STUDY GUIDE, CALCULUS III, 2017 SPRING


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1 TUY GUIE, ALULU III, 2017 PING ontents hpter 13. Functions of severl vribles Plnes nd surfces Grphs nd level curves Limit of function of two vribles Prtil derivtives hin rule irectionl derivtive nd the grdient Tngent plnes nd liner pproximtion Optimiztion problems 4 hpter 14. Multiple integrtion ouble integrls over rectngulr regions ouble integrls over generl regions ouble integrls in polr coordintes Triple integrls Triple integrls in cylindricl nd sphericl coordintes hnge of vribles 6 hpter 15. Vector clculus Vector fields Line integrls onservtive vector fields Green s Theorem ivergence nd curl urfce integrls tokes Theorem ivergence Theorem 11 Abstrct. The following list includes the key topics nd importnt homework questions. Going through them helps your preprtion of the Finl. It is, however, by no mens exhustive. You re entitled to rewrd of 2 points towrd Test if you re the first person to report bonfide mthemticl mistke (i.e. not including lnguge typos nd grmmticl errors.) emrk. There re no choice questions in the Tests nd Finl. o when you review the choice questions in the online homework, understnd the correct nswer nd mke sure you cn nswer the question without ll the choices being given. hpter 13. Functions of severl vribles (i). ompute the derivtives, including pplictions of chin rules nd implicit differentition. (ii). Grph the six qudric surfces in 3 ; drw the level curves nd contour mp on 2 ; derive the eqution of the tngent plnes to these surfces using prtil derivtives. (iii). ompute the limit of function t point if the point is in the domin of the function (nd therefore the limit exists); prove the limit does not exist by choosing two pths. 1
2 2 TUY GUIE, ALULU III, 2017 PING (iv). ompute the directionl derivtive u f, where u is unit directionl vector nd u f is rte of chnge; compute the grdient vector f nd use it to find the mximl nd miniml rtes of chnge nd the corresponding directions u mx nd u min. (v). Use firstorder prtil derivtive to find the liner pproximtion of function. (vi). Use firstorder prtil derivtive to find the criticl points nd use secondorder prtil derivtives to clssify them s locl mximum, locl minimum, sddle points, or inconclusive Plnes nd surfces. Equtions of plnes in 3. Grphs of qudric surfces through intercepts nd trces: ellipsoid, elliptic prboloid, hyperboloid of one sheet, hyperboloid of two sheets, elliptic cone, nd hyperbolic prboloid. Importnt online questions: 2, 3, 9, 10, 12, 14, 17, Grphs nd level curves. omin nd rnge of function f(x, y) with two independent vribles x nd y. Level curves nd contour mp of function. The contour mps consisting of the level curves re on the xyplne, not in the xyzspce! Importnt online questions: 1, 4, 5, Limit of function of two vribles. If (, b) is in the domin of the function f nd f is continuous t (, b), then lim f(x, y) = f(, b). (x,y) (,b) The polynomil, rtionl, exponentil, rdicl, trigonometric functions re ll continuous in their nturl domins. This mens tht, if the point is in the domin of such function, then the limit of the function pproching the point is simply the function vlue t this point. To prove lim f(x, y) does not exist, (x,y) (,b) one needs to construct two pths of (x, y) (, b) such tht there re two distinct limits long the two pths. Importnt online questions: 1, 4, 5, 7, Prtil derivtives. ompute the prtil derivtives. lirut Theorem: f xy = f yx. Importnt online questions: 1, 5, 6, 10, hin rule. hin rule with one independent vrible: z = f(x, y), x = x(t), nd y = y(t). Then dz dt = z dx x dt + z dy y dt. hin rule with two independent vribles: z = f(x, y), x = x(s, t), nd y = y(s, t). Then z s = z x x s + z y nd z y s t = z x x t + z y y t. hin rule cn be generlized to more independent vribles.
3 TUY GUIE, ALULU III, 2017 PING 3 Implicit differentition: If F (x, y, z) = 0, then z x = F x nd z F z y = F y. F z emember the minus sign in the bove formuls! Importnt online questions: 3, 4, 7, 8, 13, irectionl derivtive nd the grdient. Let f = f(x, y) be differentible t (, b) nd u = u 1, u 2 be unit vector in 2. Then the directionl derivtive of f t (, b) in the direction of u is u f(, b) = u 1 f x (, b) + u 2 f y (, b) = f x (, b), f y (, b) u 1, u 2 = f u. The directionl derivtive is sclr! The grdient of f t (, b) is f(, b) = f x (, b), f y (, b) = f x (, b) i + f y (, b) j. The grdient is vector nd it is lwys orthogonl to the level curves of the function. Let f = f(x, y, z) be differentible t (, b, c) nd u = u 1, u 2, u 3 be unit vector in 3. Then the directionl derivtive of f t (, b, c) in the direction of u is in which u f(, b, c) = f x (, b, c), f y (, b, c), f z (, b, c) u 1, u 2, u 3 = f u, f(, b, c) = f x (, b, c), f y (, b, c), f z (, b, c) = f x (, b, c) i + f y (, b, c) j + f z (, b, c) k. At (, b), the mximl directionl derivtive is chieved when the directionl vector f(, b) u mx = f(, b), nd umx f(, b) = f(, b), while the miniml directionl derivtive is chieved when the directionl vector f(, b) u min = f(, b), nd umin f(, b) = f(, b). Keep in mind tht u mx nd u min re two (unit) directionl vectors while umx f(, b) nd umin f(, b) re two sclrs! If u is prllel to f, then u f = 0 nd there is no chnge of the function in the direction of u. Importnt online questions: 3, 5, 9, 10, 12, 15, 18, 21, Tngent plnes nd liner pproximtion. The tngent plne to F (x, y, z) = 0 t (, b, c) hs the eqution F x (, b, c)(x ) + F y (, b, c)(y b) + F z (, b, c)(z c) = 0. The tngent plne to z = f(x, y) t (, b, f(, b)) hs the eqution z = f x (, b)(x ) + f y (, b)(y b) + f(, b).
4 4 TUY GUIE, ALULU III, 2017 PING Around the bse point (, b), the function f(x, y) cn be pproximted by the liner pproximtion f(x, y) L(x, y) = f x (, b)(x ) + f y (, b)(y b) + f(, b). Keep in mind tht the prtil differentition f x nd f y here re evluted t the bse point (, b) nd remember the term f(, b). The totl differentil dz = f x (, b)dx + f y (, b)dy. Importnt online questions: 1, 2, 3, 4, 5, 8, 9, Optimiztion problems. The criticl points (, b) of f(x, y) stisfies f x (, b) = 0 nd f y (, b) = 0. econd derivtive test: Let (x, y) = f xx f yy f 2 xy. uppose tht (, b) is criticl point of f. If (, b) > 0 nd f xx (, b) < 0, then f hs locl mximum vlue t (, b); If (, b) > 0 nd f xx (, b) > 0, then f hs locl minimum vlue t (, b); If (, b) < 0, then f hs sddle point t (, b); If (, b) = 0, then the test is inconclusive. Find the bsolute mximl nd miniml vlues of function f on bounded domin : (1) etermine the vlues of f t ll criticl points in ; (2) Find the mximum nd miniml vlues of f on the boundry of ; (3) The bsolute mximl vlue is the gretest vlue in the bove two steps, nd the bsolute miniml vlue is the lest vlue in the bove two steps. Importnt online questions: 1, 3, 4, 9, 12, 18, 20. (i). Evlute double integrls hpter 14. Multiple integrtion f(x, y) da, where is rectngulr, ysimple, xsimple, polr rectngulr, rsimple in polr coordintes. In prticulr, use polr if the function f contins x 2 + y 2 or the the domin is piece of disc. (ii). The re of region in 2 is 1 da. (iii). Evlute triple integrls f(x, y, z) dv, where is cubic, zsimple, ysimple, xsimple, in cylindricl coordintes (r, θ, z), in sphericl coordintes (ρ, ϕ, θ). In prticulr, use cylindricl if the function f contins x 2 + y 2 ; use sphericl is the function f contins x 2 + y 2 + z 2 or the domin is piece of bll. (iv). The volume of solid in 3 is 1 dv. (v). The extr fctor in polr nd cylindricl is r; while the extr fctor in sphericl is ρ 2 sin ϕ. (vi). Use chnge of vribles to evlute double integrl nd remember the bsolute vlue of the Jcobin.
5 TUY GUIE, ALULU III, 2017 PING ouble integrls over rectngulr regions. If is rectngulr, i.e. = [, b] [c, d] = {(x, y) x b, c y d}, then d b b d f(x, y) da = f(x, y) dxdy = f(x, y) dydx. c If the region is rectngulr, then one cn chnge the order of integrtion freely s long s one keeps the bounds of x nd y correspondingly. Importnt online questions: 4, 7, 8, ouble integrls over generl regions. If is ysimple, i.e. = {(x, y) x b, g 1 (x) y g 2 (x)}, then b g2 (x) f(x, y) da = f(x, y) dydx. g 1 (x) If the region is ysimple, then do yintegrtion first! If is xsimple, i.e. = {(x, y) c y d, h 1 (y) x h 2 (y)}, then d h2 (y) f(x, y) da = f(x, y) dxdy. c h 1 (y) If the region is xsimple, then do xintegrtion first! One cn chnge the order of integrtion between ysimple nd xsimple. Alwys sketch the integrting region nd find ll the required informtion. The re of region in 2 is Are () = 1 da. The volume of solid in 3 bove region in 2 nd is bounded bove by z = h 1 (x, y) nd bounded below by z = h 2 (x, y) is [h 1 (x, y) h 2 (x, y)] da. Importnt online questions: 9, 13, 15, 17, 18, ouble integrls in polr coordintes. If is polr rectngulr, i.e. = {(r, θ) α θ β, 0 r b}, then β b f(x, y) da = f(r cos θ, r sin θ)r drdθ. α emember the extr fctor r; use polr coordintes if the integrting function f contins x 2 + y 2 or the integrting region is piece of disc. If is rsimple in polr coordintes, i.e. = {(r, θ) α θ β, 0 g 1 (θ) r g 2 (θ)}, then β g2 (θ) f(x, y) da = f(r cos θ, r sin θ)r drdθ. α g 1 (θ) Importnt online questions: 1, 2, 3, 5, 8, 12, 14. c
6 6 TUY GUIE, ALULU III, 2017 PING Triple integrls. If is cubic, i.e. = [, b] [c, d] [p, q] = {(x, y, z) x b, c y d, p z q}, then q d b f(x, y, z) dv = f(x, y, z) dxdydz =. p c If the region is cubic, then one cn chnge within the six orders of integrtion freely s long s one keeps the bounds of x, y, nd z correspondingly. If is zsimple, i.e. = {(x, y) (x, y), h 1 (x, y) z h 2 (x, y)}, then h2 (x,y) f(x, y, z) dv = f(x, y, z) dzda. h 1 (x,y) If the region is zsimple, then do zintegrtion first! There re lso ysimple nd xsimple regions. The volume of solid is 1dV. Importnt online questions: 4, 6, 9, Triple integrls in cylindricl nd sphericl coordintes. If = {(r, θ, z) α θ β, 0 r b, h 1 (r, θ) z h 2 (r, θ)}, then β b h2 (r,θ) f(x, y, z) dv = f(r cos θ, r sin θ, z) dzr drdθ. α h 1 (r,θ) emember the extr fctor r; use cylindricl coordintes if the integrting function f contins x 2 + y 2 nd cylindricl is simply polr plus the zvrible. In sphericl coordintes, point P in 3 cn be represented s P = (ρ, ϕ, θ). ρ is the distnce from the origin to P, hence ρ 0; ϕ is the ngle between the positive zxis nd the ry OP, hence, 0 ϕ π; θ is the ngle tht mesures the rottion with respect to the zxis, hence, 0 θ 2π. If is sphericl rectngle, i.e. = {(ρ, ϕ, θ) α θ β, p ϕ q, ρ b}, then β q b f(x, y, z) dv = f(ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ)ρ 2 sin ϕ dρdϕdθ. α p emember the extr fctor ρ 2 sin ϕ; use sphericl coordintes if the integrting function f contins x 2 + y 2 + z 2 or the integrtion region is piece of bll. If is ρsimple in sphericl coordintes, i.e. = {(ρ, ϕ, θ) α θ β, p ϕ q, h 1 (ϕ, θ) ρ h 2 (ϕ, θ)}, then β q h2 (ϕ,θ) f(x, y, z) dv = f(ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ)ρ 2 sin ϕ dρdϕdθ. α p h 1 (ϕ,θ) Importnt online questions: 3, 4, 5, 10, hnge of vribles. If x = x(u, v) nd y = y(u, v), then f(x, y) da = f(x(u, v), y(u, v)) (x, y) (u, v) da,
7 TUY GUIE, ALULU III, 2017 PING 7 in which (x, y) (u, v) = x x u v y y u v is the Jcobin determinnt. Keep in mind the extr fctor is the bsolute vlue of the Jcobin determinnt, nd you need to find the correct integrting domin in (u, v). If you hve u nd v s functions of x nd y, then you need to solve for x nd y in order to compute the Jcobin! Importnt online questions: 3, 6, 7, 12, hpter 15. Vector clculus Prmetric eqution of curve is vector function r(t). (i). : r(t) = x(t), y(t), where t b, defines curve in 2 with n orienttion. (ii). : r(t) = x(t), y(t), z(t), where t b, defines curve in 3 with n orienttion. The rclength of is b r (t) dt. Importnt curves: stright line, circle, ellipse, helix, etc. Two types of vector field: F = f(x, y), g(x, y) in 2 nd F = f(x, y, z), g(x, y, z), h(x, y, z) in 3. We cn discuss two topics for both of them: (iii). Is F conservtive? There re severl equivlent criteri. If it is, then find its potentil function ϕ such tht ϕ = F. (iv). Evlute the line integrls of F in the circultion form long curve : F T ds = F r ds = F d r. Then we discuss the fundmentl theory of clculus in 2 nd in 3, seprtely: (vi). Fundmentl theory of clculus for vector fields in 2 : A closed curve enclosing region in 2, Green s Theorem connects double integrl in the interior nd line integrl on the boundry. ircultion form: F d r = f dx + g dy = (g x f y ) da. In the specil cse, when F is conservtive, g x = f y so F d r = 0. Flux form: F n ds = f dy g dx = (f x + g y ) da. (vii). Prmetric eqution of surfce in 3 : The re of is r(u, v) = x(u, v), y(u, v), z(u, v), where (u, v). 1 d = r u r v da. Importnt surfces: plne, cylinder, cone, sphere, prboloid, etc.
8 8 TUY GUIE, ALULU III, 2017 PING (viii). Fundmentl theory of clculus for vector fields in 3. I: A closed curve s the boundry of surfce in 3, tokes Theorem connects flux surfce integrl of the curl in the interior nd circultion line integrl of the vector field on the boundry: F d r = ( F ) n d. (ix). Fundmentl theory of clculus for vector fields in 3. II: A closed surfce enclosing solid in 3, ivergence Theorem connects triple integrl of the divergence in the interior nd outwrd flux surfce integrl of the vector field on the boundry: F n d = F dv Vector fields. A vector field in 2 : F (x, y) = f(x, y), g(x, y) = f(x, y) i + g(x, y) j. A vector field in 3 : F (x, y, z) = f(x, y, z), g(x, y, z), h(x, y, z) = f(x, y, z) i + g(x, y, z) j + h(x, y, z) k. ecll tht the grdient vector field of ϕ(x, y) is defined s ϕ = ϕ x, ϕ y = ϕ x i + ϕ y j, nd the grdient vector field of ϕ(x, y, z) is defined s ϕ = ϕ x, ϕ y, ϕ z = ϕ x i + ϕ y j + ϕ z k. Importnt online questions: 4, 5, 6, 7, Line integrls. A curve in 2 is vector function r(t) = x(t), y(t) = x(t) i + y(t) j where t b. The bove prmetric form of the curve defines n orienttion long the curve s the direction when t increses from to b. Prmetrize stright line, circle, ellipse, etc. The line integrl of sclr function f(x, y) long curve is b f ds = f(x(t), y(t)) r (t) dt, in which The rclength of is r (t) = x (t), y (t) nd r (t) = (x (t)) 2 + (y (t)) 2. b r (t) dt. The line integrl of vector field F = f(x, y), g(x, y) long curve hs two forms:
9 ircultion form: F T ds = = b Flux form: TUY GUIE, ALULU III, 2017 PING 9 F r ds = f(t), g(t) x (t), y (t) dt = = F n ds = b F d r = b f dx + g dy f(t)x (t) + g(t)y (t) dt. f dy g dx f(t)y (t) g(t)x (t) dt. Importnt online questions: 1, 4, 5, 8, 14, onservtive vector fields. The following sttements re equivlent. A vector field F = f, g is conservtive. F hs potentil function, i.e. F = ϕ for some sclr function ϕ. F d r = ϕ(b) ϕ(a) if is curve from initil point A to terminl point B. Tht is, the line integrl from A to B of conservtive vector field is pth independent. F d r = 0 if is closed curve. f y = g x, i.e. curl F = 0 nd F is irrottionl. Find the potentil function if vector field is conservtive. Importnt online questions: 1, 2, 3, 6, Green s Theorem. Green s Theorem hs two forms. Let be closed curve enclosing region in 2. ircultion form: F d r = f dx + g dy = (g x f y ) da. In the specil cse, when F is conservtive, g x = f y so F d r = 0. Flux form: F n ds = f dy g dx = (f x + g y ) da. The key observtion of Green s Theorem is tht the double integrl in the interior equls the line integrl on the boundry. An ppliction of Green s Theorem is to use circultion to compute the re of : Are () = 1 da = x dy = y dx = 1 (x dy y dx). 2 Importnt online questions: 10, 11, 13, 14, 15.
10 10 TUY GUIE, ALULU III, 2017 PING ivergence nd curl. Let F be vector field in 3 : F (x, y, z) = f(x, y, z), g(x, y, z), h(x, y, z) = f(x, y, z) i + g(x, y, z) j + h(x, y, z) k. Then the divergence of F is div F = f x + g y + h z = x, y, z f, g, h = F, nd the curl of F is curl F i j k = x y z f g h = ( h y g ) ( f i + z z h ) ( g i + x x f ) i = F y. The divergence is sclr nd the curl is vector. If div F = 0, then F is incompressible (i.e. sourcefree); if curl F = 0, then F is irrottionl. The following sttements re equivlent. A vector field F = f, g, h is conservtive. F hs potentil function, i.e. F = ϕ for some sclr function ϕ. F d r = ϕ(b) ϕ(a) if is curve from initil point A to terminl point B. Tht is, the line integrl from A to B of conservtive vector field is pth independent. F d r = 0 if is closed curve. f y = g x, f z = h x, nd h y = g z, i.e. curl F = 0 nd F is irrottionl. Find the potentil function of vector field if it is conservtive. how tht the curl of grdient vector field is zero, i.e. the vector field conservtive: curl ( ϕ) = ( ϕ) = 0. how tht the divergence of curl is zero: Importnt online questions: 4, 5, 10, urfce integrls. A surfce in 3 is vector function div (curl F ) = ( F ) = 0. r(u, v) = x(u, v), y(u, v), z(u, v) = x(u, v) i + y(u, v) j + z(u, v) k, where (u, v). Prmetrize plne, cylinder, cone, sphere, prboloid, etc. The surfce integrl of sclr function f(x, y, z) on surfce is f d = f(x(u, v), y(u, v), z(u, v)) r u r v da, in which The re of is r u (t) = x u, y u, z u nd r v(t) = x v, y v, z v. Are () = r u r v da.
11 TUY GUIE, ALULU III, 2017 PING 11 The surfce integrl of vector field F = x(u, v), y(u, v), z(u, v) on surfce is F d = F n d = F (x(u, v), y(u, v), z(u, v)) ( r u r v) da. Notice tht one hs to choose between r u r v nd r u r v so tht it is consistent with the norml vector n. Importnt online questions: 1, 2, 3, 6, 7, 9, 10, 15, tokes Theorem. Let be surfce in 3 with boundry closed curve nd their orienttions re consistent by the righthnd rule. Then ( ) F d r = ( F ) n d = ( F ) d. The key observtion of tokes Theorem is tht the flux surfce integrl of the curl in the interior equls circultion line integrl of the vector field on the boundry. Also keep in mind tht the curve c : r(t) = x(t), y(t), z(t) here is in 3. A specil cse when F is conservtive, i.e. F = 0, so the surfce integrl is lwys zero nd hence the circultion line integrl of conservtive vector field long ny closed curve is zero. Importnt online questions: 1, 2, 3, 4, 7, 8, ivergence Theorem. Let be closed surfce enclosing solid region in 3. Then ( ) F d = F n d = F dv. The key observtion of ivergence Theorem is tht the triple integrl of the divergence in the interior equls outwrd flux surfce integrl of the vector field on the boundry. A specil cse when F is incompressible, i.e. F = 0, so the triple integrl is lwys zero nd hence the flux surfce integrl of n incompressible vector field on ny closed surfce is zero. Importnt online questions: 1, 3, 5, 8, 9, 10, 12, 14.
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