Exercise 6.1 · Q14: Sand cone with height $\tfrac16$ of the radius: how fast does it rise?

Basics you need

Solution

1. Name the variables

   At time $t$ seconds, let the sand cone have base radius $r$ cm, height $h$ cm and volume $V$ cm$^3$⁠. All the sand that falls stays in the cone, so the volume grows at the rate the sand arrives:

   $$\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}.$$

   The shape condition holds at every moment:

   $$h = \frac{r}{6}, \qquad\text{equivalently}\qquad r = 6h.$$
   The pipe's rate is a rate of volume, so it is $\frac{dV}{dt}$⁠. The question wants the rate of the height, $\frac{dh}{dt}$⁠. The shape condition is what lets us connect the two without needing to know $\frac{dr}{dt}$ separately.

2. Write the volume in terms of $h$ alone

   Substitute $r = 6h$ into the cone formula:

   $$\begin{aligned} V &= \frac13\pi r^2 h = \frac13\pi(6h)^2 h \\ &= \frac13\pi\cdot 36h^2\cdot h = 12\pi h^3. \end{aligned}$$
   With one variable instead of two, a single application of the chain rule finishes the job. We eliminate $r$⁠, not $h$⁠, because $h$ is the quantity whose rate we want and whose value (4 cm) we are given.
   Cone volume $\frac13\pi r^2h$⁠; $(6h)^2 = 36h^2$⁠. Basics: area and volume formulas

3. Differentiate with respect to time

   $V$ depends on $t$ through $h$⁠. By the chain rule,

   $$\begin{aligned} \frac{dV}{dt} &= \frac{d}{dh}\left(12\pi h^3\right)\cdot\frac{dh}{dt} \\ &= 36\pi h^2\,\frac{dh}{dt}. \end{aligned}$$
   The relation $V = 12\pi h^3$ holds at every moment, so it can be differentiated with respect to $t$⁠. Notice $36\pi h^2 = \pi(6h)^2 = \pi r^2$⁠, the area of the cone's base: a thin extra layer of sand, $dh$ thick when measured vertically, covers the base area seen from above and so adds roughly (base area) $\times\, dh$ of volume.
   Chain rule (Theorem 4, Section 5.3.1), $\frac{d}{dh}\left(h^3\right) = 3h^2$ (Table 5.3). Compare Example 2, Section 6.2. Basics: related rates (chain rule)

4. Substitute and solve for $\frac{dh}{dt}$

   At the instant, $h = 4$ and $\frac{dV}{dt} = 12$⁠:

   $$\begin{aligned} 12 &= 36\pi(4)^2\,\frac{dh}{dt} \\ &= 576\pi\,\frac{dh}{dt}, \\ \frac{dh}{dt} &= \frac{12}{576\pi} \\ &= \frac{1}{48\pi} \text{ cm/s}. \end{aligned}$$

   As a decimal, $\frac{1}{48\pi} \approx 0.00663$ cm/s.

   $36\times 16 = 576$ and $\frac{12}{576} = \frac{1}{48}$⁠. The units: cm$^3$/s divided by $\pi h^2$ in cm$^2$ leaves cm/s, a speed for the height, as required.

5. Interpret, and check by the product rule

   $\frac{dh}{dt} > 0$⁠, so the height is increasing, but very slowly. The reason: at $h = 4$ cm the radius is $r = 24$ cm, so the base area is $576\pi \approx 1809.56$ cm$^2$⁠, and 12 cm$^3$ per second spread over that area is a very thin layer.

   A second route, without eliminating $r$ first: differentiate $V = \frac13\pi r^2h$ by the product rule, using $\frac{dr}{dt} = 6\,\frac{dh}{dt}$ (from $r = 6h$⁠):

   $$\begin{aligned} \frac{dV}{dt} &= \frac13\pi\left(2rh\,\frac{dr}{dt} + r^2\,\frac{dh}{dt}\right) \\ &= \frac13\pi\left(12rh + r^2\right)\frac{dh}{dt}. \end{aligned}$$

   At $r = 24$⁠, $h = 4$⁠: $\frac13\pi(1152 + 576) = 576\pi$⁠, so again $12 = 576\pi\,\frac{dh}{dt}$ and $\frac{dh}{dt} = \frac{1}{48\pi}$⁠.

   The second route keeps both variables and uses the product rule; it must agree with the first, and it does. Seeing the same factor $576\pi$ both ways confirms the substitution in Step 2.
   Product rule (Section 5.3, rule (2)). Lesson: rate of change of quantities

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