Chapter 6 · Application of Derivatives

Example 2: Cube volume grows at 9 cm³/s: surface-area rate at edge 10 cm

ExamplesRate of change of quantities · Section 6.2

The question

The volume of a cube is increasing at a rate of 9 cubic centimetres per second. How fast is the surface area increasing when the length of an edge is 10 centimetres?

Exam answer

Let the edge be $x$ cm, the volume $V$ cm $^3$ and the surface area $S$ cm $^2$ at time $t$ s. Given $\frac{dV}{dt} = 9$ cm $^3$ /s.

V = x^3, \quad S = 6x^2

Differentiating w.r.t. $t$ ⁠,

\begin{aligned} \frac{dV}{dt} &= 3x^2\frac{dx}{dt} \\ \Rightarrow\ 9 &= 3x^2\frac{dx}{dt} \\ \Rightarrow\ \frac{dx}{dt} &= \frac{3}{x^2} \end{aligned}

\begin{aligned} \frac{dS}{dt} &= 12x\frac{dx}{dt} \\ &= 12x\cdot\frac{3}{x^2} = \frac{36}{x} \end{aligned}

At $x = 10$ ⁠, $\frac{dS}{dt} = \frac{36}{10} = 3.6$ ⁠.

Hence the surface area is increasing at the rate of 3.6 cm $^2$ /s.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The volume and the surface area are both formulas in one quantity, the edge $x$ ⁠, and all three change with time. The chain rule turns the known rate $\frac{dV}{dt}$ into the rate of the edge, $\frac{dx}{dt}$ ⁠, and turns that into the rate we want, $\frac{dS}{dt}$ ⁠. The edge is the bridge between the two rates.

Given, to find, and the plan

Given: A cube whose volume increases at $\frac{dV}{dt} = 9$ cm $^3$ /s. The instant of interest is when the edge is $x = 10$ cm.
To find: $\dfrac{dS}{dt}$ ⁠, the rate at which the surface area changes, at that instant.
Plan: Write $V = x^3$ and $S = 6x^2$ ⁠. Differentiate $V$ with respect to time to get $\frac{dx}{dt}$ in terms of $x$ ⁠. Differentiate $S$ with respect to time and substitute. Put $x = 10$ only at the end.

Basics you need

1Volume and surface area of a cube

A cube of edge $x$ has volume $V = x^3$ ⁠, and six square faces of area $x^2$ each, so surface area $S = 6x^2$ ⁠. Basics: area and volume formulas

2Related rates (chain rule)

If $V$ is a function of $x$ and $x$ is a function of time $t$ ⁠, then $\dfrac{dV}{dt} = \dfrac{dV}{dx}\cdot\dfrac{dx}{dt}$ (Section 6.2; Theorem 4, Section 5.3.1). Basics: related rates (chain rule)

3A rate of change is a derivative

"The volume is increasing at 9 cm $^3$ /s" means $\frac{dV}{dt} = 9$ ⁠; a positive rate means the quantity increases (Note after Example 3, Section 6.2). Basics: rate of change as a derivative

Solution

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 so also $V$ and $S$ ⁠, are functions of $t$ ⁠. We know $\frac{dV}{dt} = 9$ and want $\frac{dS}{dt}$ when $x = 10$ ⁠.

A cube of edge $x$ ⁠. The volume $x^3$ and the surface area $6x^2$ both depend on the one length $x$ ⁠, which changes with time.

A cube stays a cube as it grows, so its whole size is described by the single number $x$ ⁠. Writing both $V$ and $S$ in terms of $x$ shows that $x$ links what we know to what we want.

The surface is six equal squares, each $x$ by $x$ ⁠: $S = 6 \times x^2$ ⁠. Basics: area and volume formulas
Differentiate $V = x^3$ with respect to time

By the chain rule, with outer function $x^3$ and inner function $x(t)$ ⁠:
$\frac{dV}{dt} = \frac{d}{dx}\left(x^3\right)\cdot\frac{dx}{dt} = 3x^2\,\frac{dx}{dt} .$
Put in $\frac{dV}{dt} = 9$ and divide by $3x^2$ ⁠:
$9 = 3x^2\,\frac{dx}{dt} \;\Longrightarrow\; \frac{dx}{dt} = \frac{3}{x^2} . \qquad (1)$
The given rate is about $V$ ⁠, but $S$ is a formula in $x$ ⁠, so we first need how fast the edge grows. Dividing by $3x^2$ is allowed because $x > 0$ ⁠. Note that $\frac{dx}{dt}$ depends on $x$ ⁠: the volume grows steadily, yet the edge grows more slowly as the cube gets bigger.

Chain rule: $\frac{d}{dt}f(x) = f'(x)\,\frac{dx}{dt}$ ⁠; here $f(x) = x^3$ ⁠, $f'(x) = 3x^2$ ⁠. Basics: related rates (chain rule)
Differentiate $S = 6x^2$ with respect to time

By the chain rule again, then using (1):
$\begin{aligned} \frac{dS}{dt} &= \frac{d}{dx}\left(6x^2\right)\cdot\frac{dx}{dt} \\ &= 12x \cdot \frac{3}{x^2} = \frac{36}{x} . \end{aligned}$
This is where the two rates meet: $\frac{dS}{dt}$ needs $\frac{dx}{dt}$ ⁠, and (1) supplies it. One factor $x$ cancels (allowed, $x \ne 0$ ⁠). The formula $\frac{36}{x}$ holds at every instant, whatever the edge is then.

$\frac{d}{dx}(6x^2) = 6 \cdot 2x = 12x$ by the power and constant multiple rules (Section 5.3).
Substitute the instant $x = 10$
$\left.\frac{dS}{dt}\right|_{x=10} = \frac{36}{10} = 3.6 .$
So when the edge is 10 cm, the surface area is increasing at 3.6 cm $^2$ /s. At that instant the edge itself grows at $\frac{3}{10^2} = 0.03$ cm/s, by (1).

$x = 10$ describes one instant, so it is put in only after all the differentiation. The answer is positive, which matches "increasing". Units: $S$ in cm $^2$ and $t$ in seconds give cm $^2$ /s.

Putting $x = 10$ into $S = 6x^2$ before differentiating would make $S$ the constant 600 and give the false rate 0. Lesson: rate of change of quantities

Result

$\dfrac{dS}{dt} = \dfrac{36}{x}$ ⁠, so when the edge is 10 cm the surface area is increasing at

\frac{dS}{dt} = 3.6\ \text{cm}^2/\text{s}.

This agrees with the book's solution.

Check it with numbers

A second route. Write $S$ in terms of $V$ ⁠: $x = V^{1/3}$ ⁠, so $S = 6V^{2/3}$ and $\frac{dS}{dV} = 4V^{-1/3} = \frac{4}{x}$ ⁠. At $x = 10$ this is 0.4, and $\frac{dS}{dt} = \frac{dS}{dV}\cdot\frac{dV}{dt} = 0.4 \times 9 = 3.6$ ⁠. Same answer.

One second later. Start the clock when $x = 10$ ⁠, so $V = 1000$ ⁠. After 1 s, $V = 1009$ ⁠, the edge is $1009^{1/3} \approx 10.0299$ cm and $S = 6 \times 1009^{2/3}$ ⁠. The surface area has grown by about 3.595 cm $^2$ ⁠, very close to the instantaneous rate 3.6 (slightly less, because $\frac{36}{x}$ falls as $x$ grows).

Common mistakes

Treating 9 as the rate of the edge. 9 cm $^3$ /s is $\frac{dV}{dt}$ ⁠; the unit cm $^3$ says it is a volume rate.
Leaving out the chain-rule factor. $\frac{dS}{dt} = 12x$ forgets $\frac{dx}{dt}$ and gives 120, which is $\frac{dS}{dx}$ ⁠, a rate with respect to the edge, not time.
Substituting $x = 10$ too early, which turns $V$ and $S$ into constants with zero derivative.
Wrong surface area. A cube has six faces: $S = 6x^2$ ⁠, not $4x^2$ or $x^2$ ⁠.

What to take away

When two quantities depend on a common variable, differentiate both with respect to time and connect them through that variable's rate.
A steady volume rate does not give a steady surface-area rate: here $\frac{dS}{dt} = \frac{36}{x}$ falls as the cube grows.
Differentiate first, substitute the instant last.

Every fact used, and where it comes from

Fact	Used in	Source
$V = x^3$ ⁠, $S = 6x^2$ for a cube	Step 1	Mensuration (Basics: area and volume formulas)
Chain rule for rates with respect to time	Steps 2, 3	Section 6.2; Theorem 4, Section 5.3.1 (Basics: related rates)
$\frac{d}{dx}(x^n) = nx^{n-1}$ ⁠; constant multiple rule	Steps 2, 3	Table 5.3 and the algebra of derivatives, Section 5.3
Positive rate means increasing	Step 4	Note after Example 3, Section 6.2 (Lesson: rate of change of quantities)
Values 3.6, 0.03, 10.0299, 3.595	Step 4; Check	Arithmetic, recomputed in the site's verification script

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Example 2: Cube volume grows at 9 cm³/s: surface-area rate at edge 10 cm

Basics you need

Solution

Name the quantities and write the relations

Differentiate V=x3V = x^3V=x3 with respect to time

Differentiate S=6x2S = 6x^2S=6x2 with respect to time

Substitute the instant x=10x = 10x=10

Every fact used, and where it comes from

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

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

Substitute the instant $x = 10$