How Volume and Water Depth Are Related
A cylindrical tank has the same circular cross section at every height. That simple geometric fact makes its water-level rate especially direct to calculate: the volume of water is proportional to its depth.
Let \(r\) be the tank’s fixed radius and let \(h(t)\) be the depth of the water at time \(t\). The volume \(V(t)\) of water is the area of the circular base multiplied by the water depth. For a vertical cylinder, this gives a relationship between \(V\) and \(h\) that can be differentiated to connect their rates of change.
This model applies while the water occupies a cylindrical portion of the tank. It assumes the tank has a vertical axis and constant circular cross section. The depth must also remain within the tank’s physical height; the formula does not describe water overflowing or a tank with a changing cross section.
Deriving the Water-Level Rate
Because the radius is fixed, the base area \(\pi r^2\) is constant. Differentiating the volume equation with respect to time therefore gives a direct relation between the volume rate and the depth rate.
The left side measures how quickly the amount of water changes. The right side is the tank’s cross-sectional area multiplied by how quickly the water surface rises or falls. Solving for the depth rate gives the main calculation rule.
The rule follows from the chain rule: \(V\) depends on the time-varying depth \(h(t)\), and \(V=\pi r^2h\). Since \(\pi r^2\) is constant, the derivative is \(\pi r^2\frac{dh}{dt}\). This also explains why the depth rate in an ideal cylinder does not depend on the current depth: the cross-sectional area is the same at every height.
If the tank receives water at rate \(Q_{\text{in}}\) and loses water at rate \(Q_{\text{out}}\), both expressed as nonnegative rates, then the volume changes at the net rate \(\frac{dV}{dt}=Q_{\text{in}}-Q_{\text{out}}\). A positive difference means water is accumulating; a negative difference means it is decreasing. Use this net volume rate in the rate rule.
Worked Examples
Worked Example: Filling a Cylindrical Tank
A vertical cylindrical tank has radius \(3\) ft. Water flows in at \(18\pi\) ft\(^3\)/min, with no outflow. Find the instantaneous rate at which the water depth is rising.
Step 1: Write the volume-depth relationship. The base area is \(\pi(3)^2=9\pi\) ft\(^2\), so \(V=9\pi h\).
Step 2: Differentiate with respect to time. The radius is constant, so \(\frac{dV}{dt}=9\pi\frac{dh}{dt}\).
Step 3: Substitute the given volume rate. With \(\frac{dV}{dt}=18\pi\) ft\(^3\)/min, \(18\pi=9\pi\frac{dh}{dt}\). Dividing by \(9\pi\) gives \(\frac{dh}{dt}=2\) ft/min.
Check: Substituting \(2\) ft/min into the differentiated equation gives \(9\pi(2)=18\pi\) ft\(^3\)/min, the stated inflow rate. The positive sign confirms that the level is rising.
Worked Example: Falling Water Level During Outflow
A cylindrical tank has radius \(4\) dm and is draining at \(8\pi\) L/min. Find the rate of change of the water depth. Use \(1\) L \(=1\) dm\(^3\).
Step 1: Assign a signed volume rate. The tank is losing water, so \(\frac{dV}{dt}=-8\pi\) dm\(^3\)/min.
Step 2: Find the cross-sectional area. \(\pi(4)^2=16\pi\) dm\(^2\). Thus \(V=16\pi h\), and differentiation gives \(\frac{dV}{dt}=16\pi\frac{dh}{dt}\).
Step 3: Solve for the depth rate. \(-8\pi=16\pi\frac{dh}{dt}\), so \(\frac{dh}{dt}=\frac{-8\pi}{16\pi}=-\frac{1}{2}\) dm/min.
Check: The differentiated volume equation gives \(16\pi(-\frac{1}{2})=-8\pi\) dm\(^3\)/min, matching the signed loss rate. The negative depth rate means the level is falling at \(0.5\) dm/min.
Worked Example: Combining Inflow and Outflow
A tank has a circular base of radius \(3\) dm. Water enters at \(25\) L/min while water leaves at \(7\) L/min. Find the instantaneous rate of change of the water depth. Use \(1\) L \(=1\) dm\(^3\).
Step 1: Calculate the net volume rate. \(\frac{dV}{dt}=25-7=18\) dm\(^3\)/min.
Step 2: Relate volume and depth. The base area is \(\pi(3)^2=9\pi\) dm\(^2\), so differentiating \(V=9\pi h\) gives \(\frac{dV}{dt}=9\pi\frac{dh}{dt}\).
Step 3: Solve. \(18=9\pi\frac{dh}{dt}\), so \(\frac{dh}{dt}=\frac{18}{9\pi}=\frac{2}{\pi}\) dm/min. To three decimal places, this is approximately \(0.637\) dm/min.
Check: Using the exact rate, \(9\pi(\frac{2}{\pi})=18\) dm\(^3\)/min, exactly the net volume rate. Its positive sign shows that the level is rising.
Why the Cross-Sectional Area Matters
The rate rule has a useful interpretation. For the same volume of water added per unit time, a tank with a smaller base area has a faster-rising water level. The volume is spread over less area. A larger base area spreads the same added volume more thinly, so the level rises more slowly.
The units show the same relationship. If \(r\) is measured in feet, then \(\pi r^2\) has units of square feet. Dividing a volume rate in cubic feet per minute by a base area in square feet gives feet per minute:
A mismatch of units can make a correct-looking calculation invalid. Convert liters to cubic decimeters when using a radius measured in decimeters, for example, rather than combining liters directly with a formula in cubic feet.
A Common Mistake: Using the Wrong Area
The relevant area is the horizontal circular area of the water’s surface, \(\pi r^2\). It is not the tank’s curved side area, nor does it include the depth as a factor. For example, suppose \(r=3\) ft and the inflow is \(18\pi\) ft\(^3\)/min. If the water depth is \(4\) ft, using the curved side area \(2\pi rh\) would give this incorrect calculation:
This is wrong because the volume is not changing by the curved side area times the depth rate. The correct volume relationship is \(V=\pi(3)^2h=9\pi h\), so \(18\pi=9\pi\frac{dh}{dt}\) and \(\frac{dh}{dt}=2\) ft/min. Checking the correct rate gives \(9\pi(2)=18\pi\) ft\(^3\)/min, as required.
Another sign error is to enter an outflow as a positive volume rate. If water leaves and there is no inflow, \(\frac{dV}{dt}\) is negative. Keeping that sign makes the resulting depth rate negative as well, which correctly represents a falling level.
Summary
For a vertical cylinder with fixed radius, the water volume equals the circular base area multiplied by the water depth. Differentiating this relationship converts a volume rate into a depth rate. When both inflow and outflow are present, first find the signed net volume rate, then divide by the base area. Interpret the sign and check that the units are consistent.
Check Your Understanding
For each question, keep track of the volume-rate sign and use consistent units.
- A cylindrical tank has radius \(2\) ft and receives water at \(12\pi\) ft\(^3\)/min. What is the rate at which its water depth rises?
- A tank with radius \(5\) m loses water at \(10\pi\) m\(^3\)/min. What is \(\frac{dh}{dt}\), including its sign?
- Water enters a tank at \(14\) L/min and leaves at \(20\) L/min. Is the water level rising or falling? What is the net volume rate?
- Why does the water-level rate in a vertical cylinder not depend on the current depth?
- What area belongs in the formula for converting volume rate to depth rate, and why is the curved side area inappropriate?
