Related Rates – Example #6 (homework problem #19 in section 3.9)
Water is leaking out of an inverted conical tank at a rate of 10,000 cubic centimeters per minute. At the
same time, water is being pumped into the tank at a constant rate. The tank has height 6 meters and the
diameter at the top is 4 meters. If the water level is rising at a rate of 20 centimeters per minute when the
height of the water is 2 meters, find the rate at which water is being pumped into the tank.
Idea – water flows in at top and out at bottom, so volume changes in 2 ways
TOTAL Change in Volume = ( added water at top ) − ( leaking water at bottom )
dh
Given info:
= 20 cm
Water in
min
dt
Diameter top
4m = 400 cm
h = 2m = 200cm
10, 000 cm
3
min
= rate at which water leaks out bottom
Find: constant rate at which water is being added at top, C
Solution: We need an equation that relates given info and what we are
to find. Start with an equation that relates the Volume of the cone to the
height.
1
V = π r2h
3
Water out
ht of tank 6m
= 600 cm
The given info is about “height” and “volume” --- so we want to make
this equation in terms of only “height” and “volume”.
200 cm
Can we find an expression for “r” in terms of “h”? Using
similar triangles we get:
r
600 cm
h
200 r
1 r
h
=
→→→
= →→→ = r
600 h
3 h
3
So, instead of writing “r” in the Volume equation, we can put what “r”
is equivalent to,
h
. We do this below (*).
3
(*)
1
V = π r2h
3
2
1 ⎛h⎞
1 h2
1
V = π⎜ ⎟ h= π
h=
π h3
3 ⎝3⎠
3 9
27
V =
π
h3
27
Now the equation is in terms of only “height” and “volume”.
The next step is to take the derivative of the equation with
respect to time.
d
d⎛π
⎞
(V ) = ⎜ h3 ⎟
dt
dt ⎝ 27 ⎠
(continued in next column)
dV π
dh
=
3h 2
(cancel the 3 and the 27)
dt 27
dt
dV π 2 dh
= h
dt
dt
9
plug in givens from above
dV π
2
= ( 200 ) ( 20 )
dt
9
3
dV
= 279, 253 cm
min
dt
This means that the “TOTAL change” in
3
volume is at a rate of 279,253 cm
.
min
To finish the problem we have to start with the original idea:
TOTAL Change in Volume = ( added water at top ) − ( leaking water at bottom )
dV
dt
=
C
−
10, 000
279, 253
=
C
−
10, 000
279, 253 + 10, 000
289, 253
=
C
= C
3
So, water is coming into the tank at a constant rate of 289,253 cm
min
.
End of problem