Water fills a cone at 3 cm³/s — how fast is depth increasing

Question

Water is poured into an inverted conical tank at a rate of 3 cm³/s. The cone has a height of 10 cm and a radius of 5 cm. How fast is the depth of water increasing when the depth is 4 cm?

Solution — Step by Step

Given:

Rate of volume increase: $\dfrac{dV}{dt} = 3$ cm³/s
Cone dimensions: height $H = 10$ cm, radius $R = 5$ cm
At the moment of interest: depth $h = 4$ cm

Find: $\dfrac{dh}{dt}$ (rate of increase of depth)

As water fills the cone, the water surface forms a smaller cone similar to the full cone. By similar triangles:

\frac{r}{h} = \frac{R}{H} = \frac{5}{10} = \frac{1}{2}

Therefore: $r = \dfrac{h}{2}$

This is the key step — it eliminates $r$ and expresses everything in terms of $h$ alone.

Volume of a cone: $V = \dfrac{1}{3}\pi r^2 h$

Substitute $r = \dfrac{h}{2}$ :

V = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3}\pi \cdot \frac{h^2}{4} \cdot h = \frac{\pi h^3}{12}

Differentiate both sides with respect to $t$ :

\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{4} \cdot \frac{dh}{dt}

Substitute $\dfrac{dV}{dt} = 3$ and $h = 4$ :

3 = \frac{\pi (4)^2}{4} \cdot \frac{dh}{dt} = \frac{16\pi}{4} \cdot \frac{dh}{dt} = 4\pi \cdot \frac{dh}{dt}

\frac{dh}{dt} = \frac{3}{4\pi} \text{ cm/s}

\boxed{\frac{dh}{dt} = \frac{3}{4\pi} \approx 0.239 \text{ cm/s}}

Why This Works

The rate of depth increase slows as the cone gets wider. When $h$ is small, the cross-section is small — the same volume fills a narrow area, so depth rises quickly. When $h$ is large, the cross-section is larger — the same volume spreads over more area, so depth rises slowly.

Mathematically: $\dfrac{dh}{dt} = \dfrac{4 \cdot dV/dt}{\pi h^2}$ . As $h$ increases, the denominator $\pi h^2$ increases, so $\dfrac{dh}{dt}$ decreases. The depth rate is inversely proportional to $h^2$ .

Alternative Method — Check with Instantaneous Numbers

At $h = 4$ cm: $r = 2$ cm (from similar triangles).

Cross-sectional area of water surface = $\pi r^2 = 4\pi$ cm².

Rate of depth increase = Rate of volume / Area = $3 / (4\pi)$ cm/s. ✓

This makes physical sense: pouring 3 cm³ per second into a cross-section of $4\pi$ cm² raises the level at $3/(4\pi)$ cm/s.

Common Mistake

Substituting $h = 4$ into the volume formula before differentiating. If you compute $V = \pi(4)^3/12 = 16\pi/3$ (a constant) and then differentiate, you get $dV/dt = 0$ — which is wrong. Always differentiate the general expression in terms of variables first, then substitute the specific value.

Question

Solution — Step by Step

Why This Works

Alternative Method — Check with Instantaneous Numbers

Common Mistake

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