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標題:
Application of Differentiation
發問:
Find equations of tangent from point ( 3, - 3) to the curve 4x^2 + 9y^2 = 36. 更新: To : 香港的士司機 Please notice that (3, - 3) is NOT on the curve!!! 更新 2: To : 翻雷滾天 風卷殘雲 (3, -3) is NOT on the curve, can you solve it ? 更新 3: To : Godfrey (3, -3) is NOT on the curve, can you solve it ?
最佳解答:
To 香港的士司機: The first step should be: 8x + 18y(dy/dx) = 0 as any constant function has its derivative = 0 2011-12-16 14:04:54 補充： Let (h, k) be the point on the curve where the tangent from (3, -3) touches the curve, then: dy/dx = -4x/(9y) So at (h, k), dy/dx = -4h/(9k) We have: -4h/(9k) = (k + 3)/(h - 3) 12h - 4h2 = 9k2 + 27k 4h2 + 9k2 - 12h + 27k = 0 36 - 12h + 27k = 0 12 - 4h + 9k = 0 h = 3(3k + 4)/4 Sub this relation into: 4h2 + 9k2 = 36: 9(3k + 4)2/4 + 9k2 = 36 9(3k + 4)2 + 36k2 = 144 (3k + 4)2 + 4k2 = 16 13k2 + 24k + 16 = 16 13k2 + 24k = 0 k(13k + 24) = 0 k = 0 or -24/13 h = 3 or -30/13 N.B. Since (3, -3) is in quadrant II, the tangents are expected to touch the curve in quadrant II or III. So for the one touching at (3, 0), the equation is x = 3 For the one touching at (-30/13, -24/13) the slope is -5/9, making up the equation: (y + 3)/(x - 3) = -5/9 9y + 27 = 15 - 5x 5x + 9y - 12 = 0
其他解答:
Application of Differentiation
發問:
Find equations of tangent from point ( 3, - 3) to the curve 4x^2 + 9y^2 = 36. 更新: To : 香港的士司機 Please notice that (3, - 3) is NOT on the curve!!! 更新 2: To : 翻雷滾天 風卷殘雲 (3, -3) is NOT on the curve, can you solve it ? 更新 3: To : Godfrey (3, -3) is NOT on the curve, can you solve it ?
最佳解答:
To 香港的士司機: The first step should be: 8x + 18y(dy/dx) = 0 as any constant function has its derivative = 0 2011-12-16 14:04:54 補充： Let (h, k) be the point on the curve where the tangent from (3, -3) touches the curve, then: dy/dx = -4x/(9y) So at (h, k), dy/dx = -4h/(9k) We have: -4h/(9k) = (k + 3)/(h - 3) 12h - 4h2 = 9k2 + 27k 4h2 + 9k2 - 12h + 27k = 0 36 - 12h + 27k = 0 12 - 4h + 9k = 0 h = 3(3k + 4)/4 Sub this relation into: 4h2 + 9k2 = 36: 9(3k + 4)2/4 + 9k2 = 36 9(3k + 4)2 + 36k2 = 144 (3k + 4)2 + 4k2 = 16 13k2 + 24k + 16 = 16 13k2 + 24k = 0 k(13k + 24) = 0 k = 0 or -24/13 h = 3 or -30/13 N.B. Since (3, -3) is in quadrant II, the tangents are expected to touch the curve in quadrant II or III. So for the one touching at (3, 0), the equation is x = 3 For the one touching at (-30/13, -24/13) the slope is -5/9, making up the equation: (y + 3)/(x - 3) = -5/9 9y + 27 = 15 - 5x 5x + 9y - 12 = 0
其他解答:
此文章來自奇摩知識+如有不便請留言告知
4x^2+9y^2=36 ----------(1) Differentiate the equation (1), dy/dx = -4x/9y Slope between the required (x,y) & (3,-3) = (y+3)/(x-3) (y+3)/(x-3)=-4x/9y After solving, x = 3(4+3y)/4 Substitute this into 1, after solving y = 0 or y=-24/13 the corresponding values of x are 3 and -15/13 respectively (15/13 rejected) Thus, x=3 in the first condition. dy/dx| (-15,13, -24,13) = -5/18 y+24/13 = (-5/18)(x+15/13) 65x+234y+507=0|||||I think i will first let the tangents' equation be y = mx+c, then use the point (3,-3) and discriminant to solve m and c|||||Differentiate 36 (a constant) with respect to x = 0 Final answer: Equation of tangent y = (4/9) x – 13/3|||||By implicit differentiation 8x + 18y(dy/dx) = 36 36 - 8x = 18y(dy/dx) dy/dx = (18 - 4x)/9y dy/dx | x = 3, y = -3 = [18 - 4(3)]/9(-3) = (18 - 12)/-27 = -2/9 Equation of tangent: y + 3 = -2/9(x - 3) y = -2/9x + 2/3 - 9/3 y = -2/9x - 7/3 Did it within a minute, i don't know if it is right or not please tell me if there's any mistake
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