Why a Ledge Changes the Water-Level Rate
A tank can widen or narrow abruptly at a horizontal ledge. If water is added at a steady rate, the water level does not generally rise at the same speed throughout the tank: the same added volume must spread over different horizontal areas below and above the ledge.
To describe this precisely, let \(h\) denote the water depth measured upward from the bottom. Suppose the tank has constant horizontal cross-sectional area \(A_1\) from the bottom to height \(H\), and constant horizontal cross-sectional area \(A_2\) above height \(H\). The change at \(H\) is the ledge. The areas need not be equal.
The central task is to build a volume formula that accounts for the two regions without leaving a gap or counting any part twice. Once that formula is established, differentiating it within each region relates the inflow rate to the water-level rate.
Volume as a Function of Depth
Below the ledge, the water occupies a column of height \(h\) and area \(A_1\), so its volume is \(A_1h\). At and above the ledge, the entire lower region is already full. It contributes \(A_1H\). The water above it has height \(h-H\) and area \(A_2\), giving the additional volume \(A_2(h-H)\).
The expression \(h-H\) is the height of the water above the ledge, not its total depth from the tank bottom. For example, if \(h=45\text{ cm}\) and \(H=30\text{ cm}\), that upper height is \(45\text{ cm}-30\text{ cm}=15\text{ cm}\).
The two formulas agree at the ledge. Substituting \(h=H\) gives \(A_1H\) from the lower formula and \(A_1H+A_2(H-H)=A_1H\) from the upper formula. Thus the volume is continuous as the water reaches the ledge, even when the cross-sectional area changes there.
Worked Example: Finding the Volume on Both Sides of a Ledge
Worked Example: Finding the Volume on Both Sides of a Ledge
A tank has cross-sectional area \(120\text{ cm}^2\) below a ledge at height \(30\text{ cm}\), and area \(200\text{ cm}^2\) above it. Find the water volume at depths \(18\text{ cm}\), \(30\text{ cm}\), and \(45\text{ cm}\).
Here \(A_1=120\text{ cm}^2\), \(A_2=200\text{ cm}^2\), and \(H=30\text{ cm}\). At depth \(18\text{ cm}\), the water is below the ledge, so
At the ledge, the lower region is full:
At depth \(45\text{ cm}\), the lower region still contributes \(3600\text{ cm}^3\), and the upper water height is \(45\text{ cm}-30\text{ cm}=15\text{ cm}\). Therefore,
Check: The volume increases from \(3600\text{ cm}^3\) at the ledge to \(6600\text{ cm}^3\) at \(45\text{ cm}\), a difference of \(3000\text{ cm}^3\). The added height is \(15\text{ cm}\), and \((200\text{ cm}^2)(15\text{ cm})=3000\text{ cm}^3\), as required.
Relating Inflow to the Water-Level Rate
Let \(t\) denote time, and suppose water enters at a constant volumetric rate \(q\), measured in volume per unit time. While the water level stays within one region, the appropriate volume formula is differentiable. Applying the chain rule to \(V(h(t))\) gives the relationship between the rate of change of volume and the rate of change of depth.
For the lower region, \(V=A_1h\), so differentiating with respect to time gives \(\dfrac{dV}{dt}=A_1\dfrac{dh}{dt}\). For the upper region, \(V=A_1H+A_2(h-H)\). The quantities \(A_1\), \(A_2\), and \(H\) are fixed, so \(\dfrac{dV}{dt}=A_2\dfrac{dh}{dt}\). In either region, solving for the level rate gives the stated formula.
The equation also has a direct interpretation: during a short time interval, a level change of \(\Delta h\) in a region of area \(A_i\) corresponds to a volume change of \(A_i\Delta h\). A fixed volume added in that interval produces a larger depth change when the area is smaller.
Worked Example: Finding the Level Rate Below and Above the Ledge
Worked Example: Finding the Level Rate Below and Above the Ledge
Use the tank from the previous example. Water enters at \(600\text{ cm}^3/\text{min}\), with no outflow. Find the water-level rate when the water is at depth \(18\text{ cm}\) and when it is at depth \(45\text{ cm}\).
At \(18\text{ cm}\), the level is below the \(30\text{ cm}\) ledge. Use \(A_1=120\text{ cm}^2\):
At \(45\text{ cm}\), the level is above the ledge. Use \(A_2=200\text{ cm}^2\):
Check: Below the ledge, \((120\text{ cm}^2)(5\text{ cm}/\text{min})=600\text{ cm}^3/\text{min}\). Above the ledge, \((200\text{ cm}^2)(3\text{ cm}/\text{min})=600\text{ cm}^3/\text{min}\). Both calculations recover the stated volume rate. The level rises more slowly in the wider upper part because more volume is needed for each centimetre of rise.
Why the Rate Changes at the Ledge
The volume-depth function joins continuously at \(h=H\), but its slope with respect to depth changes there. Immediately below the ledge, the slope is \(A_1\); immediately above it, the slope is \(A_2\). Because the time rate of volume is fixed, the corresponding one-sided level rates are \(q/A_1\) and \(q/A_2\).
The volume-depth slope is \(A_1\), so the level rises at rate \(q/A_1\) as the water approaches \(H\).
The volume-depth slope is \(A_2\), so the level rises at rate \(q/A_2\) after the water passes \(H\).
If \(q\neq 0\) and \(A_1\neq A_2\), the one-sided level rates differ, so there is no single instantaneous level rate at the exact crossing time.
To see the last point more explicitly, suppose the water reaches \(H\) at time \(t_0\). For a positive constant inflow, the level is below \(H\) just before \(t_0\), and above \(H\) just after it. The rate from the left is \(q/A_1\), while the rate from the right is \(q/A_2\). The ordinary derivative at \(t_0\) exists only if these one-sided rates agree. With \(q\neq 0\), that requires \(A_1=A_2\). The volume remains continuous at the crossing, but the water-level graph has a corner if the areas differ.
Worked Example: Time to Reach a Depth Above the Ledge
Worked Example: Time to Reach a Depth Above the Ledge
A tank has area \(80\text{ cm}^2\) below a ledge at \(25\text{ cm}\) and area \(160\text{ cm}^2\) above it. It fills from empty at a steady rate of \(400\text{ cm}^3/\text{min}\). Find the time to reach depth \(40\text{ cm}\), and state the level rate in each region.
The volume at \(40\text{ cm}\) is the volume of the full lower region plus the volume above the ledge. The upper water height is \(40\text{ cm}-25\text{ cm}=15\text{ cm}\):
Starting empty, the required time is total volume divided by the constant volume rate:
The level rate below the ledge is \(\dfrac{400\text{ cm}^3/\text{min}}{80\text{ cm}^2}=5\text{ cm}/\text{min}\). Above the ledge it is \(\dfrac{400\text{ cm}^3/\text{min}}{160\text{ cm}^2}=2.5\text{ cm}/\text{min}\).
Check: The time to fill the lower region is \(\dfrac{2000\text{ cm}^3}{400\text{ cm}^3/\text{min}}=5\text{ min}\). The time to add the upper \(2400\text{ cm}^3\) is \(\dfrac{2400\text{ cm}^3}{400\text{ cm}^3/\text{min}}=6\text{ min}\). Together, \(5\text{ min}+6\text{ min}=11\text{ min}\), agreeing with the calculation from the total volume.
A Common Mistake: Using the Wrong Upper Height
A frequent error is to multiply the upper area \(A_2\) by the total depth \(h\) after the water has passed the ledge. That assigns the upper area to the lower part of the tank as well, even though the lower region has area \(A_1\). The upper contribution must use only the height above the ledge, \(h-H\).
For the first tank at depth \(45\text{ cm}\), the incorrect calculation \( (200\text{ cm}^2)(45\text{ cm})=9000\text{ cm}^3 \) treats the entire water column as if its area were \(200\text{ cm}^2\). The corrected calculation separates the regions:
Another related error is using the lower area to calculate the level rate above the ledge. The area in the rate equation must match the region where the water level currently lies. Above the ledge in this tank, the correct rate is \(600/200=3\text{ cm}/\text{min}\), not \(600/120=5\text{ cm}/\text{min}\).
Why the Piecewise Model Is Useful
The piecewise volume-depth relationship separates two questions that are easy to confuse. First, how much water is in the tank at a given depth? Second, how fast is the depth changing at that level for a given volume rate? The first uses the volumes of the regions already filled; the second uses the area of the region currently being filled.
Before calculating, identify the ledge height and the current depth. If the depth is below the ledge, use the lower formula and area. If it is above, include the full lower volume and add the upper-region contribution. At the ledge itself, either volume formula gives the same volume, but the left and right level rates may differ.
Check Your Understanding
Answer each question using the piecewise volume and rate relationships.
- A tank has area \(90\text{ cm}^2\) below a ledge at \(20\text{ cm}\) and area \(150\text{ cm}^2\) above it. Write the volume-depth formula for both regions.
- For that tank, calculate the volume when the water depth is \(32\text{ cm}\), showing the upper water height explicitly.
- If the tank fills at \(450\text{ cm}^3/\text{min}\), what is the water-level rate below the ledge? What is it above?
- Explain why the volume is continuous at the ledge even when the areas differ.
- For positive constant inflow, when does the water-level derivative exist at the exact time the level crosses the ledge?
