F x = 2x λy 2 z 3 = 0 (1) F y = 2y λ2xyz 3 = 0 (2) F z = 2z λ3xy 2 z 2 = 0 (3) F λ = (xy 2 z 3 2) = 0. (4) 2z 3xy 2 z 2. 2x y 2 z 3 = 2y 2xyz 3 = ) 2

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Maryann Martin
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1 0 微甲 07- 班期中考解答和評分標準 5%) Fid the poits o the surfce xy z = tht re closest to the origi d lso the shortest distce betwee the surfce d the origi Solutio Cosider the Lgrge fuctio F x, y, z, λ) = x + y + z λxy z ) The criticl poits of F x, y, z, λ) stisfy F x = x λy z = 0 ) F y = y λxyz = 0 ) F z = z λxy z = 0 ) F λ = xy z ) = 0 4) 8pts) To solve these equtios, sice x 0, y 0, z 0 o the surfce xy z =, from ), ), d ), we get λ = x y z = y xyz = z xy z Secod equlity gives x = y x = ± y Third equlity gives z = y z = ± 6 y We put x = ± d z = ± 6 y ito 4) d get ) ) ) 6 ± y y ± y = y 6 = y = y = ± 4 y 4pts) So we get four criticl poits ) 4, 4, 4, ) 4, 4, 4, ) 4, 4, 4, ) 4, 4, 4 pts) These four criticl poits hve the sme distce to the origi: d = ) ) = + ) + = = 4 pts) Solutio Aother method to solve the system of equtios ) 4) is comprig 6x ), y ), d z ), the we get x = 6y = 4z So we lso fid reltios x = y d z = y Solutio Isted of fidig the mximum or miimum vlues of the fuctio fx, y, z) = x + y + z, we cosider its squre fuctio f x, y, z) = x + y + z becuse they both tti mximum or miimum t the sme plces Sice xy z =, we get y = xz, so the questio reduce to fid the bsolute miimum of the followig fuctio of two vribles: The criticl poits of fx, z) stisfy fx, z) = x + xz + z f x = x x z = x z x z = x z ) x z = 0 f z = xz5 + z = xz4 xz 4 = + xz5 ) xz 4 = 0 From f x = 0, we get xz) = xz = From f z = 0, we get xz 5 = z 4 = z = ± 4 Pge of 5

2 ) If z = 4, the x = 4, d y = xz = y = ± 4 At P = 4, 4, 4 ), dp, O) = = 4 At P = 4, 4, 4 ), dp, O) = = 4 b) If z = 4, the x = 4, d y = xz = y = ± 4 At P = 4, 4, 4 ), dp, O) = = 4 At P 4 = 4, 4, 4 ), dp 4, O) = = 4 These four criticl poits hve the shortest distce betwee the surfce d the origi %) Fid ll the criticl poits of fx, y) = 4 + x + y xy The determie which gives locl mximum or locl miimum or sddle poit fx, y) = 4 + x + y x y = f x, f y ) = x y, y x) f x : %, f y : %) Solve { fx = x y = 0 f y = y x = 0 to obti x, y) = 0, 0),, ) Therefore the criticl poits re x, y) = 0, 0),, ) f x = 0 : %, f y = 0 : %, solvig: %) f x, f y ) = x y, y x) = f xx = 6 x, f xy = f yx =, f yy = 6 y, d Dx, y) = f xxx, y) f xy x, y) f yx x, y) f yy x, y) = 6 x 6 y = 6 x y 9 f xx : %, f xy : %, f yx : %, f yy : %, Dx, y) : %) D0, 0) = 9 < 0 = 0, 0) is sddle poit %) f xx, ) = 6 > 0 d D, ) = 7 > 0 = f, ) is locl miimum %) %) Let the uit vectors u d be respectively the tget directio d the orml directio with positive x-compoets) of the circle x + y x = 0 t the poit, ) Let fx, y) = t ) y x Fid f, ), D u f, ) d D f, ) f x x, y) = + y x f y x, y) = x x +y x y x ) = y x +y poits) poits) fx, y) = y x +y i + x x +y j = f, ) = i + j poit) The equtio x + y x = 0 c be rewritte s x ) + y =, which represets circle cetered t, 0) Let F x, y) = x + y x The orml directio of the circle t x, y) is F = x, y) = x, y) poits) The tget directio of the circle t x, y) is y, x + ) poit) At, ), the orml directio with the positive x-compoet is =, ) poit) Pge of 5

3 At, ), the tget directio with the positive x-compoet is u =, ) poit) D f, ) =, ), ) = poit) D u f, ) =, ), ) = poit) 4 ) 0%) Fid the 4-th degree McLuri polyomils of sec x d of x ) 5% ech) b) 4%) Fid lim sec x x ) x 0 x 4 ) The 4-th degree McLuri polyomil of sec x c be derived from the cosie fuctio: sice cos x = x + x4 4! +, the McLuri polyomil of sec x c be obtied usig log divisio or by comprig the coefficiets i The we hve = cos x) sec x) = x + x4 4! + ) 0 + x + 4 x 4 + ) sec x = + x x4 + Sice oly fiite umber of terms re required, you my lso use the defiitio of McLuri polyomil: fx) = f0) + f 0)! x + f 0)! x + d perform the required differetitio to get the swer) O the other hd, by the biomil expsio x ) = =0 ) x ) = + x + 8 x4 + Grdig policy: 5 poits for ech polyomil Three poits re credited if oly two terms re correct b) From the results i prt ) sec x x ) lim + = lim x x4 + ) + x + 8 x4 + ) 6 = lim x4 + = 6 4 poits) 5 ) %) Fid the rdius of covergece d the itervl of covergece of the power series fx) = ) b) %) Evlute f ) 0) = x 4 l ) By Rtio Test, fx) coverges bsolutely if lim + < 4%) + x lim + )+ 4 = lim + l + ) x = lim 4 l Sice lim l y y l y + = lim y + = y y Hece, the rdius of covergece is 4 x l 4 l + = x 4 < %) Pge of 5

4 for x = 4, fx) = ) l coverges = Sice i) lim l = 0, ii) l > coverges by Altertig Series Test %) l + for x = 4, fx) = l > diverges Sice = = = is p-series of p = diverges %) b) f x) = ) x 4 l = f x) = ) x 4 )%) l = f ) x) = ) x 4 ) )%) l = f ) 0) = ) 4 l = 6 4 l %) < Solutio > f ) 0) %) = coefficiet of x = )! 4 l %) f ) 0) = 6 4 l 6 0%) Suppose tht z = fx, y) is smooth fuctio d let x = uv, d y = v u Express z x, y, f x, f y, f xx, f xy, d f yy u v i terms of The chi rule gives z v x x v + f y y v x u + f y ) = u f x + f y poits) poits) Apply the product rule d gi the chi rule, d lso ote tht u u) =, z u v = u f u x + f ) y x + u f u x + f u y x + u f x x x + u Sice f is smooth fuctio, f xy = f yx Therefore u + f y y x u ) f x v + f y x ) + ) f + x y = uv f x + v f x y u f y x f y + f x x u + f y y u ) f x y v + f y ) z u v = uvf xx + v u)f xy f yy + f x 4 poits) = xf xx + yf xy f yy + f x poit) Becuse f stisfies the coditio of Clirut s theorem, you c first clculte z u d z uv, d the clim tht z uv = z vu This pproch yields the sme solutio s bove) ) 7 0%) Fid the equtio of the tget ple to the elliptic prboloid z c = x + y b t the poit, b, c) Pge 4 of 5

5 Let fx, y, z) = x + y b z c the f x = x, f y = y b, f z = c t, b, c), we hve f x =, f y = b, f z = c Hece the tget ple is x ) + b x b) c x c) = 0 f 對 x 和 y 和 z 的偏微分各分把點帶進去 4 分切平面方程式分 8 ) %) Prmetrize the curve of itersectio of the prbolic cyliders x = y d z = x by settig t = y b) 0%) Fid the uit tget T d the curvture κ t the poit,, ) ) Sice x = y,z = x, d y = t, we c see xt) = t d zt) = t 4 x = t y = t, t R z = t 4 or write s rt) = t, t, t 4 ), t R Although I do ot deductio y poits, you should still wirte dow rge of t b) By formul Tt) = r t) r t), d esy to kow r t) = t,, 4t ), so r ) =,, 4), d r t) = So we get T) =,,4) If you do perfect, you get 5 poits If you compute some error, you will get from to 4 poits, depedig your swer If you use wrog formul, you will get 0 or poit By formul κt) = r t) r t) r t), so we oly eed to compute κ) = r ) r ) r ) First, r t) =, 0, t ), so r ) r ) =,, 4), 0, ) =, 6, ) So r ) r ) = 404 = 0, d r ) = We get the swer is κ = 0 If you do perfect, you get 5 poits If you compute some error, you will get from to 4 poits, depedig your swer If you use wrog formul, you will get 0 or poit Pge 5 of 5

1. (25 points) Use the limit definition of the definite integral and the sum formulas to compute. [1 x + x2

1. (25 points) Use the limit definition of the definite integral and the sum formulas to compute. [1 x + x2 Mth 3, Clculus II Fil Exm Solutios. (5 poits) Use the limit defiitio of the defiite itegrl d the sum formuls to compute 3 x + x. Check your swer by usig the Fudmetl Theorem of Clculus. Solutio: The limit