3.8.2 Integration, Long Questions (Question 3 & 4)


Question 3:
The gradient function of a curve which passes through P(2, 14) is 6x² – 12x
Find
(a) the equation of the curve,
(b) the coordinates of the turning points of the curve and determine whether each of the turning points is a maximum or a minimum.

Solution:
(a)
Given gradient function of a curve  dy dx =6 x 2 12x The equation of the curve, y= ( 6 x 2 12x )  dx y= 6 x 3 3 12 x 2 2 +c y=2 x 3 6 x 2 +c 14=2 ( 2 ) 3 6 ( 2 ) 2 +c, at point P ( 2,14 ) 14=8+c c=6 y=2 x 3 6 x 2 6


(b)
dy dx =6 x 2 12x At turning points,  dy dx =0 6 x 2 12x=0 6x( x2 )=0 x=0x=2 x=0,  y=2 ( 0 ) 3 6 ( 0 ) 2 6=6 x=2,  y=2 ( 2 ) 3 6 ( 2 ) 2 6=14 d 2 y d x 2 =12x12 When x=0 d 2 y d x 2 =12( 0 )12=12 <0 ( 0,6 ) is a maximum point. When x=2 d 2 y d x 2 =12( 2 )12=12 >0 ( 2,14 ) is a minimum point.


Question 4:
Diagram below shows a curve x = y2 – 1 which intersects the straight line 3y = 2x at point Q.
Calculate the volume generated when the shaded region is revolved 360o about the y-axis.


Solution:

x= y 2 1( 1 ) 3y=2x x= 3 2 y( 2 ) Substitute (2) into (1), 3 2 y= y 2 1 2 y 2 3y2=0 ( 2y+1 )( y2 )=0 y= 1 2    or   y=2


When y=2,x= 3 2 ( 2 )=3, Q=( 3, 2 ) I 1 ( Volume of cone ) = 1 3 π r 2 h= 1 3 π ( 3 ) 2 ( 2 ) =6π  unit 3 I 2 ( Volume of the curve ) π 1 2 x 2 dy π 1 2 ( y 2 1 ) 2 dy π 1 2 ( y 4 2 y 2 +1 )dy =π [ y 5 5 2 y 3 3 +y ] 1 2 =π[ ( 2 5 5 2 ( 2 ) 3 3 +2 )( 1 5 5 2 ( 1 ) 3 3 +1 ) ] =π( 46 15 8 15 ) = 38 15 π  unit 3  Volume generated = I 1 I 2                                  =6π 38 15 π                                  = 52 15 π  unit 3

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