Why a Conical Tank Needs a Geometry Model
A conical tank narrows toward its bottom. As water enters, its depth increases, and the radius of its circular surface increases too. The volume formula for a cone involves both radius and height, so finding how quickly the water level rises requires accounting for how these dimensions change together.
Assume the tank is an upright, rigid cone with its vertex at the bottom. Let \(H\) and \(R\) denote the tank’s fixed full height and top radius. At time \(t\), let \(h\) be the water depth and \(r\) the radius of the water’s circular surface. The goal is to use the tank’s fixed dimensions to relate \(r\) and \(h\), and then relate the volume rate to the depth rate.
Relating Water Depth and Surface Radius
A vertical cross-section through the tank’s axis contains two similar triangles: the triangle representing the full tank and the smaller triangle occupied by the water. Their corresponding horizontal radii and vertical heights are proportional.
This equation shows that \(r\) is not fixed while the tank fills. It grows in direct proportion to \(h\). The constant ratio \(R/H\) depends only on the tank’s shape.
Deriving a Volume Model in Terms of Depth
The volume of a cone with radius \(r\) and height \(h\) is \(V=\frac{1}{3}\pi r^2h\). In the water-filled part of the tank, both \(r\) and \(h\) describe the smaller cone of water. Substituting the similar-triangle relationship gives a volume formula involving depth alone.
The cubic dependence matters: doubling the depth multiplies the water volume by eight, not by two. Near the bottom, a small increase in depth corresponds to a relatively small volume increase; higher in the tank, more volume is needed for the same additional depth.
Finding the Water-Level Rate
Both \(V\) and \(h\) change with time. Differentiating the volume model with respect to \(t\) relates their instantaneous rates. The tank dimensions \(R\) and \(H\) are constants, so they remain in the coefficient.
When \(h>0\), this equation can be solved for the water-level rate:
Here \(dV/dt\) is measured in volume per unit time, and \(dh/dt\) in length per unit time. If the tank is filling, \(dV/dt>0\), so the formula gives \(dh/dt>0\) whenever \(h>0\). The factor \(h^2\) in the denominator also means the water-level rate generally changes as the depth changes, even if the inflow rate stays constant.
Why the Differentiation Works
The volume model already incorporates the geometric constraint. Thus, differentiating it with respect to time automatically accounts for both the changing depth and the changing surface radius. The following derivation shows this directly from the two-variable cone formula.
Since \(r\) and \(h\) both depend on time, differentiate \(V=\frac{1}{3}\pi r^2h\) using the product rule and chain rule: $$ \frac{dV}{dt} =\frac{\pi}{3}\left(2rh\frac{dr}{dt}+r^2\frac{dh}{dt}\right). $$
The relation \(r=\frac{R}{H}h\) has constant \(R/H\), so $$ \frac{dr}{dt}=\frac{R}{H}\frac{dh}{dt}. $$ Also, \(r/h=R/H\), or \(h/r=H/R\), for positive depth.
Using \(dr/dt=(r/h)(dh/dt)\) in the differentiated volume formula gives $$ \frac{dV}{dt} =\frac{\pi}{3}\left(2rh\frac{r}{h}\frac{dh}{dt} +r^2\frac{dh}{dt}\right) =\frac{\pi}{3}\left(2r^2+r^2\right)\frac{dh}{dt} =\pi r^2\frac{dh}{dt}. $$ Since \(r=(R/H)h\), this becomes $$ \frac{dV}{dt} =\frac{\pi R^2}{H^2}h^2\frac{dh}{dt}, $$ which is the same rate equation obtained by first expressing \(V\) in terms of \(h\).
This verifies why the one-variable model works: it includes the effect of the changing radius through the similarity relation. Differentiating the cone formula while holding the radius fixed would omit that effect.
Worked Examples
Worked Example: A Tank with a Six-Foot Water Depth
An upright conical tank is \(12\) ft high and has a top radius of \(4\) ft. Water enters at \(2\) ft\(^3\)/min. Find the rate at which the water depth is increasing when \(h=6\) ft.
Step 1: Relate radius to depth. Similar triangles give $$ \frac{r}{h}=\frac{4}{12}=\frac{1}{3}, \qquad r=\frac{h}{3}. $$
Step 2: Write volume in terms of depth. $$ V=\frac{1}{3}\pi r^2h =\frac{1}{3}\pi\left(\frac{h}{3}\right)^2h =\frac{\pi h^3}{27}. $$
Step 3: Differentiate with respect to time. $$ \frac{dV}{dt} =\frac{\pi}{9}h^2\frac{dh}{dt}. $$
Step 4: Substitute the instantaneous data. At \(h=6\) ft and \(dV/dt=2\) ft\(^3\)/min, $$ 2=\frac{\pi}{9}(6)^2\frac{dh}{dt} =4\pi\frac{dh}{dt}. $$ Therefore, $$ \frac{dh}{dt}=\frac{2}{4\pi}=\frac{1}{2\pi}\text{ ft/min}. $$
Check: Substitution into the differentiated model gives $$ \frac{\pi}{9}(6)^2\left(\frac{1}{2\pi}\right) =4\pi\left(\frac{1}{2\pi}\right) =2\text{ ft}^3/\text{min}, $$ the stated inflow rate. The water-level rate is positive, as expected for a filling tank.
Worked Example: A Tank with a Three-Meter-Per-Minute Inflow
A conical tank is \(10\) m high and has a top radius of \(5\) m. Its water volume is increasing at \(3\pi/2\) m\(^3\)/min. How quickly is the water depth increasing when \(h=4\) m?
Step 1: Find the radius-depth relation. $$ \frac{r}{h}=\frac{5}{10}=\frac{1}{2}, \qquad r=\frac{h}{2}. $$
Step 2: Substitute into the volume formula. $$ V=\frac{1}{3}\pi\left(\frac{h}{2}\right)^2h =\frac{\pi h^3}{12}. $$
Step 3: Differentiate. $$ \frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}. $$
Step 4: Use \(h=4\) m and the given volume rate. $$ \frac{3\pi}{2} =\frac{\pi(4)^2}{4}\frac{dh}{dt} =4\pi\frac{dh}{dt}. $$ Thus, $$ \frac{dh}{dt} =\frac{3\pi/2}{4\pi} =\frac{3}{8}\text{ m/min}. $$
Check: The rate equation gives $$ \frac{\pi(4)^2}{4}\left(\frac{3}{8}\right) =4\pi\cdot\frac{3}{8} =\frac{3\pi}{2}\text{ m}^3/\text{min}, $$ which matches the given volume rate.
Worked Example: A Narrow Cone Filling Quickly
A conical tank is \(8\) ft high and has a top radius of \(2\) ft. Water enters at \(3\pi\) ft\(^3\)/min. Find the water-level rate when the water is \(2\) ft deep.
Step 1: Use similar triangles. $$ \frac{r}{h}=\frac{2}{8}=\frac{1}{4}, \qquad r=\frac{h}{4}. $$
Step 2: Express volume using \(h\). $$ V=\frac{1}{3}\pi\left(\frac{h}{4}\right)^2h =\frac{\pi h^3}{48}. $$
Step 3: Differentiate with respect to time. $$ \frac{dV}{dt}=\frac{\pi h^2}{16}\frac{dh}{dt}. $$
Step 4: Substitute the given values. $$ 3\pi=\frac{\pi(2)^2}{16}\frac{dh}{dt} =\frac{\pi}{4}\frac{dh}{dt}. $$ So $$ \frac{dh}{dt}=\frac{3\pi}{\pi/4}=12\text{ ft/min}. $$
Check: $$ \frac{\pi(2)^2}{16}(12) =\frac{\pi}{4}(12) =3\pi\text{ ft}^3/\text{min}. $$ This equals the inflow rate. The comparatively large level rate is consistent with the narrow water cross-section at this shallow depth.
A Common Mistake: Treating the Radius as Constant
In the first example, when \(h=6\) ft, the water-surface radius is \(r=6/3=2\) ft. It might seem convenient to substitute \(r=2\) into the cone formula first, writing \(V=\frac{1}{3}\pi(2)^2h=\frac{4\pi}{3}h\), and then differentiate. That calculation incorrectly gives $$ \frac{dV}{dt}=\frac{4\pi}{3}\frac{dh}{dt}, \qquad 2=\frac{4\pi}{3}\frac{dh}{dt}, \qquad \frac{dh}{dt}=\frac{3}{2\pi}\text{ ft/min}. $$
The error is that \(r=2\) ft holds only at the instant \(h=6\) ft. It does not hold as the water level changes. The radius must remain connected to depth by \(r=h/3\) before differentiating. Using the correct model gives \(dV/dt=(\pi/9)h^2(dh/dt)\), and therefore \(dh/dt=1/(2\pi)\) ft/min at that instant. Substitution into the correct rate equation verifies the given \(2\) ft\(^3\)/min inflow.
Check Your Understanding
For each problem, identify the similar-triangle relation before differentiating.
- A conical tank is \(9\) ft high with top radius \(3\) ft. Write the water-surface radius \(r\) in terms of depth \(h\).
- For a conical tank with full height \(H\) and top radius \(R\), derive \(V(h)=\frac{\pi R^2}{3H^2}h^3\).
- In a cone with \(H=8\) m and \(R=4\) m, water enters at \(2\pi\) m\(^3\)/min. Find \(dh/dt\) when \(h=4\) m.
- Why is it incorrect to substitute the instantaneous value of \(r\) into the volume formula and then treat that radius as constant during differentiation?
- What prevents the rearranged water-level-rate formula from being used directly at \(h=0\)?
