F.4 Maths M1 differentiation

2014-06-08 9:51 pm

Consider the curve C: y=x(x-2)^(1/3) and the straight line L that passes through the origin and is parallel to the tangent to C at x=3.
a) find the equation of L
b) find the x-coordinate of the two intersecting points of C and L.

x=3 means the slope =0? Or I need to calculate dy/dx at x=3 to find the slope?
Thx!!!

回答 (1)

用戶26240

2014-06-08 10:21 pm

✔ 最佳答案

(a)
y = x(x - 2)^(1/3)
dy/dx = x*(1/3)(x - 2)^(-2/3) + (x - 2)^(1/3)

Slope of the tangent to C at (x = 3)
= dy/dx |x=3
= 3*(1/3)(3 - 2)^(-2/3) + (3 - 2)^(1/3)
= 2

Slope of L = 2 and L passes through the origin (0, 0).

Equation of L :
(y - 0) = 2(x - 0)
y = 2x

(b)
L: y = 2x ...... [1]
L: y = x(x - 2)^(1/3) ...... [2]

[1] - [2] :
2x - x(x - 2)^(1/3) = 0
x[2 - (x - 2)^(1/3)] = 0
x = 0 or 2 - (x - 2)^(1/3) = 0
x = 0 or (x - 2)^(1/3) = 2
x = 0 or x - 2 = 8
x = 0 or x = 10

The x-coordinate of interesting points :
x = 0 or x = 10

參考: 土扁

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