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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 dydx=6x212xThe equation of the curve,y=(6x212x) dxy=6x3312x22+cy=2x36x2+c14=2(2)36(2)2+c, at point P (2,14)14=8+cc=6y=2x36x26


(b)
dydx=6x212xAt turning points, dydx=06x212x=06x(x2)=0x=0x=2x=0,  y=2(0)36(0)26=6x=2,  y=2(2)36(2)26=14d2ydx2=12x12When x=0d2ydx2=12(0)12=12 <0(0,6) is a maximum point.When x=2d2ydx2=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=y21(1)3y=2xx=32y(2)Substitute (2) into (1),32y=y212y23y2=0(2y+1)(y2)=0y=12   or   y=2


When y=2,x=32(2)=3, Q=(3, 2)I1(Volume of cone)=13πr2h=13π(3)2(2)=6π unit3I2(Volume of the curve)π21x2dyπ21(y21)2dyπ21(y42y2+1)dy=π[y552y33+y]21=π[(2552(2)33+2)(1552(1)33+1)]=π(4615815)=3815π unit3 Volume generated =I1I2 =6π3815π =5215π unit3

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