Re: [考古] 幾題計算題請問
※ 引述《flygey (努力達成目標)》之銘言：
: 1.Show that the parabola y=-x^2 and the line x-4y-18=0 intersect at right
: angle at one of their points of intersection(於兩個交點之一以直角相交)
: (請問這題方程式如何設)
: 2.Find the volume of the solid obtained by rotating the region bounded by
: y=x-x^2 and y=0 about the line x=2
: (想請問這題界線是設 0-1 還是 0-2)
: 3.The plane x+y+z=1 cuts the cylinder x^2+y^2=1 in an ellipse.Use Lagrange
: multipliers to find the points on the ellipse that lie closest to and
: farthest from the origin.
: (這題方程式是 f(x,y,z)=x^2+y^2-1+入(x+y+z-1) 這樣設嗎
: 如果方程式求出xyz後要如何求最近最遠點 )
: 請版友指導
: 感謝
3. Let f(x,y,z) = x^2 + y^2 + z^2
g(x,y,z) = x + y + z - 1
h(x,y,z) = x^2 + y^2 - 1
F(x,y,z) = f(x,y,z) + (λ)(g(x,y,z)) + (μ)(h(x,y,z))
= x^2 + y^2 + z^2 + (λ)(x + y + z - 1)
+ (μ)(x^2 + y^2 - 1)
Fx = 2x + λ + 2μx = 0 ------(1)
Fy = 2y + λ + 2μy = 0 ------(2)
Fz = 2z + λ = 0 ------(3)
(1) - (2) => 2x - 2y + 2μx - 2μy = 0
=> 2x(1 + μ) - 2y(1 + μ) = 0
=> 2(x - y)(1 + μ) = 0
x = y 或 μ = -1
當 x = y 時
x^2 + y^2 = 1 => x^2 + x^2 = 1
=> (2)(x^2) = 1
1
=> x^2 = ---
2
1 1 -1 -1
(x , y , z) = (----- , ----- , 1 - √2) , (----- , ----- , 1 + √2)
√2 √2 √2 √2
當 μ = -1 時
代入(1)和(2)得 λ = 0 代入(3)得 2z = 0 => z = 0
x + y = 1 ------(4)
x^2 + y^2 = 1 ------(5)
(4) => y = 1 - x 代入(5)得
x^2 + (1 - x)^2 = 1
x^2 + x^2 - 2x + 1 = 1
(2)(x^2) - 2x = 0 => x^2 - x = 0 => (x)(x - 1) = 0
x = 0 或 1
(x , y , z) = (0 , 1 , 0)和(1 , 0 , 0)
1 1 1 1
f(---- , ---- , 1 - √2) = --- + --- + 3 - (2)(√2) = 4 - (2)(√2)
√2 √2 2 2
-1 -1 1 1
f(---- , ---- , 1 + √2) = --- + --- + 3 + (2)(√2) = 4 + (2)(√2)
√2 √2 2 2
f(0,1,0) = 1
f(1,0,0) = 1
-1 -1
當 (x,y,z) = (---- , ---- , 1 + √2) 時
√2 √2
有最遠距離 √(4 + (2)(√2))
當 (x,y,z) = (0 , 1 , 0) 和 (1 , 0 , 0) 時
有最短距離 1
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07/05 23:53, , 1F
07/05 23:53, 1F
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