Re: 青松考古題 2

[steve1012 (steve)]

※ 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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推 ss355227:這題他好像還沒教到拉 不會的不用緊張喔
推 phishingphi:C is decided by the I.C. 才是吧?
推 phishingphi:然後如果題目所給的函數是F 求線積分 從a積到b F dot
→ phishingphi:dr，必須先驗證del cross F = 0 才知道F是保守場之後
→ phishingphi:才能帶起點終點( F(b)-F(a) )
→ steve1012:無旋場就是保守場 所以crossF=0就是保守場
→ steve1012:另外有del的反函數跟cross F=0是等價的