Differentiation (Paper 2), Question 2 & 3


Question 2:
Given the equation of a curve is:
y = x2 (x – 3) + 1
(a) Find the gradient of the curve at the point where x = –1.
(b) Find the coordinates of the turning points.

Solution:
(a)
y= x 2 ( x3 )+1 y= x 3 3 x 2 +1 dy dx =3 x 2 6x When x=1 dy dx =3 ( 1 ) 2 6( 1 )      =9 Gradient of the curve is 9.

(b)
At turning points, dy dx =0
3x2 – 6x = 0
x2 – 2x = 0
x (x – 2) = 0
x = 0, 2

y = x2 (x – 3) + 1
When x = 0, y = 1
When x = 2,
y = 22 (2 – 3) + 1
y = 4 (–1) + 1 = –3
Therefore, coordinates of the turning points are (0, 1) and (2, –3).


Question 3:
It is given the equation of the curve is y = 2x (1 – x)4 and the curve pass through T(2, 4).
Find
(a) the gradient of the curve at point T.
(b) the equation of the normal to the curve at point T.

Solution:
(a)
y=2x ( 1x ) 4 dy dx =2x×4 ( 1x ) 3 ( 1 )+ ( 1x ) 4 ×2     =8x ( 1x ) 3 +2 ( 1x ) 4 At T( 2,4 ),x=2. dy dx =16( 1 )+2( 1 )     =16+2     =18

(b)
Equation of normal: y y 1 = 1 dy dx ( x x 1 ) y4= 1 18 ( x2 ) 18y72=x+2 x+18y=74

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