SME
Calculus I · Problem type
Filling a cone tank
Use this for “Water pours at 0.03 m³/s into a cone tank, point down, 1.5 m in rim radius and 3 m deep. How fast does the level rise when the water is 1 m deep?”
Diagram
Similar
triangles:
r
=
R·
h
H
=
2
×
2
4
=
1
dh/dt
=
dV/dt
πr²
=
0.05
π ×
1
2
=
0.01592
This is an example. Type your own values to start a new problem.
r
=
R (m)
×
h (m)
÷
H (m)
=
r (m)
dh
dt
=
dV/dt (m³/s)
÷
(π
(
r (m)
)
2
)
=
dh/dt (m/s)
Formulas
r
=
R
×
h
H
1
=
2
×
2
4
dV/dt
=
π
×
r
2
×
(dh/dt)
0.05
=
π
×
1
2
×
0.01592
Assumptions
•
The tank is a cone standing on its point, so the water is a smaller cone of the same shape: r ÷ h = R ÷ H by similar triangles.
•
The water’s volume is V = ⅓πr²h. Putting in r = Rh ÷ H and differentiating in time t gives dV/dt = πr²·(dh/dt).
•
So the inflow spreads over the water’s surface: the same inflow raises the level fast near the point and slowly near the rim.
•
Lengths are in m, the inflow in m³/s and the rise in m/s.
Step-by-step
WE KNOW
R = 2 m, H = 4 m, h = 2 m, dV/dt = 0.05 m³/s
FIND
radius of the water’s surface (r), rate the water level rises (dh/dt)
Step 1 · Find radius of the water’s surface (r)
Use r = R × h ÷ H
Similar triangles: the surface radius is to the depth as the rim radius is to the height.
r
=
2
×
2
4
r
=
4
4
r
=
1
m
Step 2 · Find rate the water level rises (dh/dt)
Use dV/dt = π × r² × (dh/dt)
Divide the inflow by the area of the water’s surface, πr².
dh/dt
=
dV/dt
π ×
r
2
dh/dt
=
0.05
π ×
1
2
dh/dt
=
0.05
π
dh/dt = 0.01592 m/s
Check
Put the numbers back in. Both sides should match.
1
=
2
×
2
4
✓
0.05
=
π
×
1
2
×
0.01592
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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