# Chapter 12 Vectors and the Geometry of Space 12.1 Three-dimensional Coordinate systems

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1 hpter 12 Vectors nd the Geometry of Spce 12.1 Three-dimensionl oordinte systems A. Three dimensionl Rectngulr oordinte Sydstem: The rtesin product where (x, y, z) isclled ordered triple. B. istnce: R 3 = R R R = {(x, y, z) : x, y, z R}, The distnce P 1 P 2 between two points P 1 = (x 1, y 1, z 1 ) nd P 2 = (x 2, y 2, z 2 ) is. Sphere: P 1 P 2 = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 + (z 2 z 1 ) 2. An eqution of sphere with center (h, k, l) nd rdius r is 12.2 Vectors (x h) 2 + (y k) 2 + (z l) 2 = r 2. A. Vector: quntity tht hs both of mgnitude nd direction. B. (Re)presenttion nd Nottion:, < 1, 2, 3 >, AB. efinition 1. A two dimensionl vector is n ordered pir =< 1, 2 > with 1 nd 2 rel numbers. A three dimensionl vector is n ordered triple =< 1, 2, 3 > with 1, 2 nd 3 rel numbers.. Mgnitude: Let =< 1, 2, 3 >. Then, the mgnitude of is = Multipliction of vector by sclr: Let =< 1, 2, 3 > nd c be sclr (rel number). Then, c =< c 1, c 2, c 3 > nd c = c. E. Vector ddition/difference : Let =< 1, 2, 3 > nd b =< b 1, b 2, b 3 >. Then, + b =< 1 + b 1, 2 + b 2, 3 + b 3 >,

2 b =< 1 b 1, 2 b 2, 3 b 3 > F. Stndrd Bsis vectors : In R 2, In R 3, i =< 1, 0 >, j =< 0, 1 >. i =< 1, 0, 0 >, j =< 0, 1, 0 >, k =< 0, 0, 1 >. If =< 1, 2, 3 >, then = 1 i + 2 j + 3 k. G. Unit Vector : A vector whose length is 1. If 0, then the unit vector tht hs the sme direction s is u = The ot Product efinition 2. If =< 1, 2, 3 > nd b =< b 1, b 2, b 3 >, the dot product of nd b is the number b given by b = 1 b b b 3. Note 1: A result of the dot product of two vectors is sclr (not vector) Note 2: The dot product is sometimes clled inner product or sclr product. A. Properties of the dot product (1) = 2 (2) b = b (3) ( b + c ) = b + c (4) (c ) b = c( b ) = (c b ) (5) 0 = 0 Theorem 1. Let =< 1, 2, 3 > nd b =< b 1, b 2, b 3 >. If θ is the ngle between the vectors nd b, then b = b cos θ. 2

3 orollry 1. Let =< 1, 2, 3 > nd b =< b 1, b 2, b 3 >. If θ is the ngle between the vectors nd b, then b cos θ = b. B. irection Angles nd irection osine: Let the ngles betwees nd x-xis, y-xis, nd z-xis re α, β, nd γ respectively. Then, if =< 1, 2, 3 >, Then,. Projection : cos α = 1, cos β = 2 =< cos α, cos β, cos γ >. cos γ = Sclr Projection : component of b long.2. Vector Projection: comp b = b cos θ = b proj b = comp b = ( b. ) The ross Product efinition 3. If =< 1, 2, 3 > nd b =< b 1, b 2, b 3 >, the cross product of nd b is the vector b given by b =< 2 b 3 3 b 2, 3 b 1 1 b 3, 1 b 2 2 b 1 >. Theorem 2. The cross product b is orthogonl to both of =< 1, 2, 3 > nd b =< b1, b 2, b 3 >. Note 1: A result of the cross product of two vectors is vector (Not sclr). So, the cross product is sometimes clled vector product. 3

4 Theorem 3. If θ, 0 θ π, is the ngle between =< 1, 2, 3 > nd b =< b 1, b 2, b 3 >, then b = b sin θ. Note 2: the length of the cross product b is equl to the re of the prllelogrm determined by nd b. A. Sclr Triple Product : The volume of the prllelopiped determined by the vectors, b, nd c is the mgnitude of the sclr triple product ( b c ) Equtions of Lines nd Plnes A. Eqution of line L : A line in spce is determined by point P 0 (x 0, y 0, z 0 ) vector n tht is prllel to the line. Let v =<, b, c >, r =< x, y, z >, r 0 =< x 0, y 0, z 0 >, nd t is sclr. A.1. Vector Eqution : r = r 0 + t v. A.2. Prmetric Eqution : x = x 0 + t y = y 0 + bt z = z 0 + ct. A.3. Symmetric Eqution : x x 0 = y y 0 b = z z 0. c B. Eqution of Plne : A plne in spce is determined by point P 0 (x 0, y 0, z 0 ) vector n (norml vector) tht is orthogonl to the plne. Let r =< x, y, z >, r 0 =< x 0, y 0, z 0 >, nd n =<, b, c >. B.1. Vector Eqution of Plne: n ( r r 0 ) = 0. B.2.. Sclr Eqution of plne : (x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0 (or x + by + cz + d = 0 with d = x 0 by 0 cz 0 ). 4

5 . istnce from point P 1 (x 1, y 1, z 1 ) to the plnce x + by + cz + d = 0: = x 1 + by 1 + cz 1 + d 2 + b 2 + c /7 ylinder nd ylindricl oordintes Identify nd sketch the surfces (1) x 2 + y 2 = 1 (2) y 2 + z 2 = 1. To convert from cylindricl to rectngulr oordinte, x = r cos θ y = r sin θ z = z. To convert from rectngulr to cylindricl oordinte, r 2 = x 2 + y 2 tn θ = y x z = z. hpter 13 Vector Functions 13.1 Vector Functions nd Spce urves A. Vector Function: A vector (vlued) function (e.g, in R 3 ) is of the form r (t) =< f(t), g(t), h(t) >, where ll the component functions f, g, nd h re rel vlued function. B. Limit of vector function r : If r (t) =< f(t), g(t), h(t) >, then lim r (t) =< lim f(t), lim g(t), lim h(t) > t t t t provided the limits of f(t),g(t), nd h(t) exist. In prticulr, vector function r is continuous t t = lim r (t) = r (). t. Spce urve: Suppose tht f, g, nd h re continuous functions on n intervl I. Let = {(x, y, z) : x = f(t) y = g(t) z = h(t)}, where t vries through the intervl I, is clled spce curve. The equtions x = f(t), y = g(t), z = h(t) re clled prmetric equtions of nd t is clled prmeter. 5

6 13.2 erivtives nd Integrls of Vector Functions A. erivtive : The derivtive of vector (vlued) function r is defined by if the limit exists. d r dt = r (t + h) r (t) r (t) = lim h 0 h The vector r (t) is clled the tngent vector of r, nd it unit tngent vector is given by r (t) T(t) = r (t). Theorem 4. If r (t) =< f(t), g(t), h(t) >, where f, g, nd h re differentible functions, then r (t) =< f (t), g (t), h (t) >. Theorem 5. Suppose u nd v re differentible vector functions, c is sclr, nd f is rel vlued function. Then, (1) d dt [ u (t) + v (t)] = u (t) + v (t) (2) d dt [c u (t)] = c u (t) (3) d dt [f(t) u (t)] = f (t) u (t) + f(t) u (t) (4) d dt [ u (t) v (t)] = u (t) v (t) + u (t) v (t) (5) d dt [ u (t) v (t)] = u (t) v (t) + u (t) v (t) (6) d dt [ u (f(t))] = f (t) u (f(t)) (hin Rule) See Exmple 1-5 B. Integrls : If r (t) =< f(t), g(t), h(t) >, where f, g, nd h re integrble in [, b], then the definite integrl if the vector function r (t) cn be defined by b ( b ) ( i b ) ( j b k r (t)dt = f(t)dt + g(t)dt + h(t)dt). See Exmple 6 6

7 13.3 Arc Length nd urvture A. Arc Length : Let t b, nd let r (t) =< f(t), g(t), h(t) > where f, g, nd h re continuous on I. Then, the length of the spce curve (rc length) from t = to t = b is defined by b L = r (t) 2 dt = = b b [f (t)] 2 + [g (t)] 2 + [h (t)] 2 dt (dx ) 2 + dt ( ) 2 dy + dt ( ) 2 dz dt dt See Exmple Motion in Spce :Velocity nd Accelertion A. Velocity vector : Suppose prticle moves through spce so tht its position vector t time t is r (t). The the velocity vector t time t is defined by v (t) = lim h 0 r (t + h) r (t) h = r (t). The velocity vector v (t) is lso the tngent vector nd points in the direction of the tngent line. Further, the speed of the prticle t time t is v (t) = r (t). B. Accelertion vector : The ccelertion of the prticle t time t is (t) = v (t) = r (t). See Exmples Newton s Second Lw of Motion : If, t ny time t, force F(t) cts on n object of mss m producing n ccelertion (t), then See Exmples 4 nd 5. F(t) = m (t). 7

8 hpter 14 Prticl derivtives 14.1 Functions of Severl Vribles A. Functions of two vrilbes: efinition 4. A function f of two vribles is rule of the form f : (x, y) z = f(x, y). Here the set is the domin of f nd its rnge is the set {f(x, y) : (x, y) }. See Exmples 1, 4 B. Grph : Let f is function of two vribles with domin. Then, the grph of f is See Exmples 6, Limits nd ontinuity Let s think bout the two limits A. Limit : {(x, y, z) R 3 : z = f(x, y), lim lim x 2 y 2 nd lim x 0 y 0 x 2 + y 2 (x, y) }. x 2 y 2 x 0 x 2 + y. 2 y 0 lim efinition 5. Let f be function of two vribles with domin nd (, b) closure(). We sy tht the limit of f(x, y) s (x, y) (, b) is L, i.e., lim f : (x, y) = L (x,y) (,b) if for ny number ɛ > 0 there is number δ > 0 such tht f(x, y) L < ɛ whenever (x, y) (, b) < δ nd (x, y). Remrk: Let f(x, y) L 1 s (x, y) (, b) long pth 1 nd f(x, y) L 2 s (x, y) (, b) long pth 2. Thn, if L 1 L 2, the limit lim (x,y) (,b) f : (x, y) does not exist. See Exmples

9 B. ontinuity : efinition 6. Let f be function of two vribles with domin nd (, b) closure(). The function f(x, y) is clled continuous t (, b) if lim f : (x, y) = f(, b). (x,y) (,b) We sy f(x, y) is continuous on if f is continuous t every points (, b) in Prtil erivtives A. efinition : Let f be function of two vribles. Its prtil derivtives re functions f x nd f y defined by B. Nottions: If z = f(x, y), f x (x, y) = f(x + h, y) f(x, y) lim h 0 h f y (x, y) = f(x, y + h) f(x, y) lim h 0 h f x (x, y) = f x = f x = z f(x, y) = x x = xf f y (x, y) = f y = f y = z f(x, y) = y x = zf See Exmples 1-5. Theorem 6. Let f be function of two vribles nd defined on. Let (, b). Then, if f xy nd f yx re both continuous on, f xy (, b) = f yx (, b) Tngent Plnes nd Liner Approximtion A. efinition : Let f be function of two vribles, nd ssume tht f x nd f y re continuous. An eqution of tngent plne to the surfce z = f(x, y) t P (x 0, y 0, z 0 ) is z z 0 = f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ). efine L(x, y) = f(x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ). Then L(x, y) is clled lineriztion of f t (x 0, y 0 ). Also, the pproximtion f(x, y) L(x, y) is clled the liner pproximtion nd the thngent plne pproximtion. 9

10 efinition 7. Let z = f(x, y) nd z = f(x 0 + x, y 0 + y). Then the function f is differentible t (x 0, y 0 ) if z cn be written z = f x (x 0, y 0 ) x + f y (x 0, y 0 ) y + ɛ 1 δx + ɛ 2 δxy, where ɛ 1, ɛ 2 0 s δx, δy 0. Theorem 7. Asume tht f x nd f y exist ner (x 0, y 0 ) nd re continuous t (x 0, y 0 ). Then f is differentible t (x 0, y 0 ). Totl differentil: The differentil or totl differentil is is defined by dz = f x (x, y)dx + f y (x, y) = z z dx + x y dy. See Exmples 1,2, The hin Rule The hin Rule (se 1): Assume tht z = f(x, y) is differentible function of x nd y, where x = g(t) nd y = h(t) re both differentible of t. Then dz dt = f dx x dt + f dy y dt. The hin Rule (se 2): Assume tht z = f(x, y) is differentible function of x nd y, where x = g(s, t) nd y = h(s, t) re both differentible functions of s nd t. Then z s = z dx x ds + f dy dy ds z t = z x See Exmples 1,3, 5( For Generl Version) z x = dx dt + f dy Implicit ifferentition : Assume tht z = f(x, y) is given implicitly s function of the form F (x, y, z) = 0. If F nd f re differentible nd F 0, then z F F x z y See Exmples 9 F z dy dt. y = F z 10

11 14.6 irectionl erivtives nd the Grdient Vector A. efinition : Let f be function of two vribles. The directionl derivtives of f(x 0, y 0 ) in the direction of unit vector u =<, b > is f(x 0 + h, y + hb) f(x 0, y 0 ) u f(x 0, y 0 ) = lim h 0 h if the limit exists. In the cse of three vrible function, we cn define the directionl derivtives in similr mnner. Theorem 8. Let f be differentilbe function of x nd y. Then f hs directionl derivtives in the direction of unit vector u =<, b > nd u f(x, y) = f x (x, y) + f y (x, y)b. The Grdient Vector : Let f be function of severl (sy three) vribles. The Grdient of f is the vector function f fdefined by f(x, y, z) =< f x (x, y, z), f y (x, y, z), f z (x, y, z) >= f x Note tht for ny u =<, b, c >, u f(x, y, z) = f(x, y, z) u. See Exmples 2, 3, Mximum nd Minimum Vlues i + f x f j k z efinition 8. Let f be function of two vribles. Then, f(, b) is clled locl mximum vlue if f(, b) f(x, y) when (x, y) is ner (, b). Also, f(, b) is clled locl minimum vlue if f(, b) f(x, y) when (x, y) is ner (, b). Theorem 9. If f hs locl mximum or minimum vlue t (, b) nd f x nd f y exist, then f x (, b) = 0 nd f y (, b) = 0. A point (, b) is clled criticl point of f if f x ), b) = 0 nd f y (, b) = 0. Second erivtives Test: Assume tht the second prtil derivtives of f re continuous on disk with center (, b), nd ssume tht f x (, b) = 0 nd f y (, b) = 0. Let = f xx (, b)f yy (, b) [f xy (, b)] 2. () If > 0 nd f xx (, b) > 0, then f(, b) is locl minimum. (b) If > 0 nd f xx (, b) < 0, then f(, b) is locl mximum. 11

12 (c) If < 0, then f(, b) is not locl mximum or minimum. (the point (,b) is clled sddle point of f). See Exmples 1, 2, 3, 6. Theorem 10. If f is continuous on closed, bounded set in R 2, then f ttins n bsolute mximum vlue f(x 1, y 1 ) nd n bsolute minimum vlue f(x 2, y 2 ) t some points (x 1, y 1 ) nd (x 2, y 2 ) in. To find n bsolute mximum nd minimum vlues of f on closed, bounded set : (1) Find the vlues of f t the criticl points of f in. (2) Find the extreme vlues of f on the boundry of. (3) The lrgest of the vlues from steps 1 nd 2 is the bsolute mximum vlue; The smllest of the vlues is the bsolute minimum vlue. See Exmple Lgrnge Multipliers Method of Lgrnge Multifiers : In order to find the mximum nd minimum vlues of f(x, y, z) subject to the constrint g(, y, z) = k () Find ll vlues of x, y, z, nd λ such tht f(x, y, z) = λ g(x, y, z) g(, y, z) = k (b) Evlute f t ll the points (x, y, z) tht results from step (). The lrgest of the vlues is the bsolute mximum vlue; The smllest is the bsolute minimum vlue of f. See Exmples 2, 3 12

13 hpter 15 Multiple Integrls 15.1 ouble Integrls over Rectngles efinition 9. The double integrl of f over R = [, b] [c, d] is f(x, y)da = f(x i, y j ) A (1) if the limit exists, where (x i, y j ) is in R lim m,n i=1 j=1 R ij = [x i 1, x i ] [y j 1, y j ]. Here the right-hnd side of (1) is clled s double Riemnn sum. If f(x, y) 0, the volume V of the solid tht lies bove the rectngle R nd below the surfce z = f(x, y) is V = f(x, y)da Iterted Integrls Theorem 11. (Fubini) Let f be continuous function on R = [, b] [c, d]. b d d b f(x, y)da = f(x, y)dydx = f(x, y)dxdy. R c R The two integrls in the right-hnd side of the bove identity re clled iterted integrls. More generlly, this theorem is true if f is bounded on R, f is discontinuous only on finite number if snmooth curves, nd the iterted integrls exist. Specil ses : If f(x, y) = g(x)h(y) on R = [, b] [c, d], b d [ d ] [ b f(x, y)da = f(x, y)dydx = h(y)dy R See Exmples ouble Integrls over Generl Regions Type I : Let f be continuous on type I region such tht then c = {(x, y) x b, R f(x, y)da = b g2 (x) g 1 (x) 13 c c g 1 (x) y g 2 (x)} f(x, y)dydx. ] g(x)dx.

14 Type II : Let f be continuous on type II region such tht then See Exmples 1-5. = {(x, y) c y d, R f(x, y)da = d h2 (x) c h 1 (x) x h 2 (x)} h 1 (x) f(x, y)dxdy. Properties of ouble Integrls : Assume tht ll of the following integrls exist. Then, (1) [f(x, y) + g(x, y)]da = f(x, y)da + [g(x, y)]da (2) cf(x, y)da = c [f(x, y)]da (3) f(x, y)da [g(x, y)]da, if f(x, y) g(x, y) (4) 1dA = A() (4) f(x, y)da [g(x, y)]da, + 1 [g(x, y)]da, 2 if = 1 2. Here 1 nd 2 don t overlp except (perhps) on the boundry. Also, if m f(x, y) M for ll (x, y), then ma() f(x, y)da MA() ouble Integrls over Polr oordintes hnge to Polr oordintes in ouble Integrl : Let f be continuous on polr rectngle R = {(r, θ) 0 r b, α θ β} then See Exmples 1,2. R 15.7 Triple Integrls f(x, y)da = β b α f(r cos θ, r sin θ)rdrdθ). 14

15 Theorem 12. (Fubini s Theorem for Triple Integrls) Let f be continuous function on B = [, b] [c, d] [r, s]. s d b d b f(x, y, z)dv = f(x, y, z)dxdydz = f(x, y)dxdy. B r c c Triple Integrls over Generl Regions: Type I : Let f be continuous on region (type I or type II in double integrl) such tht E = {(x, y, z) (x, y) u 1 (x, y) z u 2 (x, y)}, then [ ] u2 (x,y) f(x, y, z)dv = f(x, y, z)dz da. E u 1 (x,y) Type II : Let f be continuous on region (type I or type II in double integrl) such tht E = {(x, y, z) (y, z) u 1 (y, z) z u 2 (y, z)}, then [ ] u2 (y,z) f(x, y, z)dv = f(x, y, z)dx da. See Exmples 1-3. E u 1 (y,z) 15.8 Triple Integrls in ylindricl nd Sphericl oordintes Formul for triple integrtion in cylindricl coordintes: u2 (r cos θ,r sin θ) f(x, y, z)dv = f(r cos θ, r sin θ, z)rdzdrdθ. See Exmples 1,2. E u 1 (r cos θ,r sin θ) 15

16 hpter 16 Vector lculus 16.1 Vector Fields efinition 10. Let E be subset of R 3. A vector field on R 3 is function F tht ssigns to ech (x, y, z) E three-dimensionl vector F(x, y, z). We cn write F s follows: F(x, y, z) = P (x, y, z) i + Q(x, y, z) j + R(x, y, z) k. Grdient Fields: If f is sclr function of three (or two) vribles, its grdient is vector field on R 3 given by See Exmples 1, 2, Line Integrls f(x, y, z) = f x (x, y, z) i + f y (x, y, z) j + f z (x, y, z) k. efinition 11. Let be smooth curve given by the prmetric eqution x = x(t) y = y(t), t b. If f is defined on the curve, then the line integrl of f long is defined by b (dx ) 2 ( ) 2 dy f(x, y)ds = f(x(t), y(t)) + dt. dt dt Remrk: If is piecewise-smooth curve, tht is, is finite union of smooth curves 1, n, then f(x, y)ds = f(x, y)ds f(x, y)ds. n Line integrl of f long with respect to x nd y: b f(x, y)dx = f(x(t), y(t))x (t)dt See Exmples 1, 2, 4. f(x, y)dy = b f(x(t), y(t))y (t)dt Line integrls in Spce: Suppose tht is smooth curve given by the prmetric eqution x = x(t) y = y(t) z = z(t), t b. 16

17 If f is defined on the curve, then the line integrl of f long is defined by b (dx ) 2 ( ) 2 ( ) 2 dy dz f(x, y)ds = f(x(t), y(t), z(t)) + + dt. dt dt dt ompct Nottion : f(x, y, z)ds = b f( r (t)) r (t) dt. See Exmples 5, 6. Line integrls of Vector Fields: Let F be continuous vector field defined on smooth curve given by vector function r (t), t b. Then, the line integrl of F long is F d b r = F( r (t)) r (t)dt = F T ds, where T (x, y, z) is the unit tngent vector t the point (x, y, z). See Exmples 7, The Fundmentl Theorem for Line Integrls Theorem 13. Let be smooth curve given by the vector function r (t), t b. Let f be continuous function nd its f is continuous on. Then, f d r = f( r (b)) f( r ()). Note: We cn evlute f d r by knowing the vlue of f t the end of points of. efinition 12. A vector field F is clled conservtive vector field if there is sclr function f such tht F = f. Here f is clled potentil function of F. Note: Theorem 14. Line integrls of conservtive vector fields re independent of pth. every closed pth in. f d r is independent of pth in domin iff f d r = 0 for 17

18 Theorem 15. Let F = P i + Q j be conservtive vector field, where ll the prtil derivtives re continuous. Then, P y = Q, in. x Theorem 16. Let F = P i + Q j be vector fields on n open simply-connected region. Supppose tht ll the prtil derivtives re continuous nd P y = Q in. x Then F is conservtive Green s Theorem Theorem 17. Let be positively oriented, piecewise-smooth, simple closed curve in the plne nd let be the region bounded by. Supppose tht ll the prtil derivtives of P nd Q re continuous on n open region contins, then P dx + Qdy = ( Q x P y ) da Appliction: Line The formuls to find the re of : A = xdy = ydx = 1 xdy ydx. 2 See Exmples 1,2, url nd ivergence url: If F = P i + Q j + R k is vector field on R 3 nd prtil derivtives of P, Q, nd R ll exist, then the curl of F is the vector field defined by curl F = ( F R = y Q ) ( i R z x P ) ( j Q + z x P ) k y Theorem 18. If f is function of three vribles tht hs continuous second-order prtil derivtives, then curl ( f) = 0. Theorem 19. If F is vector field defined on ll of R 3 whose component functions hve continuous prtil derivtives nd curl F = 0, then F is conservtive vector field. 18

19 See Exmples 1,2,3 ivergence: If F = P i + Q j + R k is vector field on R 3 nd prtil derivtives of P, Q, nd R ll exist, then the divergence of F is the function of three vribles defined by divf = F = P x + Q y + R z Theorem 20. If F = P i + Q j + R k is vector field defined on ll of R 3 nd P, Q, nd R hve continuous second-order prtil derivtives, then See Exmples 4,5 div curl F = 0. 19

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