Calculus Applications of Differentiation

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# h3cadiap1

Water is poured into an inverted circular cone of base radius 5 cm and height 15 cm at the rate of 10 cm3•s-1  Calculate; a)the rate of increase of the height of water level. b)the rate of increase of the surface area of the water,when the water level is 4 cm high. (Leave answers in terms of π where applicable.)

(Problem credited by O-Level --Marshal Cavendish)

Strategy the rule "Must have" of Mr.Zhang ® Give the water level be h cm high after time t seconds, and the radius of the water surface r cm at this moment as the figure 1 which is a cross-section of the cone.

1. Similar triangle,find r in terms of h
2. Water volume       (V)   =		1 3	πr2h → (1)

3.  Take ;

dy

dx

   both side to get the rate of increase.

Solution
Take similar triangle ;
	r h	=	5 15
	r	=	h 3	→ (2)
From equation(1)	v	=	1 3	π r2h → (1)
Take r from equation(2) substitued to (1)	v	=	1 3	π	(	h 3	)2	h
⇒	v	=	π	h3 27
Take Diff	dV dh	=	π	h2 9
From	dV dt	=		dV dh	x	dh dt

dV

dt

=	10 cm3•s-1 and h = 4 , then ;

	10	=	π 42 9	x	dh dt
	10x9 16 π	=	dh dt
	45 8π	=	dh dt

Ans.   a) the rate of increase of the height when the height is 4 cm, is

45 8π

cm•s-1

b) The water surface is circular,therefore, the surface area

	A	=	πr2	at any instant
From equation(2),	r	=	h 3
Hence	A	=	πh2 9
Take Diff	dA dh	=	2πh 9
And	dA dt	=	dA dh	x	dh dt

Substitue h=4 and

dh dt

=

45 8π

,

we get

dA dt

=

2π(4) 9

x

45 8π

⇒

dA dt

=

5 cm2•s-1

Ans.   the surface rate of the water is increasing at the rate of 5 cm2•s-1 when the water level height is 4 cm.

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