Example 31: Conical tank with $\alpha=\tan^{-1}(0.5)$⁠: how fast the water level rises at depth 4 m

Basics you need

Solution

1. Name the changing quantities

   At time $t$ hours after some moment, let

   • $h$ = depth of the water (m), measured up the axis from the vertex,
   • $r$ = radius of the water surface (m),
   • $V$ = volume of water in the tank (m³).

   We know $\frac{dV}{dt} = 5$ m³/h at every instant, and we want $\frac{dh}{dt}$ at the instant when $h = 4$ m.

   All three quantities change as water is poured in, so they are functions of $t$⁠. Writing their units now tells us what unit the answer must have: $\frac{dh}{dt}$ will be in metres per hour.
   "Level rising" means the depth $h$ increasing, so the rate asked for is $\frac{dh}{dt}$⁠. Basics: rate of change as a derivative

2. Link $r$ and $h$ using the angle

   The water is a cone with the tank's vertex and axis, so its semi-vertical angle is also $\alpha$⁠. In its right triangle (Basic 2),

   $$\tan\alpha = \frac{r}{h}.$$

   Since $\alpha = \tan^{-1}(0.5)$⁠, we have $\tan\alpha = 0.5$⁠, so

   $$\frac{r}{h} = 0.5, \qquad r = \frac{h}{2}.$$

   This holds at every instant, whatever the depth.

   The volume formula contains two changing lengths, $r$ and $h$⁠, but we are told about $dV/dt$ and asked about $dh/dt$ only. The fixed shape of the tank lets us remove $r$⁠. Note the order of the ratio: $r$ is opposite $\alpha$⁠, $h$ is adjacent.
   $\tan(\tan^{-1}y) = y$ for every real $y$ (Chapter 2, principal value branch of $\tan^{-1}$⁠).

3. Volume in terms of the depth alone

   $$\begin{aligned} V &= \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\left(\frac{h}{2}\right)^2 h \\ &= \frac{1}{3}\pi\cdot\frac{h^2}{4}\cdot h = \frac{\pi h^3}{12}. \end{aligned}$$
   Now $V$ is a function of the single variable $h$⁠, which in turn is a function of $t$⁠. That is exactly the set-up the chain rule needs.
   Cone volume $\frac{1}{3}\pi r^2 h$⁠; $\left(\frac{h}{2}\right)^2 = \frac{h^2}{4}$⁠. Basics: area and volume formulas

4. Differentiate with respect to time (chain rule)

   $$\begin{aligned} \frac{dV}{dt} &= \frac{d}{dh}\left(\frac{\pi h^3}{12}\right)\cdot\frac{dh}{dt} \\ &= \frac{3\pi h^2}{12}\cdot\frac{dh}{dt} = \frac{\pi h^2}{4}\cdot\frac{dh}{dt}. \end{aligned}$$

   Here the outer function is $h \mapsto \frac{\pi h^3}{12}$ and the inner function is $t \mapsto h(t)$⁠.

   This equation holds at every instant. It is the bridge between the rate we know ($dV/dt$⁠) and the rate we want ($dh/dt$⁠). We differentiate before substituting $h = 4$⁠: if we put $h = 4$ first, $V$ would become the constant $\frac{16\pi}{3}$⁠, whose derivative is 0, and all information about the change would be lost.
   Chain rule: $\frac{dV}{dt} = \frac{dV}{dh}\cdot\frac{dh}{dt}$⁠; power rule $\frac{d}{dh}h^3 = 3h^2$⁠. Basics: related rates

5. Substitute the instant and solve

   At the instant, $\frac{dV}{dt} = 5$ and $h = 4$⁠:

   $$5 = \frac{\pi (4)^2}{4}\cdot\frac{dh}{dt} = 4\pi\,\frac{dh}{dt},$$ $$\frac{dh}{dt} = \frac{5}{4\pi}\ \text{m/h}.$$
   Only now, with the derivative relation in hand, do we use the particular depth. The equation is linear in $\frac{dh}{dt}$⁠, so we just divide by $4\pi$⁠.
   $\frac{\pi\cdot 16}{4} = 4\pi$⁠.

6. Put a number on it

   Exactly, $\frac{dh}{dt} = \frac{5}{4\pi} \approx 0.3979$ m/h. The textbook uses $\pi = \frac{22}{7}$⁠:

   $$\frac{5}{4\cdot\frac{22}{7}} = \frac{5\cdot 7}{88} = \frac{35}{88} \approx 0.3977\ \text{m/h}.$$

   The rate is positive, so the level is rising at $\frac{5}{4\pi}$ m/h, that is $\frac{35}{88}$ m/h with $\pi = \frac{22}{7}$, roughly 40 cm per hour.

   The two decimals differ only in the fourth place because $\frac{22}{7}$ is a close approximation to $\pi$ ($\frac{22}{7} \approx 3.1429$⁠, $\pi \approx 3.1416$⁠). The book's answer $\frac{35}{88}$ uses this approximation; $\frac{5}{4\pi}$ is the exact value.
   A positive rate of change means the quantity is increasing at that instant. Basics: rate of change as a derivative

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