F.5 AM

The question is extremely bad and I just do what I have done as below
(a)
To find b
b=√(225-25*5)/9=10/3
sub y=15/2, and b=10/3 into y=(√5/b)c, give c=5√5
(b)
I don't know what h exactly means, if you want to get the result, h means that the height between the deapth between the water level from B
So the volume generated by AB
=π∫(225-25y)/9 dx [from -3 to √5]
=(π/9)[225y-(25/3)y^3]
=(π/9)[450+(550/3)√5]
The volume generated by BC
=π∫(100/45)y^2 dx [from √5 to h]
=(20π/27)[y^3]
=(20π/27)[h^3-5√5]
Combine, we have
(π/9)[450+(550/3)√5]+ (20π/27)[h^3-5√5]
=[(50+ (50/3)√5+ (20π/27)h^3]π
(c)(i)
dV/dt=π/9
d/dt[(50+ (50/3)√5+ (20π/27)h^3]π=π/9
(3h^2)dh/dt=1/9
when the deapth of water is 9
Then h=2
So dh/dt=1/108
(ii)when the deapth of water is 4, then we can do nothing

2008-12-15 13:18:45 補充：
修正
h means that the height between the deapth between the water level from y=0

(c)(i) sub h=6 (ii)

the volume generated by AB

=π∫(225-25y^2)/9 dx [from -3 to h]

=(π/9)[225y-(25/3)y^3]

=(π/9)[-675+225-225h+(25/3)h^3]

2008-12-15 13:20:54 補充：
dV/dt=π/9

d/dt(π/9)[-675+225-225h+(25/3)h^3]=π/9

-225+(25h^2)dh/dt=1

when the deapth of water is 4

Then h=1

So dh/dt=226/25