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a-maths~~~~~~
發問:
已知曲綫C:y=x3-3x2+2x-1。A(1,-1)、B(a,b)為曲綫上的兩點,而O為原點。 (a)證明曲綫在A點的切綫通過O點。 (b)若曲綫在B點的切綫通過O點。證明2a3-3a2+1=0。 (c)由此,試求從O點到曲綫的切綫方程。 更新: (c)part ans 是 y=-x,y=23/4 x
y = x3 - 3x2 + 2x - 1 y` = 3x2 - 6x + 2 slope = 3(1)2 - 6(1) + 2 = -1 切綫 : y - (-1) = (-1)(x - 1) x - 1 + y + 1 = 0 x + y = 0 好明顯 : O點 係 切綫上 -------------------------------------------------- slope = 3(a)2 - 6(a) + 2 = 3a2 - 6a + 2 切綫 : y - b = (3a2 - 6a + 2)(x - a) B點的切綫通過O點, 0 - b = (3a2 - 6a + 2)(0 - a) b = a(3a2 - 6a + 2) B(a,b)為曲綫上的點, b = a3 - 3a2 + 2a - 1 a(3a2 - 6a + 2) = a3 - 3a2 + 2a - 1 2a3 - 3a2 + 1= 0 -------------------------------------------------- 2a3 - 3a2 + 1 = (a - 1)(2a2 - a - 1) = (a - 1)2 (2a + 1) 2a3 - 3a2 + 1= 0 a = 1 or a = -1/2 a = -1/2, b = (-1/2)3 - 3(-1/2)2 + 2(-1/2) - 1 = -23/8 切綫, y - (-23/8) = [ 3(-1/2)2 - 6(-1/2) + 2 ] [ x - (-1/2) ] y + 23/8 = [ 23/4 ] [ x + 1/2 ] 8y + 23 = [ 23 ] [ 2x + 1 ] 46x - 8y = 0 23x - 4y = 0 2008-06-08 12:29:42 補充： 投我一票哦
其他解答:
y=x^3 -3x^2 +2x -1. dy/dx = 3x^2 -6x + 2 . For point A(1,-1), dy/dx = 3 -6 +2 = -1. Therefore, tangent passing through A is y+1 = -1(x -1). y= -x + 1 -1 = -x. which passes the origin. For point B(a,b), dy/dx = 3a^2 - 6a + 2 . Therefore, tangent passing through B is y-b = (3a^2 -6a +2)(x -a). If it passes through (0,0), that is -b = -a(3a^2 -6a +2) -b = 6a^2 -3a^3 -2a.......................(1) Since B is on the curve, therefore, b=a^3 -3a^2 +2a -1 ................(2) (1) + (2) We get 0=6a^2 -3a^3 -2a + a^3 -3a^2 +2a -1 That is 2a^3-3a^2 +1 = 0. 2008-06-08 12:36:10 補充： Substitute (1) into equation of tangent, we get y+6a^2 -3a^3 -2a = (3a^2 -6a +2)(x -a) =(3a^2 -6a +2)x -3a^3 +6a^2 -2a. That is y = (3a^2 -6a +2)x is the tangent from origin to curve.
a-maths~~~~~~
發問:
已知曲綫C:y=x3-3x2+2x-1。A(1,-1)、B(a,b)為曲綫上的兩點,而O為原點。 (a)證明曲綫在A點的切綫通過O點。 (b)若曲綫在B點的切綫通過O點。證明2a3-3a2+1=0。 (c)由此,試求從O點到曲綫的切綫方程。 更新: (c)part ans 是 y=-x,y=23/4 x
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最佳解答:y = x3 - 3x2 + 2x - 1 y` = 3x2 - 6x + 2 slope = 3(1)2 - 6(1) + 2 = -1 切綫 : y - (-1) = (-1)(x - 1) x - 1 + y + 1 = 0 x + y = 0 好明顯 : O點 係 切綫上 -------------------------------------------------- slope = 3(a)2 - 6(a) + 2 = 3a2 - 6a + 2 切綫 : y - b = (3a2 - 6a + 2)(x - a) B點的切綫通過O點, 0 - b = (3a2 - 6a + 2)(0 - a) b = a(3a2 - 6a + 2) B(a,b)為曲綫上的點, b = a3 - 3a2 + 2a - 1 a(3a2 - 6a + 2) = a3 - 3a2 + 2a - 1 2a3 - 3a2 + 1= 0 -------------------------------------------------- 2a3 - 3a2 + 1 = (a - 1)(2a2 - a - 1) = (a - 1)2 (2a + 1) 2a3 - 3a2 + 1= 0 a = 1 or a = -1/2 a = -1/2, b = (-1/2)3 - 3(-1/2)2 + 2(-1/2) - 1 = -23/8 切綫, y - (-23/8) = [ 3(-1/2)2 - 6(-1/2) + 2 ] [ x - (-1/2) ] y + 23/8 = [ 23/4 ] [ x + 1/2 ] 8y + 23 = [ 23 ] [ 2x + 1 ] 46x - 8y = 0 23x - 4y = 0 2008-06-08 12:29:42 補充： 投我一票哦
其他解答:
y=x^3 -3x^2 +2x -1. dy/dx = 3x^2 -6x + 2 . For point A(1,-1), dy/dx = 3 -6 +2 = -1. Therefore, tangent passing through A is y+1 = -1(x -1). y= -x + 1 -1 = -x. which passes the origin. For point B(a,b), dy/dx = 3a^2 - 6a + 2 . Therefore, tangent passing through B is y-b = (3a^2 -6a +2)(x -a). If it passes through (0,0), that is -b = -a(3a^2 -6a +2) -b = 6a^2 -3a^3 -2a.......................(1) Since B is on the curve, therefore, b=a^3 -3a^2 +2a -1 ................(2) (1) + (2) We get 0=6a^2 -3a^3 -2a + a^3 -3a^2 +2a -1 That is 2a^3-3a^2 +1 = 0. 2008-06-08 12:36:10 補充： Substitute (1) into equation of tangent, we get y+6a^2 -3a^3 -2a = (3a^2 -6a +2)(x -a) =(3a^2 -6a +2)x -3a^3 +6a^2 -2a. That is y = (3a^2 -6a +2)x is the tangent from origin to curve.
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