標題:

Integrals 問題X2

發問:

1)The equation of the tangent to the curve at (3,1) is y = 4 - x and at each point (x,y) on the curve,d^2 y/dx^2 = x^2 - 2. Find the equation of the curve.Ans:y=1/12x^4 - x^2 - 4x + 61/42)It is given that dy/dx = x - k and du/dx =y,where k is a constant.When x=k,y=0 and u=0.(a)Express y in terms of x and... 顯示更多 1)The equation of the tangent to the curve at (3,1) is y = 4 - x and at each point (x,y) on the curve,d^2 y/dx^2 = x^2 - 2. Find the equation of the curve. Ans:y=1/12x^4 - x^2 - 4x + 61/4 2)It is given that dy/dx = x - k and du/dx =y,where k is a constant.When x=k,y=0 and u=0. (a)Express y in terms of x and k. (b)Hence,prove that u=1/6(x-k)^3. 求詳細列式 .. THANKS ^^

最佳解答:

1. d^2y/dx^2 = x^2 - 2 dy/dx = int (x^2 - 2) = x^3/3 - 2x + c where c is a constant dy/dx│x = 3 = -1 (3)^3/3 - 2(3) + c = -1 c = -4 dy/dx = x^3/3 - 2x - 4 y = int (x^3/3 - 2x - 4) y = x^4/12 - x^2 - 4x + c' c' is a constant Since the graph passes through (3 , 1) 1 = (3)^4/12 - (3)^2 - 4(3) + c' c' = 61/4 So, equation of the graph: y = 1/12 x^4 - x^2 - 4x + 61/4 2.a. dy/dx = x - k y = int(x - k) = x^2/2 - kx + c When x = k, y = 0 0 = k^2/2 - k^2 + c c = k^2/2 So, y = 1/2 x^2 - kx + k^2/2 = 1/2 (x - 2kx + k^2) = 1/2 (x - k)^2 b. du/dx = y du/dx = 1/2 (x - k)^2 u = int[1/2 (x - k)^2] = 1/6 (x - k)^3 + c When x = k, u = 0 c = 0 So, u = 1/6 (x - k)^3

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