A cylindrical tank with radius 6 m is being filled with water at a rate of 4 m3/min. How fast is the height of the water increasing

Determine the rate of change of the height of water,
d
h
d
t
, in the cylindrical tank. We do this by first expressing the volume of the cylindrical tank,

V
=
π
r
2
h

where r is the radius and h is the height. Now, we must take note that the radius of the water in the tank is constant. We determine the answer by differentiating the equation with respect to time and then plugging in the given values:

d
V
d
t
=
4

m
3
/
m
i
n
r
=
6

m
We proceed with the solution.

V
=
π
r
2
h
d
V
d
t
=
d
d
t
π
r
2
h
d
V
d
t
=
π
r
2
d
h
d
t
1
π
r
2
d
V
d
t
=
d
h
d
t
1
π
(
6

m
)
2
(
4

m
3
/
m
i
n
)
=
d
h
d
t
1
9
π

m
/
m
i
n
=
d
h
d
t

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