Why Cone Filling Needs a Geometric Constraint
A cone’s water level does not rise through a container of constant width. As the depth increases, the water’s horizontal surface gets wider too. To find the rate at which the water level rises, we must account for both changes in a single equation.
Similar triangles provide the needed geometric relationship: at every depth, the water occupies a smaller cone that has the same shape as the container. This lets us express the water-surface radius using the water depth, then differentiate a volume equation with just one changing geometric variable.
Set Up the Cone and the Water
Consider a rigid conical container with its vertex at the bottom and its circular opening at the top. Let \(H\) be the container’s full height and \(R\) its radius at the opening. These are fixed dimensions. At time \(t\), let \(h(t)\) be the water depth measured up from the vertex, and let \(r(t)\) be the radius of the water’s horizontal surface. Let \(V(t)\) denote the volume of water.
Assume the cone has straight sides, the water surface is horizontal, and the water occupies the cone from its vertex up to that surface. The smaller cone occupied by the water and the full container are therefore similar. We consider the filling stage \(0<h<H\), so the water cone is smaller than the full cone.
Use Similar Triangles to Relate Radius and Depth
A vertical cross-section through the axis of the cone gives two similar triangles. Their corresponding vertical sides are \(H\) and \(h\), and their corresponding horizontal half-widths are \(R\) and \(r\). The ratios of corresponding sides are equal:
Solving for the radius of the water surface gives
This relation is the essential geometric constraint. It says that the water radius changes in proportion to the water depth. For example, when the water reaches half the container’s height, its surface radius is half the opening radius.
Rewrite the Volume in Terms of Depth
The volume of a cone with base radius \(r\) and height \(h\) is \(V=\frac{1}{3}\pi r^2h\). Apply this formula to the water, then substitute the similar-triangles relation \(r=\frac{R}{H}h\):
Because \(R\) and \(H\) are constant, this equation expresses the water volume entirely in terms of the changing depth. It also shows why volume is not proportional to depth: the water radius grows with depth, so the volume is proportional to \(h^3\).
Differentiate with Respect to Time
Differentiate the volume equation with respect to time. The coefficient \(\frac{\pi R^2}{3H^2}\) is constant, while \(h=h(t)\) depends on time. By the chain rule, \(\frac{d}{dt}(h^3)=3h^2\frac{dh}{dt}\). Therefore,
This equation relates the volume rate to the water-level rate at the same instant. If the water is being added at a known rate and \(h>0\), solve for the level rate:
These formulas require the cone dimensions to remain fixed and the water to occupy a similar cone as it rises. The rates are instantaneous: values of \(h\) and \(\frac{dV}{dt}\) used in the equation must refer to the same time. If water is being added, then \(\frac{dV}{dt}>0\); for \(h>0\), the formula gives \(\frac{dh}{dt}>0\), as expected.
Why the Differentiation Works
The derivation rests on two linked ideas. First, the cone’s fixed shape forces the surface radius to grow in proportion to depth. Second, differentiating a cubic dependence on depth produces a factor of \(3h^2\), which cancels the \(\frac{1}{3}\) in the cone-volume formula.
The factor \(h^2\) in the rate equation has a useful interpretation. At greater depths, each additional increment of depth corresponds to a wider layer of water, so a given incoming volume rate produces a smaller rise in level.
Worked Examples
Worked Example: Find the Water Radius at a Given Depth
A conical container has height \(H=10\) feet and opening radius \(R=5\) feet. Find the water-surface radius when the water depth is \(h=4\) feet. Then calculate the water volume at that depth.
Use the similar-triangles relation:
The volume is the volume of a cone with radius \(2\) feet and height \(4\) feet:
Check using the depth-only formula. Here \(\frac{\pi R^2}{3H^2}=\frac{25\pi}{300}=\frac{\pi}{12}\), so \(V=\frac{\pi}{12}(4\text{ ft})^3=\frac{16\pi}{3}\text{ ft}^3\), matching the result from the cone formula.
Worked Example: Find the Water-Level Rate
A conical container is \(12\) feet tall and has an opening radius of \(4\) feet. Water is flowing in at \(\frac{dV}{dt}=2\pi\) cubic feet per minute. How fast is the water level rising when \(h=6\) feet?
Use the rate equation with \(H=12\) feet, \(R=4\) feet, \(h=6\) feet, and \(\frac{dV}{dt}=2\pi\text{ ft}^3/\text{min}\):
Check by differentiating the volume equation directly. In this case, \(V=\frac{\pi(4\text{ ft})^2}{3(12\text{ ft})^2}h^3=\frac{\pi}{27}h^3\), so \(\frac{dV}{dt}=\frac{\pi}{9}h^2\frac{dh}{dt}\). Substituting \(h=6\) feet and \(\frac{dh}{dt}=\frac{1}{2}\text{ ft}/\text{min}\) gives \(\frac{dV}{dt}=\frac{\pi}{9}(36)(\frac{1}{2})=2\pi\text{ ft}^3/\text{min}\), as required.
Worked Example: Find the Incoming Volume Rate
A cone has height \(H=8\) metres and opening radius \(R=3\) metres. At the instant when the water is \(4\) metres deep, its level is rising at \(\frac{dh}{dt}=\frac{1}{6}\) metre per minute. Find the rate at which water is entering.
Use \(\frac{dV}{dt}=\frac{\pi R^2}{H^2}h^2\frac{dh}{dt}\):
Check by writing \(V=\frac{9\pi}{192}h^3=\frac{3\pi}{64}h^3\). Differentiation gives \(\frac{dV}{dt}=\frac{9\pi}{64}h^2\frac{dh}{dt}\). At \(h=4\) metres and \(\frac{dh}{dt}=\frac{1}{6}\) metre per minute, this is \(\frac{9\pi}{64}(16)(\frac{1}{6})=\frac{3\pi}{8}\text{ m}^3/\text{min}\), confirming the result.
A Common Mistake: Treating the Water Radius as Fixed
The opening radius \(R\) is fixed, but the water-surface radius \(r\) is not. Using the full container’s radius in the water-volume formula ignores the similar-triangles constraint.
For example, in the \(12\)-foot-high container with \(4\)-foot opening radius, suppose the depth is \(6\) feet and water enters at \(2\pi\text{ ft}^3/\text{min}\). An incorrect calculation might use \(V=\frac{1}{3}\pi R^2h\) and differentiate to get \(\frac{dV}{dt}=\frac{1}{3}\pi R^2\frac{dh}{dt}\). This would give
which is not the correct rate. At depth \(6\) feet, the water radius is \(r=(4/12)(6)=2\) feet, not \(4\) feet. Using the correct similar-triangles relation gives \(\frac{dh}{dt}=\frac{1}{2}\text{ ft}/\text{min}\). Substituting that rate into the correct volume-rate equation gives \(\frac{dV}{dt}=\frac{\pi(4)^2}{(12)^2}(6)^2(\frac{1}{2})=2\pi\text{ ft}^3/\text{min}\).
Check Units and Interpret the Result
The rate equation is also a useful unit check. The factor \(\frac{R^2}{H^2}\) has no units, while \(h^2\frac{dh}{dt}\) has units of length squared times length per time, or volume per time. Thus the right-hand side has the units of \(\frac{dV}{dt}\). Solving for \(\frac{dh}{dt}\) gives units of length per time, as a water-level rate should.
For a fixed positive volume inflow, the equation also predicts that the water-level rate decreases as \(h\) increases, because \(\frac{dh}{dt}\) is proportional to \(1/h^2\). The formula for \(\frac{dh}{dt}\) requires \(h>0\); at the exact instant \(h=0\), one cannot divide by \(h^2\) to find a finite level rate from that rearranged equation.
Check Your Understanding
Use the similar-triangles relation and the differentiated volume equation to answer these questions.
- A cone has height \(9\) metres and opening radius \(6\) metres. What is the water-surface radius when the depth is \(3\) metres?
- Starting from \(r=\frac{R}{H}h\) and \(V=\frac{1}{3}\pi r^2h\), derive a formula for \(V\) in terms of \(h\), \(R\), and \(H\).
- Differentiate the depth-only volume formula with respect to time. Which quantities are constant, and which depend on time?
- For a cone with \(H=10\) feet, \(R=5\) feet, and \(h=5\) feet, what water-level rate corresponds to an inflow of \(5\pi\text{ ft}^3/\text{min}\)?
- Why is it incorrect to use the opening radius \(R\) as the radius of the water surface at every depth?
