Amath 問題2條

1.有一直線族 (9+k)x-y-(8+k)=0,其中k為實數,設L1和L2為真線族的直線,L為直線3x-y-12=0
已知L1和L之間的銳角與L2和L之間的銳角同為 ? ,其中 tan ? =1/2 ,求L1和L2方程
3x -y - 12 = 0
slope = 3
(9+k)x -y - (8+k) = 0
slope = (9+k)
! [(9+k) - 3] / [1 + 3(9+k)] ! = 1/2
! (6+k) / (28+3k) ! = 1/2
(6+k) / (28+3k) = 1/2 or (6+k) / (28+3k) = -1/2
2(6+k) = (28+3k) or 2(6+k) = -(28+3k)
k = -16 or k = -8
L1 : -7x -y - (-8) = 0, 7x + y - 8 = 0
L2 : (1)x -y - (0) = 0, x -y = 0
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2.直線L與y軸相交於(0,11),並與C: y=8x^3+1 相交於(1,9),求兩線與y軸內圍成的面積
直線L :
(11-9)/(0-1) = (y-11)/(x-0)
x = -y/2 + 11/2
C :
y = 8x^3 + 1
x^3 = (y - 1) / 8
x = (y - 1)^(1/3) / 2
(積分 y = 1 to 9) (y - 1)^(1/3) / 2 dy + (積分 y = 9 to 11) -y/2 + 11/2 dy
= [3(y - 1)^(4/3) / 8] (y = 1 to 9)
+ (-y^2/4 + 11y/2) ( y = 9 to 11 )
= 6 + 1
= 5




2007-10-01 17:42:49 補充：
上面兩位都做錯左,支持我啊

(1)
L的斜率為3
設 L1 斜率為m

tan ? = | (m - 3) / (1 + 3m)| = 1/2
m = -7 或 m=1

9+k = -7 或 9+k = 1
k= -16 或 k= -8
L1 和 L2 為
-7x-y+8=0 和 x-y=0

(2)
L 的方程為 y= -2x + 11
INT[0,1](-2x + 11 - 8x^3-1)dx = (-x^2 + 10x - 2x^4)[0,1]= 7

m of the family:9+k m of L:3

1/2=I(9+k-3)/(1+3(9+k))l

1/2=(9+k-3)/(1+3(9+k)) or -1/2=(9+k-3)/(1+3(9+k))

28+3k=12+2k or 28+3k= -12-2k

k= -16 or -8

7x+y-8=0 and x-y=0