Integrals 問題X2

[匿名]

2009-12-07 1:09 am

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 ^^

回答 (1)

Prof. Physics

2009-12-07 1:25 am

✔ 最佳答案

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

參考: Physics king

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