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2.11.1 Differentiation SPM Practice (Paper 2 Question 1 – 5)


Question 1:
The curve y = x3 – 6x2 + 9x + 3 passes through the point P (2, 5) and has two turning points, A (3, 3) and B.
Find 
(a) the gradient of the curve at P.
(b) the equation of the normal to the curve at P.
(c) the coordinates of B and determine whether B is a maximum or the minimum point.

Solution:
(a)
y = x3 – 6x2 + 9x + 3
dy/dx = 3x2– 12x + 9
At point P (2, 5),
dy/dx = 3(2)2 – 12(2) + 9 = –3

Gradient of the curve at point P = –3.

(b)
Gradient of normal at point P = 1/3
Equation of the normal at P (2, 5):
y – y1 = m (x – x1)
y – 5 = 1/3 (x – 2)
3y – 15 = x – 2
3y = x + 13

(c) 
At turning point, dy/dx = 0.
3x2 – 12x + 9 = 0
x2 – 4x + 3 = 0
(x – 1)( x – 3) = 0
x – 1 = 0  or x – 3 = 0
x = 1  x = 3 (Point A)

Thus at point B:
x = 1
y = (1)3– 6(1)2 + 9(1) + 3 = 7

Thus, coordinates of B = (1, 7)

when  x = 1 ,   d 2 y d x 2 = 6 x − 12 d 2 y d x 2 = 6 ( 1 ) − 12 d 2 y d x 2 = − 6 < 0 Since  d 2 y d x 2 < 0 ,   B  is a maximum point .


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 ( x−3 )+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 ( 1−x ) 4 dy dx =2x×4 ( 1−x ) 3 ( −1 )+ ( 1−x ) 4 ×2 =−8x ( 1−x ) 3 +2 ( 1−x ) 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 ) y−4=− 1 18 ( x−2 ) 18y−72=−x+2 x+18y=74


Question 4:
(a) A worker is pumping air into a spherical shape balloon at the rate of  25 cm3 s-1.
[ Volume of a sphere, V= 4 3 π r 3 ]
Leaving answer in terms of π, find,
(i) rate of change of radius of the balloon when its radius is 10 cm.   [3 marks]
(ii) approximate change of volume when radius of the balloon decrease from 10 cm to  9.95 cm.    [2 marks]
(b) A curve y=h x 3 − 3 x  has turning point x = 1, find value of h.  [3 marks]

Solution:
(a)(i)
V= 4 3 π r 3 dV dr =4π r 2 dr dt = dr dV × dV dt = 1 4π r 2 ×25 = 1 4π ( 10 ) 2 ×25 = 1 16π

(a)(ii)
δV δr ≈ dV dr δV= dV dr ×δr =4π r 2 ×( 9.95−10 ) =4π ( 10 ) 2 ×( −0.05 ) =−20π

(b)
y=h x 3 − 3 x y=h x 3 −3 x −1 dy dx =3h x 2 +3 x −2 dy dx =3h x 2 + 3 x 2 At turning points,  dy dx =0 0=3h x 2 + 3 x 2 0=3h ( 1 ) 2 + 3 1 →given x=1 3h=−3 h=−1



Question 5 (SPM 2017 - 7 marks):
It is given that the equation of a curve is y= 5 x 2 .  
(a) Find the value of dy dx when x = 3.
(b) Hence, estimate the value of 5 ( 2.98 ) 2 .  

Solution:
(a)
y= 5 x 2 =5 x −2 dy dx =−10 x −3 =− 10 x 3 When x=3 dy dx =− 10 3 3 =− 10 27


(b)
δx=2.98−3=−0.02 δy= dy dx .δx =− 10 27 ×( −0.02 ) =0.007407 Values of  5 ( 2.98 ) 2 =y+δy = 5 x 2 +( 0.007407 ) = 5 3 2 +( 0.007407 ) =0.56296


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