equation & integration-urgent

(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

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)