Briggs/Cochran/Gillett Calculus for Scientists and Engineers Multivariable Study Card

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1 Briggs/ochrn/Giett cuus for cientists Engineers Mutivrie tudy rd Vectors the Geometry of pce Three-imension oordinte ystems The distnce etween P 1 (x 1, y 1, z 1 ) P (, y, z ) is ƒp 1 P ƒ = ( - x 1 ) + (y - y 1 ) + (z - z 1 ) The strd eqution for the sphere of rdius r center is Vectors et u = 8u 1, u, u 3 9, v = 8v 1, v, v 3 9, w = 8w 1, w, w 39 e vectors c scr The mgnitude or ength of vector v is = v 1 + v + v 3 The vector with initi point P = (x 1, y 1, z 1 ) termin point Q = (, y, z ) is PQ = 8 - x 1, y - y 1, z - z 1 9 pertions on Vectors (x - ) + ( y - ) + (z - c) = r Addition: u + v = 8u 1 + v 1, u + v, u 3 + v 3 9 cr mutipiction: c u = 8cu 1, cu, cu 3 9 ot Product: u # v = u 1 v 1 + u v + u 3 v 3 (,, c) The nge etween u v is u = cos -1 u # v u Becuse ƒ u ƒƒvƒ cos p we hve tht u v re orthogon or perpendicur = 0, if ony if u # v = 0 Aso, u # u = ƒ u ƒ The orthogon projection of u onto v is ross Product i j k u * v = u 1 u u 3 v 1 v v 3 = (u v 3 - u 3 v )i - (u 1 v 3 - u 3 v 1 )j + (u 1 v - u v 1 )k ross Product Properties proj v u = u # v v # v v ƒ ƒ ƒ ƒ ƒ ƒ 1 u * v = u v sin u = re of preogrm determined y u v 3 u * v = -(v * u), u * u = 0 i * j = k, j * k = i, k * i = j 4 u * v is orthogon to oth u v, therefore u # (u * v) = v # (u * v) = 0 u 1 u u 3 5 u # (v * w) = v 1 v v 3 = voume of preepiped w 1 w determined y u, v, w w 3 Vector-vued Functions Motion in pce et r(t) = f (t) i + g(t) j + h(t)k e vector function Then v(t) = dr is the veocity vector ƒ v (t) ƒ is the speed The cceertion vector is = d v = d r The unit tngent vector is T = v the ength of r(t) from t = to t = is = ines in pce Avector eqution of the ine through P 0 (x 0, y 0, z 0 ) pre to v = 8,, c9 is r(t) = r 0 + t v, or 8x, y, z9 = 8x 0, y 0, z t 8,, c9, for - q 6 t 6 q Prmetric equtions for this ine re x = x 0 + t, y = y 0 + t, z = z 0 + c t, for - q 6 t 6 q Formus urvture: d T k = ` ds ` = 1 v ` dius of urvture: r = 1 k Princip Unit Norm: N = d T/ ƒ d T/ ƒ Tngenti norm scr components of cceertion: = T T + N N, where N = k = ƒ v * ƒ T = d s = v # Unit Binorm Vector: B = T * N = v * ƒ v * ƒ Torsion: t = - d B d s # N = (v * ) # ƒ v * ƒ d T d t ` = v * v 3-8

2 Briggs/ochrn/Giett cuus for cientists Engineers Mutivrie tudy rd Pnes urfces in pce Pnes The pne through P 0 (x 0, y 0, z 0 ) norm to n = i + j + c k hs vector eqution: component eqution: (x - x 0 ) + ( y - y 0 ) + c(z - z 0 ) = 0 component eqution simpified: n P P 0 = 0 x + y + cz = d, where d = x 0 + y 0 + cz 0 Qudric urfces 1 Eipsoid: + y + z c = 1 Eiptic Prooid: z = 3 Eiptic one: + y = z c 4 Hyperooid of one sheet: 5 Hyperooid of two sheets: - - y + z c = 1 6 Hyperoic Prooid: z = - y + y + y - z c = 1 Prti erivtives To compute differentite f (x, y) treting y s constnt, To compute differentite f (x, y) treting x s constnt, Thus, if f (x, y) = y cos xy, = cos xy - xy sin xy Nottion = f x, = f y hin ue Grdient Vector = -y To find d z, strt t z red down ech rnch to t, mutipying derivtives ong the wy Then dd the products x sin xy, The grdient vector (grdient) of f (x, y) t point P 0 (x 0, y 0 ) is the vector f = 0 f 0 x i + 0 f 0 y j t dz dx z = f (x, y) otined y evuting the prti derivtives of f t P 0 dy = dx + y t dy ependent vrie Intermedite vries Independent vrie irection erivtive The direction derivtive of f t (, ) in the direction of unit vector u is Pne Tngent to urfce The tngent pne t the point P 0 (,, c) on the eve surfce F (x, y, z) = 0 is the pne through norm to F P0 Tngent Pne for F (x, y, z) = 0 t P 0 (,, c) F x (P 0 )(x - ) + F y (P 0 )( y - ) + F z (P 0 )(z - c) = 0 Pne Tngent to urfce z = f (x, y) The pne tngent to the surfce z = f (x, y) t the point (,, f (, )) is ineriztion u f (, ) = f (, ) # u z = f x (, )(x - ) + f y (, )( y - ) + f (, ) The ineriztion of function f (x, y) t point (, ) where f is differentie is (x, y) = f x (, )(x - ) + f y (, )( y - ) + f (, ) econd erivtive Test for oc Extrem uppose tht f (x, y) its first second prti derivtives re continuous throughout disk centered t (, ) tht f x (, ) = f y (, ) = 0 Then i f hs oc mximum t (, ) if f xx 6 0 f xx f yy - f xy 7 0 t (, ) ii f hs oc minimum t (, ) if f xx 7 0 f xx f yy - f xy 7 0 t (, ) P 0 iii f hs sdde point t (, ) if f xx f yy - f xy 6 0 t (, ) iv The test is inconcusive t (, ) if f xx f yy - f xy = 0 t (, ) In this cse, we must find some other wy to determine the ehvior of f t (, ) -9

3 Briggs/ochrn/Giett cuus for cientists Engineers Mutivrie tudy rd Prti erivtives (continued) grnge Mutipiers ne onstrint: uppose tht f (x, y, z) g (x, y, z) re differentie To find the oc mximum minimum vues of f suject to the constrint g (x, y, z) = 0, find the vues of x, y, z, tht simutneousy stisfy the equtions f = g Two onstrints: For constrints g (x, y, z) = 0 h (x, y, z) = 0, g h differentie, find the vues of x, y, z,, m tht simutneousy stisfy the equtions g (x, y, z) = 0 f = g + m h, g(x, y, z) = 0, h(x, y, z) = 0 Mutipe Integrs oue Integrs s Voumes When f (x, y) is positive function over region in the xy-pne, we my interpret the doue integr of f over s the voume of the 3-dimension soid region over the xy-pne ounded eow y ove y the surfce z = f (x, y) This voume cn e evuted y computing n iterted integr et f (x, y) e continuous on region 1 If is defined y x, g 1 (x) y g (x), with g 1 g continuous on [, ], then yindric oordintes (r, U, z) Equtions eting ectngur (x, y, z) yindric (r, U, z) oordintes x = r cos u y = r sin u z = z r = + y tn u = y/x g (x) f (x, y) da = g 1 (x) f (x, y) dy dx Tripe Integrs in yindric oordintes If is defined y c y d, h 1 ( y) x h ( y), with h 1 h continuous of [c, d], then d h ( y) f (x, y) da = c h 1 ( y) Are vi oue Integr A = da = dx dy = dy dx Are in Por oordintes f (x, y) dx dy f (r, u, z) dv = f (r, u, z) dz r dr d u pheric oordintes (, W, U) Equtions eting pheric oordintes to rtesin yindric oordintes x = r cos u = r sin w cos u y = r sin u = r sin w sin u z = r cos w r = r sin w r = + y + z = r + z A = r dr du Tripe Integrs in pheric oordintes Tripe Integrs f(x, y, z) dv = h(x) g(x) where = {(x, y, z): x, g(x) y h(x), G(x, y) z H(x, y)} H(x, y) G(x, y) f (x, y, z) dz dy dx, f (r, w, u) dv = f (r, w, u)r sin w dr dw du hnge of Vries Formu for oue Integrs f (x, y) dy dx = f (g (u, v), h (u, v)) J(u, v) da, where x = g(u, v), y = h(u, v) tke region onto region 0(x, y) J(u, v) = 0(u, v) = is the Jcoin determinnt 0u 0u 0v = 0v 0u 0v - 0v 0u -10

4 Briggs/ochrn/Giett cuus for cientists Engineers Mutivrie tudy rd Integrtion in Vector Fieds ine Integrs To integrte continuous function f (x, y, z) over curve : 1 Find smooth prmeteriztion of, r(t) = x (t)i + y (t)j + z (t)k, t Evute the integr s f (x, y, z) ds = f ( x (t), y (t), z (t)) ƒ v(t) ƒ, where v(t) = d r ds = ƒ v(t) ƒ Work The work done y force 8 f, g, h9 over smooth curve r(t) from t = to t = is W = F T ds = F # r (t) = F dr = f dx + g dy + h dz omponent Test for onservtive Fieds et F = 8 f, g, h9 e fied whose component functions hve continuous first prti derivtives on n open, simpy connected region Then F is conservtive if ony if Green s Theorem If is simpe cosed curve is the region encosed y then 0 f = 0g, utwrd fux 0 f = 0h, F # n ds = f dy - g dx = F F F # T ds = f dx + g dy = F F ountercockwise circution 0g = 0h 0 f + 0g da ivergence integr 0g - 0 f da ur integr ircution The circution of F on is F T ds Green s Theorem Are Formu Are of = x dy = - y dx = 1 (x dy - y dx) F F F Fux Fux of F cross = F n ds = f dy - g dx where F = 8 f, g9 F F n is outwrd pointing norm ong onservtive Vector Fied F is conservtive if F = wfor some function w(x, y, z) If F is conservtive, then F # d r A is independent of pth B F # d r = w(b) - w(a) A Aso, B urfce Integrs et r(u, v) = 8 f (u, v), y (u, v), z (u, v)9 u, c v d e prmeteriztion of surfce et The unit vector norm to the surfce is t u * t v = n ƒt u * t v ƒ Are t u = 0r 0u = h 0u, 0u, 0u i The re of the surfce is ƒt u * t v ƒ du dv c d t v = 0r 0v = h 0v, 0v, 0v i F # d r = 0 round every cosed curved in this cse -11

5 Briggs/ochrn/Giett cuus for cientists Engineers Mutivrie tudy rd Integrtion in Vector Fieds (continued) urfce Integr If f (x, y, z) is defined over then the integr of f over is f (x, y, z) d = c d f (x(u, v), y (u, v), z (u, v)) ƒt u * t v ƒ du dv ivergence Theorem The fux of vector fied F cross cosed oriented surfce in the direction of the surfce s outwrd unit norm fied n equs the integr of # F over the region encosed y the surfce: F # n d = # F dv utwrd fux ivergence integr ur of Vector Fied If F = 8 f, g, h9, then cur F = 0h - 0g i + 0 f - 0h j + 0g - 0 f k = *F ivergence of Vector Fied div F = # F = 0 f + 0g + 0h tokes Theorem The circution of vector fied F = 8 f, g, h9 round the oundry of n oriented surfce in the direction countercockwise with respect to the surfce s unit norm vector n equs the integr of *F n over Green s Theorem Its Generiztion to Three imensions ircution form of Green s Theorem: tokes Theorem: Fux form of Green s Theorem: ivergence Theorem: F # d r = F ( *F) # k da F # dr = ( *F) # n d F F # n ds = # F da F F # n d = F dv F # dr = F ountercockwise circution A *FB # n d ur integr -1