Ch.11 PDF
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The document contains handwritten mathematical equations and examples related to variational calculus. It seems to be a collection of notes, possibly from a class or textbook. It covers concepts like direct and inverse variation, as well as examples of solving problems related to these concepts.
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變分 ☆ 平 : X (1) 正變 : y= kx 立站 : X...
變分 ☆ 平 : X (1) 正變 : y= kx 立站 : X 3 反變 y 艾 : = 聯率變: e. 避隨 X 正變且 Y 變即. 9kx (波 ) z = 部分變 : e. · y - 部分常數另 部分隨 x 反變 pny g = k. 股 [ 2) example from 總複習 部分變 埗 = [a ] z = 1{ w t 以 - 4 = Ki [ 1 + ) 舞 2= 19 G1 )F + - 4 = K 1 七些 2= k 号 - 8 2 ki + kz O =3 k , fKz _ ② K≈ - 8 - 2 K 1 — ① ①代入 Q 6 = 319. 8 2K K tkz - 3= ② - ↑ - 1 K = 14 1 ① 代入 ② K1 14 代入 ① 4 2 kzfkr = 3 = - Kz = - 8 - 2 (14 ) kz= 1 Kz = - 36 Kz = 1 代入 φ = 4 - 2[1 ) 1 : : - z = 14 2 w 所 - 江 +x. w = 2 ( b) w = ki + kzx w = 5 K t kz 14 [5 P + 4 = 1 2 - 2 = K 1 = 4 - 21 ε 2 - ① = 341 飛 ∆ 十 《 、 二 、十 十 方 一 部分變 (a | k (x + 2 F + *× y = , 21 = k , (5 ) 2 + 3 kz 36 = 36 K + 4 kz ( 212251 k 1 t 3 kz - U d = qkp f kz 些 d ac 1 - - 巨 ② 代入 ① z |= 25 k 1 + 3 (a - qk ) 1 21 25 K 1217 - 27 k 1 - O =~ 2k 1 K = 3 = 3 什入 ② Kz = 9 - 9 (3 ] = - 18 y ( xt2 F -18X = 3 世 ( b) = 3 (x 2 + 4x + 4) 98x ☆ ch. 3 極 值 u y = Y= 3x = - Oa + 12 - k = 4 ( 3 ) (2 ) 比)< - — _ -4 = 9 片 二 大 小 2 3 32. = kx 尷 =X 3 之 X 款x 330 餐 = 挹 百分化 - 和 100 很 36 % 口 一一 二 二 二 怀 Ca ) f(X ) = k xz + kz × , f( 2 ) = 4k ( t 2 kz 12 = 4 Kk t 2 kz 6 = 2 K + k2 , 62 - 2 K 1— ① f ( 4) = 16 k 1 + 4 kz 10 = 16 K , t 4 kz 4 = 419, tkz - ② ① 代入 ② 4 = 4 k1 t 6 - 2 k ) - Ʃ 2 K 1 ki = - 1 ki 1 代入 ① = - Kz 6 t 2 = = 8 f (x ) 2. : = - x + 8x (b 7 ( i ) h= 上) ( ii ) 34 - = - 2 (x - 4 Rt16 少 = 4 x = 9/ l- 極 值 OABCarea = [9 + 1 ) ( 6 — / 2 2G = 250 平 單位 一 𠮩 大 方
