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equation & integration-urgent

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1. For the relation x^3-3xy + y^3 = 3 do the following. (i) Find the rst derivative y'(x) as a funtion of x and y. (ii) Find the equation of the tangent line at (x, y) =(0,3). (iii) Find any points (x, y) where the tangent is horizontal. integrate (x^2)*2^(x^3+1)dx

最佳解答:

(1)(i) 3x2 - 3xy' - 3y + 3y2y' = 0 x2 - xy' - y + y2y' = 0 y'(y2 - x) = y - x2 y' = (y - x2)/(y2 - x) (ii) At (0, 3): y' = (3 - 0)/(9 - 0) = 1/3 By point-slope form: (y - 3)/x = 1/3 3y - 9 = x x - 3y + 9 = 0 (iii) When tangent is horizontal, y' = 0, i.e. (y - x2)/(y2 - x) = 0 with restriction y2 - x being non-zero So the condition for y' = 0 is y - x2 = 0, and when substituted into the original equation: y = x2 x3 - 3xy + y3 = 0 x3 - 3x3 + x6 = 0 x3(x3 - 2) = 0 x = 0 or 3√2 y = 0 or 3√4 For (0, 0) pair, it will make y2 - x = 0 and so the only ans is x = 3√2 and y = 3√4 So at (3√2, 3√4), the tangent is horizontal. (2) ∫x2 2(x^3 + 1) dx = (1/3)∫2(x^3 + 1) d(x3 + 1) = (1/3)∫2u du, u = (x3 + 1) = 2u/(3 ln 2) + C = 2(x^3 + 1)/(ln 8) + C 2009-11-03 22:47:20 補充： To 石石 : 我的確在做 2 ^ (x^3+1), 你試將我的答案微分, 將它當 2 ^ (x^3+1) 辦就行

其他解答:

六呎將軍: 他是說(x^2)*2 ^ (x^3+1), 應該是2 to the power of (x^3+1) 而不是 2 times (x^3+1)