FINAL REVIEW. 1. Vector Fields, Work, and Flux Suggested Problems:

Transcription

1 FINAL EVIEW 1. Vector Fields, Work, nd Flux uggested Problems: { , 13, , 25, 27, 29, 36, 45 We dene vector eld F (x, y) to be vector vlued function tht mps ech point in the plne to two dimensionl vector. Exmple 1.1. Let F (x, y) = 2, 3. hen F is constnt vector eld, nd every point in the plne mps to the vector 2, 3. Drw the vector eld F. Now, suppose tht F = 2, 3 is force eld nd let = ( 1, 2 ) nd b = (b 1, b 2 ). Dene the directed line segment from to b by d = b 1 1, b 2 2 = x, y. hen the work done by F when moving prticle from to b is given by W = F d = 2 x + 3 y. herefore, ssocited with our force eld nd the directed line segment from to b is the work integrl W = b 2dx + 3dy. We cn generlize this to rbitrry force elds F (x, y) nd curves. Denition 1.2. Let F (x, y) = f(x, y), g(x, y) be ny force eld nd be ny curve. hen the ssocited work integrl is W = 2π f(x, y)dx + g(x, y)dy. Exmple 1.3. Let F (x, y) = y, x nd be ny curve. Write down the ssocited work integrl. he trick to solving the work integrl is to prmeterize the curve nd turn the work integrl into n integrl over t. Exmple 1.4. Let F (x, y) = y, x nd be the curve x = y 2 from (, ) to (1, 1). hen the work integrl is given by We cn prmeterize by r(t) = ydx + xdy. t, t, t 1. 1

2 FINAL EVIEW 2 herefore, x(t) = t dx = 1dt nd y(t) = t dy = 1 2 dt. he work integrl t becomes 1 ( ) t t W = + dt = tdt = t = 1. Exmple 1.5. Let F be the sme force eld from Exmple 1.4. ompute the work done from (, ) to (1, 1) long the following curves: is the curve y = x 2, is the line segment from (, ) to (2, ) to (1, 1). Now, lets go bck to constnt vector eld F = v 1, v 2 nd directed line segment from = ( 1, 2 ) to b = (b 1, b 2 ), given by d = b 1 1, b 2 2 = x, y. If F is constnt ow vector, then we cn consider the uniform ow of F cross d. he mount of uid tht crosses d in one unit time is equl to the re of the prllelogrm spnned by d nd F, which is given by F d = i j k v 1 v 2 x y = i() j() + k (v 1 y v 2 x) = v 1 y v 2 x. he re of the prllelogrm gives us the mount of uid tht crosses d in one unit time, but we lso need to consider the direction. A convenient nottion is to mke the sign of our ow gree with the sign of the z-component in the cross product. herefore, ssocited with our ow vector nd the directed line segment from to b is the ux integrl flux = b v 2 dx + v 1 dy. We cn generlize this to rbitrry ow vectors F (x, y) nd curves. Denition 1.6. Let F (x, y) = f(x,, y), g(x, y) be ny ow vector nd be ny curve. hen the ssocited ux integrl is flux = g(x, y)dx + f(x, y)dy Just like the work integrl, the key to solving the ux integrl is to prmeterize the curve nd turn the ux integrl into n integrl over t. Exmple 1.7. Let F (x, y) = y, x. (, ) to (1, ), for the following curves. = {( sin(t), cos(t π 2 )) : t π 2 }. is the directed line segment from (, ) to (1, ). ompute the ux cross the curve from Let us summrize our work nd ux integrls for rbitrry vector elds F (x, y) = f(x, y), g(x, y) nd curve.

3 FINAL EVIEW 3 W = f(x, y)dx + g(x, y)dy = flux = g(x, y)dx + f(x, y)dy = b b ( ) f(t)x (t) + g(t)y (t) dt, ( ) g(t)x (t) + f(t)y (t) dt, where r(t) = x(t), y(t) is prmeteriztion of, nd f(t) := f (r(t)) nd g(t) := g (r(t)). Exmple 1.8. Let F (x, y) = x 2 + y 2, 2xy nd be the upper hlf of the unit circle. ompute the work integrl nd ux integrl ssocited with F nd. 2. Grdient Fields, Potentil functions, nd conservtive vector fields uggested { Problems: , 27, 29, 31, 38, , 19, 21, 23, 26, 27, 36, 37, 41, 43, 47, 48 We dene potentil function φ(x, y) to be sclr vlued function tht mps ech point in the plne to rel number. hese functions re clled potentil functions becuse they re intended to represent the potentil energy t given point. Exmple 2.1. Let φ(x, y) = x 2 + y 2. onsider the surfce z = φ(x, y) nd drw the level surfces φ(x, y) = 1, φ(x, y) = 4. Associted with every potentil function is grdient eld nd equipotentil curves. Denition 2.2. Let φ(x, y) be some potentil function. he grdient eld is the vector eld F (x, y) = φ(x, y). he equipotentil curves re the level curves φ(x, y) = c, where c is some constnt. Exmple 2.3. Let φ(x, y) = x 2 + y 2. Find the grdient eld nd show tht the grdient eld is orthogonl to the equipotentil curves. ecll tht in ssignment 1, we proved tht for ny potentil function, the grdient eld is lwys orthogonl to the equipotentil curves. he importnce of potentil functions nd grdient elds is their use in how we dene conservtive vector elds. Denition 2.4. he vector eld F (x, y) is conservtive, if there exists potentil function φ(x, y) such tht F is the grdient eld of φ. ht is, F (x, y) = φ(x, y). Exmple 2.5. Let F (x, y) = xy, 1 2 x2. hen F (x, y) is conservtive vector eld, since φ(x, y) = 1 2 x2 y is potentil function whose grdient is F (x, y).

4 FINAL EVIEW 4 onservtive vector elds ply n extremely importnt role. For one, they re exctly the vector elds for which the fundmentl theorem of clculus holds. heorem 2.6. Let F (x, y) be conservtive vector eld (force eld) nd φ(x, y) be potentil function such tht F (x, y) = φ(x, y). Let be some curve nd r(t) be prmeteriztion of, where t b. hen, the work integrl b Proof. ince F = φ, we hve F (r(t)) r (t)dt = φ (r(b)) φ (r()). b φ (r(t)) r (t)dt. We recognize wht is inside the integrl s the derivtive of φ (r(t)) with respect to t. herefore, we hve b ( ) d dt φ (r(t)) dt. From the fundmentl theorem of clculus, we know tht b ( ) d dt φ (r(t)) dt = φ (r(b)) φ (r()). Let φ(x, y) mesure the potentil energy t ny point in the plne. hen heorem 2.6 sys tht if the vector eld (force eld) is conservtive, then the work done is equl to the chnge in potentil energy. Exmple 2.7. Let F (x, y) = y, x be the given vector eld. how tht F (x, y) is conservtive by nding potentil function φ(x, y) such tht F is the grdient eld of φ. ompute the work done by F long ny pth from the point (, ) to (1, 1). In the bove exmple, nding potentil function should be cler. However, in some cses this cn be dicult problem. Exmple 2.8. Let F (x, y) = y, x be the given vector eld. We will show tht F is not conservtive vi contrdiction. uppose tht F is conservtive, then there exists φ such tht F = φ. It follows tht φ x = y, φ y = x. herefore, we hve two eqully vlid equtions φ(x, y) = yx + f(y), φ(x, y) = xy + g(x). Let us pick on the second eqution bove, by tking the derivtive with respect to x: φ x = y + g (x).

5 FINAL EVIEW 5 We know tht the bove eqution should equl y, therefore we hve g (x) = 2y. But this is n bsurd sttement, since g should be function of x, only. his is our contrdiction, nd it follows tht F is not conservtive. he procedure outlined in the bove exmple cn lso be used to nd potentil function. Moreover, everything we've done cn be generlized to three dimensions. Exmple 2.9. Let F (x, y, z) = 2xy z 2, x 2 xz + sin(z), y cos(z) z 2 be the given vector eld. how tht F (x, y, z) is conservtive by nding potentil function φ(x, y, z) such tht F = φ. 3. Loop Integrls nd Greens heorem uggested { Problems: , 14, 15, 29, , 9, 11, 13, 21, 22, 27, 29, 33 We sy tht curve is closed if the curve strts nd ends in the sme plce. peciclly, if r(t) : t b is prmeteriztion of curve, then is closed if r() = r(b). Integrls round closed curves rise so frequently tht they hve specil nottion, for exmple the work integrl fdx + gdy. he circle on the integrl emphsizes tht is closed curve. his integrl is known s loop integrl. ometimes n rrow will pper on the circle, specifying whether the pth is to be trveled counterclockwise or clockwise. Exmple 3.1. Let F (x, y) = xy, x be the given force eld nd be the unit circle. he work done by F in moving prticle round is given by xydx + xdy. ompute the work integrl. ince the work integrl in Exmple 3.1 did not equl zero (in fct W = π), it should be cler tht the force eld given is not conservtive. heorem 3.2. Let F (x, y) be conservtive vector eld nd be ny closed curve. hen the work done by F in moving prticle round is zero. Proof. ince F is conservtive there exists potentil function φ such tht F = φ. Let r(t) : t b be prmeteriztion of the closed curve. By heorem 2.6, the work done is equl to the chnge in potentil energy, tht is W = φ (r(b)) φ (r()). ince is closed r(b) = r(), nd it follows tht W =.

6 FINAL EVIEW 6 Now, let be closed curve nd let be the region in the plne which is enclosed by. hen we cn relte the work nd ux integrls over to double integrls over. he following theorem is known s the Green's theorem. heorem 3.3. Let F (x, y) = f(x, y), g(x, y) be ny vector eld, let be closed curve nd be the region in the plne enclosed by. hen the work integrl over cn be written s ( g fdx + gdy = x f ) dxdy. y Proof. We cn pproximte by union of rectngulr blocks whose sides re prllel to the x nd y xis. he integrl over the boundry of this union is the sum of the integrls over the individul boundries. herefore, it suces to prove Green's theorem for single rectngulr region. o this end, let be the rectngle with vertices (, c), (b, c), (b, d), (, d) nd let be the perimeter of. By the fundmentl theorem of clculus we hve g(b, y) g(, y) = f(x, d) f(x, c) = b d c g x dx, f y dy. Now, we tke the work integrl nd split it into 4 pieces f(x, y)dx + g(x, y)dy = b d f(x, c)dx + c g(b, y)dy + b c f(x, d)dx + d g(, y)dy = d c b (g(b, y) g(, y)) dy + = d c b = g x dx dy + b d c b (f(x, c) f(x, d)) dx d ( g x f ) dxdy. y c f y dy dx As direct consequence of heorem 3.3 we hve wht is known s the ux form of Green's theorem. heorem 3.4. Let F,, nd be s they were in heorem 3.3. hen the ux integrl over cn be written s ( f gdx + fdy = x + g ) dxdy. y

7 FINAL EVIEW 7 he use of Green's theorem cn gretly simplify the computtion of work nd ux integrls, when the curve is closed. ecll tht we spent gret del of time discussing double integrls in chpter 13 of our book. ee the review for exm 2, or ssignments 8-1 to refresh your understnding. Exmple 3.5. Let be the rectngle with vertices (, ), (4, ), (4, 2), (, 2). Let F (x, y) = x 2 y 2, 2xy be the given force eld. Use Green's theorem to compute the ssocited work integrl. Exmple 3.6. Use Green's theorem to compute the rte t which the ow x 3, y 3 leves the unit disc D = { (x, y) : x 2 + y 2 1 }. We re interested in generlizing the results from Green's theorem to work over curves in 3 nd ux cross three dimensionl surfces. o this end, we introduce the following vocbulry. Denition 3.7. Let F = f, g, h be vector eld. = / x, / y, / z is known s the Del Opertor, f = f x, f y, f z is known s the grdient of f, F = f x + g y + h z is known the divergence of F (divf ), h F = y g z, f z h x, g x f y is known s the curl of F. With these denitions in mind we cn write Green's theorem in terms of the divergence nd curl of F. heorem 3.8. Let F = f, g be vector eld, let be closed curve nd be the region in the plne enclosed by. hen under suitble conditions the work nd ux cn be written s W = fdx + gdy = ( F ) kdxdy, flux = gdx + fdy = ( F ) dxdy. Green's theorem is useful for relting work nd ux integrls over curve in the plne. o generlize Green's theorem to three dimensions we must rst discuss integrtion over surfce integrls. 4. urfce Integrls nd 3D Flux uggested Problems: { , 38, 41, 48 We will derive the ux integrl, while lerning how to integrte over surfces. First, note tht the integrl b dx = (b ) is mpping from the intervl [, b] to the length of the intervl (b ). Now, consider the oriented tringle = [, b, c], where, b, c re the vertices of the

8 tringle. hen, the integrl FINAL EVIEW 8 dxdy is mpping from the tringle to the signed re of. Exmple 4.1. We consider positive orienttion to be counterclockwise, therefore the integrl mpping over looks like nd Now, consider tringle in 3D dxdy : [(, ), (3, ), (, 4)] 6, dxdy : [(, ), (, 4), (3, )] 6. = [, b, c] = [( 1, 2, 3 ), (b 1, b 2, b 3 ), (c 1, c 2, c 3 )]. he projection of the tringle onto the x,y plne is dened by the projection of the vertices herefore, we dene the integrl [( 1, 2, ), (b 1, b 2, ), (c 1, c 2, )]. dxdy to be the mpping from the oriented tringle to the signed re of its projection onto the x,y plne, which is the z-coordinte of 1 2 (b ) (c ). imilrly, dzdx mps the oriented tringle to the signed re of its projection onto the z,x plne, which is the y-coordinte of 1 2 (b ) (c ). Finlly, dydz mps the oriented tringle to the signed re of its projection onto the y,z plne, which is the x-coordinte of 1 2 (b ) (c ). Exmple 4.2. Let = [(1,, 2), ( 1, 1, 3), (, 2, 2)] be our oriented tringle, then 1 2 (b ) (c ) = 1 ( 2, 1, 1) ( 1, 2, ) 2 ( = 1, 1 ) 2, 3. 2 herefore, dxdy = 3 2, dzdx = 1 2, dydz = 1.

9 FINAL EVIEW 9 onsider three-dimensionl uid owing t constnt velocity, v = v 1, v 2, v 3. ince we re in three dimensions, we now consider the rte t which this uid crosses surfce. Let us tke our surfce to be the tringle = [, b, c]. he positive direction from this tringle is dened to be the direction of the norml vector n = (b ) (c ). he volume of uid crossing our tringle per unit time is the volume of the prllel prism whose bse is our tringle nd whose sides re determined by v. his is hlf the volume of the prllelepiped dened by v, b, c, which is equl to 1 2 v 1 b 1 1 c 1 1 v 2 b 2 2 c 2 2 v 3 b 3 3 c 3 3 = 1 v (b ) (c ). 2 herefore, the volume of uid crossing our tringle per unit time is v 1 dydz + v 2 dzdx + v 3 dxdy. We cn generlize these results, for ny vector eld F = f, g, h nd ny surfce we hve the ssocited ux integrl flux = fdydz + gdzdx + hdxdy. ecll, when we solved ux integrls over curve, our rst step ws to prmeterize the curve. imilrly, to compute the bove integrl, we must prmeterize the surfce. Let r(u, v) = x(u, v), y(u, v), z(u, v) be prmeteriztion of. hen you cn think of r s mpping from region (in the u,v plne) to the surfce in 3. Associted with this trnsformtion is Jcobin J = x u y u z u where x u, x v, y u, y v, z u, z v re the prtil derivtives of x, y, z with respect to u nd v. Now, we know tht dx [ ] dy du = J. dv dz It follows tht dydz = y u y v z u z v dudv, dzdx = z u z v x u x v dudv, dxdy = x u x v dudv. herefore, given vector eld F = f, g, h our ux integrl becomes f y u y v + g z u z v + h x u x v dudv. z u z v x u y u x v x v y v z v y v, y u y v

10 FINAL EVIEW 1 Let r u nd r v denote the prtil derivtives of r with respect to u nd v. reful considertion of the vector r u r v should convince you tht we cn write our ux integrl s F (r u r v ) dudv. Note tht r u r v is norml vector to the surfce. herefore, the orienttion of should be reected in the vector r u r v. onsider the following exmple. Exmple 4.3. Let F = y 2, 2z, 1 nd be the upper hlf of the unit sphere oriented so tht the norml vector points wy from the origin. Let r(θ, φ) = cos θ sin φ, sin θ sin φ, cos φ, θ 2π, φ π 2 be the prmeteriztion of. hen the ux integrl cn be written s F (r θ rφ) dθdφ, where = { (θ, φ) : θ 2π, φ π 2 }. Note tht r θ rφ = cos θ sin 2 φ, sin θ sin 2 φ, cos φ sin φ, nd F (r(θ, φ)) = sin 2 θ sin 2 φ, 2 cos φ, 1. Finlly, note tht is oriented so tht the norml vector points wy from the origin. herefore, insted of r θ r φ we wnt Now, we hve flux = π 2 2π = r φ r θ = cos θ sin 2 φ, sin θ sin 2 φ, cos φ sin φ. sin 2 θ sin 2 φ, 2 cos φ, 1 cos θ sin 2 φ, sin θ sin 2 φ, cos φ sin φ dθdφ π 2 2π cos θ sin 2 θ sin 4 φ + 2 sin θ cos φ sin 2 φ cos φ sin φdθdφ = π. Note tht if ws oriented so tht the norml vector pointed towrds the origin, then we wnt r θ r φ nd the ux would hve been equl to π. We tke the time to note tht we my be sked to integrte over surfce with respect to surfce re (d). Let r(u, v) be prmeteriztion of. Dene the unit norml vector n = r u r v r u r v, then d = r u r v dudv, nd the ux integrl becomes F (r u r v ) dudv = F nds. Moreover, this provides insight into how we cn integrte over surfce with respect to surfce re (d). onsider the following exmple.

11 FINAL EVIEW 11 Exmple 4.4. Let be the surfce of the unit sphere. hen the integrl d is mpping from to the surfce re of. Let r(θ, φ) = cos θ sin φ, sin θ sin φ, cos φ, θ 2π, φ π be prmeteriztion of. hen herefore, d = r θ r φ dθdφ = sin φdθdφ. d = 2π π sin φdφdθ = 4π. 5. tokes heorem uggested Problems: { , 13 tokes' theorem is generliztion of the curl form of Green's theorem (see heorem 3.8) for closed curves in 3. heorem 5.1. Let be n oriented surfce in 3 nd be the correspondingly oriented closed curve bounding. Let F = f, g, h be vector eld. hen under suitble conditions fdx + gdy + hdz = ( F ) nd tokes' theorem reltes the work integrl over closed curve to n integrl over surfce. Exmple 5.2. Let F = z y, x, x, where is the hemisphere x 2 +y 2 +z 2 = 4, for z, nd is the circle x 2 + y 2 = 4 oriented counterclockwise. We cn compute the work integrl over by prmeterizing with r(t) = 2 cos(t), 2 sin(t),, t 2π. hen, r (t) = 2 sin(t), 2 cos(t), nd the work integrl becomes W = 2π 2 sin(t), 2 cos(t), 2 cos(t) 2 sin(t), 2 cos(t), dt = 2π 4dt = 8π. We cn lso compute the work integrl by using tokes' theorem, which sys tht W = ( F ) nd,

12 FINAL EVIEW 12 where F =, 2, 2 nd n is the unit norml vector pointed wy from the origin. Let r(θ, φ) = 2 cos θ sin φ, 2 sin θ sin φ, 2 cos φ, θ 2π, φ π 2 be the prmeteriztion of. ince the norml vector points wy from the origin we wnt r φ r θ = 4 cos θ sin 2 φ, 4 sin θ sin 2 φ, 4 cos φ sin φ. herefore, the integrl becomes W =, 2, 2 4 cos θ sin 2 φ, 4 sin θ sin 2 φ, 4 cos φ sin φ dφdθ, where = { (θ, φ) : θ 2π, φ π 2 }. It follows tht W = 2π π 2 8 ( sin θ sin 2 φ + cos φ sin φ ) dφdθ = 8π. tokes' theorem provides dditionl insight into conservtive vector elds. uppose tht the curl F is identiclly zero in some region. hen for every closed curve in tht region the work done W = F dr = ( F ) nd =, nd it follows tht F is conservtive on the region. As n exmple of this, consider problem 4 of ssignment he Divergence theorem uggested Problems: { , 18, 23 he Divergence theorem is generliztion of the divergence form of Green's theorem (see heorem 3.8) for closed curves in 3. heorem 6.1. Let be region in 3 closed by n oriented surfce. Let F be vector eld. hen under suitble conditions F nd = ( F ) dxdydz he Divergence theorem reltes the ux integrl over surfce to n integrl of the region tht is enclosed by. As n exmple of this, consider problem 5 of ssignment 12. For now, we consider the following exmple. Exmple 6.2. Let F = x, y, z nd let be the sphere x 2 + y 2 + z 2 = 1, oriented so tht the norml vector points wy from the origin. We cn compute the ux integrl F nd by prmeterizing the surfce. Let r(θ, φ) = cos θ sin φ, sin θ sin φ, cos φ, θ 2π, φ π.

13 FINAL EVIEW 13 hen r φ r θ = cos θ sin 2 φ, sin θ sin 2 φ, cos φ sin φ nd F (r(θ, φ)) = cos θ sin φ, sin θ sin φ, cos φ. herefore, the ux integrl becomes cos 2 (θ) sin 3 φ + sin 2 θ sin 3 φ + cos 2 φ sin φdφdθ, where = {(θ, φ) : θ 2π, φ π}. o, we hve the double integrl 2π π cos 2 (θ) sin 3 φ + sin 2 θ sin 3 φ + cos 2 φ sin φdφdθ = π. We cn lso use the divergence theorem, which sys tht the ux integrl is equl to F dxdydz, where is the region enclosed by (the unit bll in this cse). Note tht It follows tht the ux is equl to 2π π 1 F = 3. (3)dxdydz = 4π.