Chapter 6 · Application of Derivatives

Exercise 6.1 · Q9: Sphere: $dV/dr$ at $r = 10$ cm

Exercise 6.1Rate of change of quantities · Section 6.2

The question

A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.

Exam answer

Let $V$ be the volume of the balloon of radius $r$ ⁠. Then

V = \frac43\pi r^3 \Rightarrow \frac{dV}{dr} = 4\pi r^2

At $r = 10$ ⁠,

\frac{dV}{dr} = 4\pi(10)^2 = 400\pi

$\therefore$ the volume is increasing at $400\pi$ cm³/cm.

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

The one idea this problem is built on

"The rate at which its volume is increasing with the radius" is $\dfrac{dV}{dr}$ ⁠: volume gained per centimetre of radius. No time is involved. For a sphere, $V = \frac{4}{3}\pi r^3$ gives $\dfrac{dV}{dr} = 4\pi r^2$ ⁠, which is exactly the surface area: growing the radius a little adds a thin shell, and a thin shell has volume (surface area) $\times$ (thickness).

Given, to find, and the plan

Given: A spherical balloon of variable radius $r$ cm. "The later" in the question means "the latter", that is, the radius: the instant of interest is $r = 10$ cm.
To find: $\dfrac{dV}{dr}$ at $r = 10$ ⁠.
Plan: Write $V = \frac{4}{3}\pi r^3$ ⁠, differentiate with respect to $r$ for a general radius, then substitute $r = 10$ ⁠. State the units: cm $^3$ of volume per cm of radius.

Basics you need

1A rate of change is a derivative

The rate of change of $V$ with respect to $r$ is $\frac{dV}{dr}$ ⁠, and its value at $r = r_0$ is the rate at that radius (Section 6.2). Basics: rate of change as a derivative

2Volume and surface area of a sphere

$V = \frac{4}{3}\pi r^3$ and surface area $S = 4\pi r^2$ ⁠. Basics: area and volume formulas

3Power rule and constant multiple

$\dfrac{d}{dr}(c\,r^3) = 3c\,r^2$ for a constant $c$ (Table 5.3 and the algebra of derivatives, Section 5.3).

Solution

Name the quantities and write the relation

Let the radius be $r$ cm ( $r > 0$ ⁠) and the volume $V$ cm $^3$ ⁠. Since the balloon is always a sphere,
$V = \frac{4}{3}\pi r^3 .$
The question compares volume with radius, so we need $V$ as a formula in $r$ ⁠. "Always remains spherical" guarantees the sphere formula holds for every radius the balloon takes.

Read "with the radius" (or "with respect to the radius") as: differentiate with respect to $r$ ⁠, not $t$ ⁠. Basics: rate of change as a derivative
Differentiate with respect to $r$
$\frac{dV}{dr} = \frac{4}{3}\pi\cdot 3r^2 = 4\pi r^2 .$
$\frac{4}{3}\pi$ is a constant factor, so it stays in front while $r^3$ is differentiated by the power rule. The 3 from the power rule cancels the 3 in $\frac{4}{3}$ ⁠. No chain rule is needed, because no time variable appears.

Power rule $\frac{d}{dr}(r^3) = 3r^2$ and constant multiple rule (Table 5.3, Section 5.3). Lesson: rate of change of quantities
Substitute $r = 10$
$\left.\frac{dV}{dr}\right|_{r=10} = 4\pi(10)^2 = 400\pi .$
So when the radius is 10 cm, the volume is increasing at $400\pi$ cm $^3$ per cm of radius, about 1256.64 cm $^3$ /cm.

The radius 10 cm is substituted only after differentiating. The answer is positive: the volume increases as the radius increases. Units: volume (cm $^3$ ⁠) per radius (cm), so cm $^3$ /cm.

A positive rate means increasing (Note after Example 3, Section 6.2). Basics: rate of change as a derivative
See why the answer is the surface area

Growing the radius from $r$ to $r + \Delta r$ adds a thin shell of volume
$\begin{aligned} \Delta V &= \tfrac{4}{3}\pi\big((r+\Delta r)^3 - r^3\big) \\ &= 4\pi r^2\Delta r + 4\pi r(\Delta r)^2 \\ &\quad + \tfrac{4}{3}\pi(\Delta r)^3 . \end{aligned}$
Dividing by $\Delta r$ and letting $\Delta r \to 0$ leaves $4\pi r^2$ ⁠. At $r = 10$ the surface area is $400\pi$ cm $^2$ ⁠, so a shell 1 mm thick ( $\Delta r = 0.1$ ⁠) holds about $400\pi \times 0.1 = 40\pi$ cm $^3$ ⁠.

A sphere of radius 10 cm (blue) and the thin gold shell added when the radius grows by a small amount $\Delta r$ (thickness exaggerated). The shell's volume is about (surface area) $\times$ $\Delta r = 400\pi\,\Delta r$ cm $^3$ ⁠.

This is the three-dimensional version of Exercise 6.1 Q1, where $\frac{dA}{dr}$ was the circumference. A thin shell is almost a sheet whose area is the surface area and whose thickness is $\Delta r$ ⁠; the leftover terms are tiny compared with $\Delta r$ ⁠.

Expansion: $(r + \Delta r)^3$ equals $r^3 + 3r^2\Delta r$ plus $3r(\Delta r)^2 + (\Delta r)^3$ ⁠; then multiply by $\frac{4}{3}\pi$ ⁠. Basics: area and volume formulas

Result

$\dfrac{dV}{dr} = 4\pi r^2$ ⁠, so when the radius is 10 cm the volume is increasing with the radius at

\frac{dV}{dr} = 400\pi\ \text{cm}^3/\text{cm} \approx 1256.64\ \text{cm}^3/\text{cm}.

(Textbook answer: $400\pi$ cm $^3$ /cm.)

Check it with numbers

Use a symmetric difference with $\Delta r = 0.001$ ⁠: $10.001^3 = 1000.300030001$ and $9.999^3 = 999.700029999$ ⁠, a difference of $0.600000002$ ⁠. Then

\frac{\frac{4}{3}\pi \times 0.600000002}{0.002} \approx 400.0000013\,\pi ,

which agrees with $400\pi$ (the tiny extra comes from the $(\Delta r)^3$ term).

Common mistakes

Looking for a time rate. No rate per second is given or asked; "with the radius" means $\frac{dV}{dr}$ ⁠.
Giving the volume. $\frac{4000}{3}\pi$ cm $^3$ is the volume at $r = 10$ ⁠, not its rate of change.
Losing the constant. $\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2$ ⁠, not $\frac{4}{3}\pi r^2$ or $3r^2$ ⁠.

What to take away

"Rate of change of $V$ with $r$ " is $\frac{dV}{dr}$ ⁠; a time rate would need $\frac{dr}{dt}$ as well (as in Exercise 6.1 Q8).
For a sphere $\frac{dV}{dr} = 4\pi r^2$ is the surface area, just as for a circle $\frac{dA}{dr} = 2\pi r$ is the circumference (Exercise 6.1 Q1).
Differentiate in general, then substitute the given radius.

Every fact used, and where it comes from

Fact	Used in	Source
Rate of change of $V$ with respect to $r$ is $\frac{dV}{dr}$	Steps 2–3	Section 6.2 (Lesson: rate of change of quantities)
$V = \frac{4}{3}\pi r^3$ ⁠, $S = 4\pi r^2$	Steps 1, 4	Mensuration (Basics: area and volume formulas)
$\frac{d}{dr}(r^3) = 3r^2$ ⁠; constant multiple rule	Step 2	Table 5.3 and algebra of derivatives, Section 5.3
Positive rate means increasing	Step 3	Note after Example 3, Section 6.2
Derivative as the limit of $\frac{\Delta V}{\Delta r}$	Step 4	Definition of the derivative, Section 5.3

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 6.1 · Q9: Sphere: dV/drdV/drdV/dr at r=10r = 10r=10 cm

Basics you need

Solution

Name the quantities and write the relation

Differentiate with respect to rrr

Substitute r=10r = 10r=10

See why the answer is the surface area

Every fact used, and where it comes from

Exercise 6.1 · Q9: Sphere: $dV/dr$ at $r = 10$ cm

Differentiate with respect to $r$

Substitute $r = 10$