Rates of Change in a Cone: Why the Radius Isn't a Constant

The question

A container is a right circular cone with its vertex at the bottom and its axis vertical. The cone has height $24\ \text{cm}$ and the radius of its open top is $8\ \text{cm}$ . Water is poured in. When the depth of the water is $h\ \text{cm}$ , the radius of the water surface is $r\ \text{cm}$ and the volume of water is $V\ \text{cm}^3$ .

An inverted cone of height 24 cm and top radius 8 cm, vertex at the bottom. Water fills the lower part of the cone to depth h, and the radius of the water surface is labelled r. — Inverted cone, height 24 cm, top radius 8 cm, with water to depth h.

(a) Show that $V = \dfrac{\pi h^3}{27}$ . [3 marks]

Water is poured in at a constant rate of $12\ \text{cm}^3$ per second.

(b) Find the rate at which the depth of the water is increasing at the instant when $h = 6$ . [4 marks]

[The volume of a cone is $\frac{1}{3}\pi r^2 h$ .]

The common wrong answer

This is the working I see most often for part (b). The student starts from the cone formula as given, with both $r$ and $h$ still in it:

V = \frac{1}{3}\pi r^2 h

Then they differentiate with respect to $h$ , treating $r$ like a number:

\frac{dV}{dh} = \frac{1}{3}\pi r^2

At $h = 6$ the water surface has radius $r = 2$ , so

\frac{dV}{dh} = \frac{1}{3}\pi (2)^2 = \frac{4\pi}{3}

\frac{dh}{dt} = 12 \times \frac{3}{4\pi} = \frac{9}{\pi} \approx 2.86\ \text{cm/s}

Every line looks reasonable, but the second line is wrong. As the water rises, the surface gets wider, so $r$ is not a constant: it changes whenever $h$ changes. Differentiating as if $r$ stays fixed misses part of the change in volume, and the answer comes out three times too big.

A second trap is in the chain rule itself: writing $\frac{dh}{dt} = \frac{dV}{dt} \times \frac{dV}{dh}$ . With the correct $\frac{dV}{dh} = 4\pi$ this gives $48\pi \approx 151\ \text{cm/s}$ , which would raise the water through the whole $24\ \text{cm}$ height of the cone in well under a second.

The correct working

(a) The water forms a smaller cone with the same shape as the container. By similar triangles,

\frac{r}{h} = \frac{8}{24} = \frac{1}{3} \quad\Rightarrow\quad r = \frac{h}{3}

Substitute into the volume formula, so that $V$ depends on $h$ only:

V = \frac{1}{3}\pi\left(\frac{h}{3}\right)^2 h = \frac{1}{3}\pi \times \frac{h^2}{9} \times h = \frac{\pi h^3}{27}

as required.

(b) Differentiate with respect to $h$ :

\frac{dV}{dh} = \frac{3\pi h^2}{27} = \frac{\pi h^2}{9}

When $h = 6$ :

\frac{dV}{dh} = \frac{\pi (6)^2}{9} = 4\pi

Now use the chain rule. We know $\frac{dV}{dt} = 12$ and want $\frac{dh}{dt}$ :

\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt} \quad\Rightarrow\quad 12 = 4\pi \times \frac{dh}{dt}

\frac{dh}{dt} = \frac{12}{4\pi} = \frac{3}{\pi}

So the depth is increasing at $\dfrac{3}{\pi} \approx 0.955\ \text{cm/s}$ .

(c) Since $r = \frac{h}{3}$ , we have $\frac{dr}{dh} = \frac{1}{3}$ , so

\frac{dr}{dt} = \frac{dr}{dh} \times \frac{dh}{dt} = \frac{1}{3} \times \frac{3}{\pi} = \frac{1}{\pi}

So the radius is increasing at $\dfrac{1}{\pi} \approx 0.318\ \text{cm/s}$ .

Why it works

To differentiate $V$ with respect to $h$ , $V$ must be written in terms of $h$ only. Any other letter left in the formula has to be a genuine constant. The radius of the water surface isn’t: the deeper the water, the wider the surface. Similar triangles give the link $r = \frac{h}{3}$ , and substituting it turns $V$ into a function of one variable, $\frac{\pi h^3}{27}$ .

You can see where the factor of $3$ comes from. Treating $r$ as fixed gives $\frac{dV}{dh} = \frac{1}{3}\pi r^2 = \frac{\pi h^2}{27}$ . Writing $V$ in terms of $h$ first gives a cube, $h^3$ , and the power rule brings down a $3$ : $\frac{dV}{dh} = \frac{\pi h^2}{9}$ . The difference is the extra volume that comes from the surface getting wider.

For the chain rule, write it so the $dh$ “cancels”: $\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$ . Then rearrange for the rate you want. Examiners have noted that candidates who don’t first express the quantity in terms of a single variable struggle to make much progress on these questions, even when they know the chain rule.

How to spot it next time

Before differentiating, look at your formula: if it has two variables that both change (here $r$ and $h$ ), use the shape (similar triangles, a given ratio) to remove one of them first.
Write the chain rule out in full with the rates in the right places, then substitute. Check the units: $\text{cm}^3/\text{s}$ divided by $\text{cm}^2$ gives $\text{cm/s}$ .
Sense-check the size. The water surface at $h = 6$ has area $\pi (2)^2 = 4\pi \approx 12.6\ \text{cm}^2$ , and $12\ \text{cm}^3$ flows in each second, so the level should rise by roughly $1\ \text{cm}$ per second, not $2.86$ .