Draining a Triangular Water Trough

Why the Water Depth Does Not Fall at a Constant Rate

A triangular trough narrows toward its bottom. When the water is deep, a small decrease in depth can correspond to a relatively large decrease in water volume. When the water is shallow, the water’s surface is narrower, so the same decrease in depth removes less volume. As a result, a constant rate of water loss generally does not produce a constant rate of decrease in depth.

To model this change, first connect the water’s depth to the width of its surface. Similar triangles provide that relationship. The cross-sectional area then gives the volume, and differentiating the volume with respect to time connects the water’s depth rate to the volume rate.

Geometry and Variables

Assume the trough is horizontal and has a constant length. Each end has the shape of an isosceles triangle, with its vertex at the bottom. Let \(H\) be the full height of an end and \(B\) its width at the top. Both dimensions are fixed. Let \(L\) be the trough’s fixed length.

As the trough drains, let \(h(t)\) denote the water depth at time \(t\). Let \(w(t)\) denote the width of the water’s surface, and let \(V(t)\) denote the water volume. The depth, surface width, and volume can change over time; \(H\), \(B\), and \(L\) do not.

Using Similar Triangles to Find the Water Width

A vertical cross-section through the trough’s end contains a large triangle representing the full end and a smaller triangle representing the water. These triangles are similar: they have the same vertex angle, and their horizontal bases are parallel. Corresponding widths and heights therefore have the same ratio.

Solving for \(w\) gives the width of the water’s surface at depth \(h\). This is the key geometric fact: the width changes with the depth, so it cannot be treated as a fixed dimension when calculating the water’s cross-sectional area.

From Cross-Sectional Area to Volume

The water cross-section is a triangle with base \(w\) and height \(h\). Its area is one-half the product of those dimensions. Substitute the similar-triangle expression for \(w\) to obtain an area formula involving only \(h\) and fixed trough dimensions.

The water occupies a length \(L\) of the trough. Because this cross-section extends uniformly along that length, its volume is \(L\) times its cross-sectional area:

A quick units check supports the model. The factor \(LB/H\) has units of length, so multiplying it by \(h^2\), which has units of length squared, gives volume units. For example, if all dimensions are measured in feet, \(V\) is measured in cubic feet.

Relating the Volume Rate to the Depth Rate

The dimensions \(L\), \(B\), and \(H\) remain constant, but \(h\) and \(V\) vary with time. Differentiate the volume model with respect to \(t\), using the chain rule for \(h(t)^2\):

The factor \(2h\) appears because the derivative of \(h^2\) with respect to time is \(2h\,dh/dt\). The resulting equation relates the instantaneous volume rate to the instantaneous depth rate at the same time.

During draining, the volume is decreasing, so \(\frac{dV}{dt}<0\). The depth also decreases, so \(\frac{dh}{dt}<0\). If a problem gives a positive outflow rate \(q\), that number describes how quickly volume leaves the trough; it is not the signed rate of change of the water volume. In that case, \(\frac{dV}{dt}=-q\).

Why the Rate Equation Works

The geometry makes volume proportional to the square of the depth: \(V=Ch^2\), where \(C=LB/(2H)\) is a fixed positive constant. A small depth change therefore does not correspond to a fixed volume change independent of depth. Differentiating this exact relationship gives the instantaneous rate connection. In particular, \(\frac{dV}{dt}=2Ch\,\frac{dh}{dt}\), which is the same equation as \(\frac{dV}{dt}=\frac{LB}{H}h\frac{dh}{dt}\).

This also explains why the water-level rate can change while the outflow rate stays constant. Solving the rate equation for \(\frac{dh}{dt}\) shows that its magnitude depends on the current depth \(h\). As \(h\) becomes smaller, a fixed amount of volume lost per unit time corresponds, in this idealized model, to a larger magnitude of depth change per unit time.

Finding the Water-Level Rate

Using a Known Depth Rate to Find the Outflow

A Common Mistake: Treating the Water Width as Fixed

A tempting but incorrect approach is to use the full top width \(B\) as the base of the water triangle at every depth, giving \(A=\frac12 Bh\). That formula would apply only if the water’s surface width actually remained equal to the full top width. In a triangular trough with the water below the top, the surface width is smaller and must be found from similar triangles.

For example, if \(B=6\) ft, \(H=4\) ft, and \(h=2\) ft, the incorrect calculation gives \(A=\frac12(6)(2)=6\text{ ft}^2\). Similar triangles instead give \(w=(6/4)(2)=3\) ft, so the correct area is \(A=\frac12(3)(2)=3\text{ ft}^2\). The incorrect area is twice the correct one. Using it would also produce an incorrect volume model and an incorrect rate equation.

Another sign error can occur when a problem states that water flows out at a rate of \(q\) units of volume per unit time. If \(q\) is stated as a positive amount leaving, the volume’s signed rate is \(-q\). Keeping the signed rate and the positive outflow magnitude distinct prevents reporting a rising water level when the trough is draining.

Limits of the Model

The rate equation applies to an instant when the water depth is positive and the geometry remains as described. At \(h=0\), there is no water depth to decrease. Moreover, the rearranged equation for \(\frac{dh}{dt}\) divides by \(h\), so it cannot be used at \(h=0\). A model that assumes a fixed, nonzero volume-loss rate all the way to an empty trough would predict an unbounded depth rate as \(h\) approaches zero. In a physical situation, the outflow rate may change as the water becomes shallow; use the rate information supplied for the particular instant in the problem.

For a well-defined calculation, match the units of \(L\), \(B\), \(H\), and \(h\), and make sure that the volume rate uses the corresponding cubic units per time. Keep the negative signs on decreasing rates until the final interpretation. A negative \(\frac{dh}{dt}\) gives the direction of change; the positive phrase “depth is falling at” describes its magnitude.