推 RogerClemens:δ 04/21 15:09
課程名稱︰微積分甲下
課程性質︰必修
課程教師︰朱樺 莊正良 (兩班用同一張考卷)
開課系所︰土木工程學系
考試時間︰03/30/06
註: 偏微分符號 partial 我不會打，以下以希臘字母 "α" 代替
試題 :
1. Find lim(x,y)→(0,0) xy^3 / (x^2 + 4y^6)
Ans:
(1) alone x-axis let y=0
lim(x,y)→(0,0) xy^3 / (x^2 + 4y^6) = lim(x,0)→(0,0) x*0^3/(x^2 + 4*0^6)
=lim x→0 [0/(x^2 + 0)] since x does not equil to 0, the limit=0
(2) alone x=y^3
lim(y^3,y)→(0,0) y^6/(y^6 + 4y^6) = lim(y→0) 1/1+4 =1/5
since 0 ≠ 1/5 , the limit does not exist
2. Let z = arctan(y/x). Find αz/αy, α^2z/αyαx.
Ans:
(1)αz/αy = 1/[1+(y/x)^2] * 1/x = x/(x^2+y^2)
(2) α^2z/αyαx = α/αy (αz/αx) = α/αy * {1/[1+(y/x)^2] * (-y)/x^2}
= α/αy [(-y)/(x^2+y^2)] = [-(x^2+y^2) + y*2y] / (x^2+y^2)^2
= (y^2-x^2)/(x^2+y^2)^2
3. Find the tangent plane of the surface x^2+y^3+2z-4x^3yz^3 = 0 at the
point (1,1,1).
Ans:
Let F(x,y,z) = x^2+y^3+2z-4x^3yz^3
▽F = < αF/αx,αF/αy,αF/αz >
= < 2x-12x^2yz^3, 3y^2-4x^3z^3, 2-3z^2*4x^3y >
substitute the point (1,1,1)
▽F = <-10,-1,-10 > = < 10,1,10 >
plane : 10(x-1) + (y-1) + 10(x-1) = 0
4. Let the function y = g(x) satisfy f(x^2+3,y) = 0 Express dy/dx in terms
of fx, fy.
Ans:
fx(x^2+3,y) * d(x^2+3)/dx + fy(x^2+3,y) * dy/dx = 0
fx(x^2+3,y) * 2x + fy(x^2+3,y) * dy/dx = 0
dy/dx = -fx(x^2+3,y) * 2x / fy(x^2+3,y)
5. Let f(x.y) = e^3x * tany and P。= (0, π/4). (a) Find Duf(P。), where
u = < 3/5,4/5 >. (b) Find the unit vector v such that Dv is maximal.
Ans:
(a) ▽f = < fx, fy > = < 3e^3x * tany , e^3x * sec^2y >
Duf(P。) = < 3e^0 * tan(π/4), e^0 * sec^2(π/4) > ‧ < 3/5,4/5 >
= < 3,2 > ‧ < 3/5,4/5 > = 9/5 + 8/5 = 17/5
p.s. subtitute P。 and dot u
(b) max unit vector = < 3/√(3^2+2^2) , 2/√(3^2+2^2) > = < 3/√13 , 2/√13 >
end the test
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 163.13.105.35
※ 編輯: RoarLiao 來自: 163.13.105.35 (04/21 01:38)
※ 編輯: RoarLiao 來自: 163.13.105.35 (04/21 01:47)