Water is leaking out of an inverted conical tank at a rate of 7,000 cm^3/min at the same time that water is being pumped into the tank at a constant rate.
Answers
Answer:
rate of inflow required to achieve the specified rate of height (depth) of water increase.
Later we'll use the fact that
Actual inflow rate
= Inflow Rate for Increased Depth + Leakage Rate
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For the given cone the ratio of r adius to h eight is
1
3
so
r
=
1
3
h
The formula for the volume of a cone:
V
=
π
r
2
h
3
becomes
V
=
π
h
3
27
d
V
d
h
=
π
h
2
9
We are interested in the change in Volume with respect to time and note that
d
V
d
t
=
d
V
d
h
⋅
d
h
d
t
Using the value we've already calculated for
d
V
d
h
and the supplied value of
20
cm/min (at a height of
h
=
200
cm)
we get:
d
V
d
t
=
π
(
200
c
m
)
2
⋅
(
20
c
m
)
9
min
=
800000
π
9
c
m
3
/min
or roughly
279
,
252.7
c
m
3
/min
This is the Inflow Rate Required to Cause Height Increase and
ignores the Rate of Leakage
The Actual Inflow Rate needs to be the sum of these two:
279
,
252.7
c
m
3
/min
+
10
,
000
c
m
3
/min
=
289
,
252.7
c
m
3
/min
i hope it's help to you
