Exercise 6.1 · Q2: Cube volume grows at 8 cm³/s: rate of surface area at edge 12 cm

Basics you need

Solution

1. Name the quantities and write the relations

   At time $t$ seconds let the edge be $x$ cm ($x > 0$⁠), the volume $V$ cm$^3$ and the surface area $S$ cm$^2$⁠. Then

   $$V = x^3, \qquad S = 6x^2 ,$$

   where $x$⁠, and therefore $V$ and $S$⁠, are functions of $t$⁠. We are given $\dfrac{dV}{dt} = 8$ and want $\dfrac{dS}{dt}$ when $x = 12$⁠.

   The cube keeps its shape as it grows, so its size is completely described by the single number $x$⁠. Writing $V$ and $S$ in terms of $x$ shows that the edge is the common link between what we know ($V$⁠) and what we want ($S$⁠).
   The surface of a cube is six equal squares, each $x$ by $x$⁠: $S = 6 \times x^2$⁠. Basics: area and volume formulas

2. Differentiate $V = x^3$ with respect to time

   By the chain rule (outer function $x^3$⁠, inner function $x(t)$⁠):

   $$\frac{dV}{dt} = \frac{d}{dx}\left(x^3\right)\cdot\frac{dx}{dt} = 3x^2\,\frac{dx}{dt} .$$

   Using $\frac{dV}{dt} = 8$⁠:

   $$8 = 3x^2\,\frac{dx}{dt} \quad\Longrightarrow\quad \frac{dx}{dt} = \frac{8}{3x^2} .$$
   The given rate is about $V$⁠, but $S$ is a formula in $x$⁠. So we first find how fast the edge grows. We may divide by $3x^2$ because $x > 0$⁠, so $3x^2 \ne 0$⁠. Note that $\frac{dx}{dt}$ depends on $x$⁠: the volume grows at a steady 8 cm$^3$/s, but the edge grows more slowly as the cube gets bigger.
   Chain rule (Theorem 4, Section 5.3.1): $\frac{d}{dt}\big(f(x)\big) = f'(x)\,\frac{dx}{dt}$⁠. Here $f(x) = x^3$ and $f'(x) = 3x^2$⁠. Basics: related rates (chain rule)

3. Differentiate $S = 6x^2$ with respect to time

   Again by the chain rule:

   $$\begin{aligned} \frac{dS}{dt} &= \frac{d}{dx}\left(6x^2\right)\cdot\frac{dx}{dt} \\ &= 12x \cdot \frac{8}{3x^2} = \frac{32}{x} . \end{aligned}$$
   This is the step that links the two rates through the edge. We substituted the expression for $\frac{dx}{dt}$ from Step 2 and cancelled one factor $x$ (allowed since $x \ne 0$⁠). The result $\frac{32}{x}$ holds at every instant, for whatever the edge happens to be then.
   $\frac{d}{dx}(6x^2) = 12x$ by the power and constant multiple rules (Table 5.3, Section 5.3). Same pattern as $\frac{dS}{dt} = 12x\cdot\frac{3}{x^2}$ in Example 2.

4. Substitute the instant $x = 12$

   $$\left.\frac{dS}{dt}\right|_{x=12} = \frac{32}{12} = \frac{8}{3} .$$

   So at the instant the edge is 12 cm, the surface area is increasing at $\frac{8}{3}$ cm$^2$/s, about 2.67 cm$^2$/s. (At that instant the edge itself grows at $\frac{8}{3\cdot 144} = \frac{1}{54}$ cm/s.)

   The value $x = 12$ describes one instant, so it goes in only after all differentiation is done. The answer is positive, which matches "increasing". Units: $S$ is in cm$^2$ and $t$ in seconds, so the rate is in cm$^2$/s.
   Substituting $x = 12$ into $S = 6x^2$ before differentiating would turn $S$ into the constant 864 and give the false rate $0$⁠. Lesson: rate of change of quantities

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