※ 引述《stan999950 (鵬鵬鵬鵬鵬鵬鵬鵬鵬鵬鵬)》之銘言：
: ※ [本文轉錄自 NTU-Exam 看板 #1CXl86h6 ]
: 作者: sr333444 (欸啥) 看板: NTU-Exam
: 標題: [試題] 98暑修 周青松 微積分甲下 期末考
: 時間: Wed Sep 8 10:16:34 2010
: 課程名稱︰微積分甲下
: 課程性質︰暑修
: 課程教師︰周清松
: 開課學院：理學院
: 開課系所︰數學系
: 考試日期（年月日）︰2010/9/9
: 考試時限（分鐘）：120分鐘
: 是否需發放獎勵金：是
: （如未明確表示，則不予發放）
: 試題 :
: It's necessary to explain all the reasons in detail and show all of your
: work on the answer sheet; Or you will NOT get any credits. If you used any
: theorems in textbook or proved in class, state it carefully and explicitly.
: 1.(a)For each integer n and r≠0, we have ▽r^n = nr^(n-2)r. Here
: r=∥r∥ and r = xi+yj+zk. Note that if n is positive and even,
: the result holds at r=0.
這題...直接微應該就可以得到
至於 hold at r=0 因為r沒有在分母了阿(n>=2) 當然可以等於0
~~~詳細的我不知道怎麼寫~~~
~~~李維峻help~~~~me
: (b)Assume that ▽f(x) exists. Prove that, for each integer n, we have
: n n-1
: ▽f (x)=nf (x)▽f(x).
也是直接微搂0.0
----------------------------不過不考證明這不會出吧XDDD------------
: 2.(a)Find the directional derivative of f(x,y)=ln(x^2+y^2) at P(0,1)
: in the direction of 8i+j.
▽f(x,y)= (2x / x^2+y^2)i + (2y / x^2+y^2)j
▽f(0,1)= 0 i + 2j
▽f(0,1)‧(8,1) = 2
......
: (b)Find the directional derivative of f(x,y)= xe^(y^2-z^2) at (1,2,-2)
: in the direction of increasing t along the path
: r(t)= ti+2cos(t-1)j-2e^(t-1)k
▽f(x,y,z)= ( e^(y^2-z^2) )i + ( 2yxe^(y^2-z^2) )j + ( -2zxe^(y^2-z^2) )k
▽f(1,2,-2)= (0,4,4)
t=1 r(t)=(1,2,-2)
v(t)= i + ( -2sin(t-1) )j + ( -2e^(t-1) )k
u = v(1) / |v(1)| =(1,0,-2)/ 5^0.5
▽f(1,2,-2)‧u = -8 / 5^0.5
乾好難算 有沒有算錯啊
: 3.(a)Use the chain rule to find the rate of change of f(x,y,z)=x^2y+zcosx
: with respect to t along the twisted cubic r(t)=ti+t^2j+t^3k
意思就x=t y=t^2 z=t^3
反正df(x,y,z)/dt = partial f/partial x * dx/dt i+ y....j + z....k
答案....
df(x,y,z)/dt = (2xy-zsinx)*1 i + x^2 * 2t j + cosx * 3t^2 k
最後算 |df(x,y,z)/dt| = .......
乾是這樣嗎 我都囧了
: (b)Find the rate of change of f(x,y,z)=ln(x^2+y^2+z^2) with respect to
: t along the twisted cubic r(t)=sinti+costj+e^(2t)k
好煩=.=
: 4.(a)Calculate by double integration the area of the bounded region determined
: by the curves x^2=4y, 2y-x-4=0.
重點就是要找出範圍 然後對dxdy積分就好
可以解出交點 x= 6 or -2 大概知道圖後 -2 <= x <= 6
x^2 /4 <= y <= 2 + x/2
一組實數，一組就有未知數這樣
積分吧 範圍就照那樣寫 先積y 有未知數
= 積分-2~6 (2 + x/2 - x^2/4) dx
= 2x + x^2 / 4 - x^3 / 12 |-2~6
= 16/3
乾好醜 我累了
: (b)Calculate the volume within the cylinder x^2+y^2=b^2 between the planes
: y+z=a and z=0 given that a>=b>0.
: 5.(a)Use triple integration to find the volume of the tetrahedron T bounded
: by x+y+z=1 in the first octant.
: (Hint: 0≦z≦1-x-y, 0≦y≦1-x, 0≦x≦1)
: (b)Calculate the mass of the solid 0≦x≦a, 0≦y≦b, 0≦z≦c, with the
: density funtion ρ(x,y,z)=xyz.
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 114.45.138.150