inner. Using implicit differentiation or from looking at the diagram, ζ achieves its maimum at y π, where it has value, and its minimum on the boundary, where it is zero. opyright c 3 Pearson Education, Inc.