微積分與統計的數學問題

x^3是位於分子抑或分母?

2011-01-08 20:06:12 補充：
1 C:y=(1/2)x^3 代x=0=>y=0﹐因此點P(0,-1)不是曲線C上的一點。

2 假定從點P到曲線C的切線方程與C相切於M(a,b)
則b=(1/2)a^3。

考慮dy/dx=(3/2)x^2。則
(b+1)/(a-0)=(3/2)a^2
b+1=(3/2)a^3
再用b=(1/2)a^3可得
(1/2)a^3+1=(3/2)a^3
a^3=1
a=1
因此b=1/2
切線斜率=3/2。從點P到曲線C的切線方程
y-1/2=(3/2)(x-1)
2y-1=3x-3
3x-2y-2=0

2.
Let the point of contact be Q.
Let the x - coordinate of Q be h, so y - coordinate of Q is h^3/2, that means the co-ordinates of Q is Q(h, h^3/2).
Slope of tangent at point Q = dy/dx = 3x^2/2 = 3h^2/2 since the x - coordinate of Q is h.
So equation of tangent is:
y - h^3/2 = (3h^2/2)(x - h)........ (1)
Since P(0, - 1) is on this tangent, sub. P into the equation we get
- 1 - h^3/2 = (3h^2/2)(0 - h)
- 1 - h^3/2 = - 3h^3/2
- 1 = - 3h^3/2 + h^3/2
- 1 = - h^3
h^3 = 1
h = 1
Sub into (1), equation of tangent is
y - 1/2 = (3/2)(x - 1)
2y - 1 = 3x - 3
2y = 3x - 2.

如果冇記錯, 應該係將x代0
c:y=1/2(0)^3
y≠-1
∴p不是c上一點

dy/dx (切線斜率) = 3/2x^2
point slope form:
y-(-1)/x-0 =3/2x^2
y=3/2x-1


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