※ 引述《ss355227 (前祐)》之銘言： : ※ [本文轉錄自 NTU-Exam 看板 #1AdpwufY ] : 作者: iwantowaylaw (我要成為太鼓達人) 看板: NTU-Exam : 標題: [試題] : 時間: Thu Sep 3 11:57:43 2009 : 課程名稱︰ 暑修微積分甲下第二次期中考 : 課程性質︰ : 課程教師︰ 周青松 : 開課學院： : 開課系所︰ : 考試日期（年月日）︰2009/9/3 : 考試時限（分鐘）：2小時 : 是否需發放獎勵金：是 : （如未明確表示，則不予發放） : ------------------------------- : 題目中好像有些筆誤 : 我先不改喔 大家應該知道~ : ------------------------------- : 試題 : : Make sure to give sufficient reason in each problem or you will NOT get any : credit for your answer. : A. : (a) : Set a = a1i + a2j + a3k : and b = b1i + b2j + b3k : Show that a x b = (a2b3-a3b2)-i(a1b3-a3b1)j+(a1b2-a2b1)k : 無言不知道要證啥 : (b) : Set a = a1i + a2j + a3k : b = b1i + b2j + b3k : c = c1i + c2j + c3k : | a1 a2 a3 | : Show that (a x b) ‧ c = | b1 b2 b3 | : | c1 c2 c3 | : p.668 Example 3 : B. : (a)Find f(t) given that f'(t) = 2costi-tsint^2+2tk and f(0)=i+3k : 應該不難吧 我算 f(t) = 2sint + 1 i + 1/2 cost^2 -1/2 j + t^2 +3 k : (b)Find f(t) given that f'(t) = ti-t(1+t^2)^(-1/2)+te^tk and f(0)=i+2j+3k : 重複 : C. : (a) : Show that if γ is a differentiable vector function of t ,then the function : r = ║γ║ is differentiable at where it is not zero and : dγ dr : γ‧ ── = r ── : dt dt : p.703 (14.2.3) 證明在p.704上方 : (b) : Show that for each integer n and all γ≠0 we have ▽r^n = nr^(n-2)γ : 未知 : D. : (xy)^2 : Set g(x,y) = ─── if (x,y) ≠ (0,0) : x^4+y^4 : 0 if (x,y) = (0,0) : dg dg : (a)Show that ─(0,0) and ─(0,0) both exist and evaluate their values : dx dy : (b)Show that lim g(x,y) doesn't exist : (x,y)→(0,0) : p.785 第29題 要練習過吧 : E. : (a) : Find all function with gradient yzi + (xz+2yz)j + (xy+y^2)k : 未知 這個很重要 在後面line integral 的時候會用到 物理裡面是保守場的條件 首先根據Gradient的定義 set there are some functions F satisfy the condition then *[1;33;40m del F = yz i + (xz+2yz) j + (xy+y^2)k*[m so we have partial F/ partial x = yz ----(1) partial F/ partial y = xz+2yz ----(2) partial F/ partial z = xy+y^2 ----(3) (1) => integrate both side with respect of x F = xyz + G(y,z) Fy = (2) = xz + Gy(y) =xz+2yz so Gy(y,z) = 2yz => G(y,z) = zy^2+c c is a constant Fz = (3) = xy+Gz(z) = xz+y^2 indeed so the general solutions of such partial differentiation equation is F = xyz + zy^2 + C the C is decided by the boundary condition btw f is a conservative force F is the potentail on the field and line integral b ∫f。dr = F(b) - F(a) (Fundamental theorem of Line Integrals) a : (b) : 2xy : Set f(x,y) = ─── if (x,y) ≠ (0,0) : x^2+y^2 : 0 if (x,y) = (0,0) : Show that f is not differentiable at (0,0) : 應該是從不同方向逼近原點的值不同 所以不連續 所以不能微分吧!? : ------------------------------------------ : 感覺這篇過多非習題題目...... : ------------------------------------------ --

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