Exercise 6.1 · Q4: Cube edge grows at 3 cm/s: rate of volume at edge 10 cm

Basics you need

Solution

1. Name the quantities and write the relation

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

   $$V = x^3, \qquad \frac{dx}{dt} = 3 .$$
   "Variable cube" means the cube changes size but stays a cube, so its edge $x$ is a function of time and $V = x^3$ holds at every instant. The rate is positive because the edge is increasing.
   Name every quantity with its unit, write the relation, and translate each given rate into a derivative with the correct sign. Basics: rate of change as a derivative

2. Differentiate with respect to time (chain rule)

   Outer function $x^3$⁠, inner function $x(t)$⁠:

   $$\frac{dV}{dt} = \frac{d}{dx}\left(x^3\right)\cdot\frac{dx}{dt} = 3x^2\,\frac{dx}{dt} .$$
   $V$ is a formula in $x$⁠, not in $t$⁠. Differentiating with respect to $x$ and multiplying by $\frac{dx}{dt}$ is exactly what the chain rule allows when $x$ itself depends on $t$⁠.
   Chain rule (Theorem 4, Section 5.3.1): $\frac{d}{dt}\big(f(x)\big) = f'(x)\,\frac{dx}{dt}$⁠. Basics: related rates (chain rule)

3. Substitute the instant: $x = 10$⁠, $\frac{dx}{dt} = 3$

   $$\frac{dV}{dt} = 3(10)^2(3) = 900 .$$

   So when the edge is 10 cm, the volume is increasing at 900 cm$^3$/s.

   The values belong to one instant, so they go in after differentiating. The result is positive, which matches "increasing". Units: cm$^3$ of volume per second.
   Substituting $x = 10$ into $V$ first would give the constant 1000 and a false rate of $0$⁠. Lesson: rate of change of quantities

4. See where $3x^2$ comes from

   If the edge grows from $x$ to $x + \Delta x$⁠, the new volume is

   $$\begin{aligned} (x+\Delta x)^3 &= x^3 + 3x^2\Delta x \\ &\quad + 3x(\Delta x)^2 + (\Delta x)^3 . \end{aligned}$$

   The main part of the gain, $3x^2\,\Delta x$⁠, is three slabs of size $x \times x \times \Delta x$⁠; the other terms are thin edge strips and a tiny corner cube, negligible when $\Delta x$ is small. At $x = 10$⁠, each slab is 100 cm$^2$ in area, so the three slabs together add about $300\,\Delta x$ cm$^3$⁠.

   This shows that $\frac{dV}{dx} = 3x^2$ is half the surface area $6x^2$⁠: growing all three dimensions together adds a layer on three faces (the other three faces stay put if one corner is held fixed). Multiplying by $\frac{dx}{dt} = 3$ cm/s turns the 300 cm$^3$ per cm of edge into 900 cm$^3$ per second.
   Binomial expansion: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$⁠. The derivative keeps only the term that is first order in $\Delta x$⁠. Basics: area and volume formulas

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