Chapter 6 · Application of Derivatives

Exercise 6.1 · Q8: Balloon filled at 900 cm³/s: rate of the radius at $r = 15$ cm

Exercise 6.1Rate of change of quantities · Section 6.2

The question

A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Exam answer

Let $r$ cm be the radius and $V$ cm³ the volume at time $t$ s. Given $\frac{dV}{dt} = 900$ cm³/s.

V = \frac43\pi r^3

Differentiating w.r.t. $t$ ⁠,

\begin{aligned} \frac{dV}{dt} &= 4\pi r^2\,\frac{dr}{dt} \\ \Rightarrow\ \frac{dr}{dt} &= \frac{900}{4\pi r^2} = \frac{225}{\pi r^2} \end{aligned}

At $r = 15$ ⁠,

\frac{dr}{dt} = \frac{225}{\pi(15)^2} = \frac{1}{\pi}

$\therefore$ the radius is increasing at $\dfrac{1}{\pi}$ cm/s.

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

The one idea this problem is built on

The gas pumped in per second is the rate at which the balloon's volume grows: $\frac{dV}{dt} = 900$ ⁠. We want the rate of the radius. Differentiate $V = \frac{4}{3}\pi r^3$ with respect to time (chain rule) to get $\frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt}$ ⁠, then solve for $\frac{dr}{dt}$ ⁠. Here the unknown rate is the one inside the chain rule, not the one we compute directly.

Given, to find, and the plan

Given: The balloon stays a sphere; gas enters at 900 cm $^3$ /s, so $\dfrac{dV}{dt} = 900$ ⁠. The instant of interest: $r = 15$ cm.
To find: $\dfrac{dr}{dt}$ at $r = 15$ ⁠.
Plan: Write $V = \frac{4}{3}\pi r^3$ ⁠, differentiate with respect to $t$ ⁠, substitute $\frac{dV}{dt} = 900$ and $r = 15$ ⁠, and solve the resulting equation for $\frac{dr}{dt}$ ⁠.

Basics you need

1Volume of a sphere

A sphere of radius $r$ has volume $V = \frac{4}{3}\pi r^3$ (and surface area $4\pi r^2$ ⁠). Basics: area and volume formulas

2Related rates (chain rule)

$\dfrac{dV}{dt} = \dfrac{dV}{dr}\cdot\dfrac{dr}{dt}$ (Section 6.2; Theorem 4, Section 5.3.1). Basics: related rates (chain rule)

3Gas pumped in is the volume rate

If 900 cm $^3$ of gas enter every second and none escapes, the volume grows by 900 cm $^3$ per second: $\frac{dV}{dt} = 900$ ⁠. Basics: rate of change as a derivative

Solution

Name the quantities and write the relation

At time $t$ seconds let the radius be $r$ cm ( $r > 0$ ⁠) and the volume $V$ cm $^3$ ⁠. Since the balloon is always a sphere,
$V = \frac{4}{3}\pi r^3, \qquad \frac{dV}{dt} = 900 .$
"Always remains spherical" is what lets us use the sphere formula at every instant; without it the volume would not be determined by one radius. The gas rate is positive because the volume increases.

The words of the question decide which formula applies; check that the shape stays the same while it changes size. Basics: area and volume formulas
Differentiate with respect to time (chain rule)

Outer function $\frac{4}{3}\pi r^3$ ⁠, inner function $r(t)$ ⁠:
$\begin{aligned} \frac{dV}{dt} &= \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)\cdot\frac{dr}{dt} \\ &= 4\pi r^2\,\frac{dr}{dt} . \end{aligned}$
The relation links $V$ and $r$ ⁠, and both change with time. Differentiating it with respect to $t$ links their rates, which is what the question is about. The factor $\frac{4}{3}\cdot 3 = 4$ comes from the power rule.

Chain rule (Theorem 4, Section 5.3.1) with $f(r) = \frac{4}{3}\pi r^3$ and $f'(r) = 4\pi r^2$ ⁠, which is the sphere's surface area. Basics: related rates (chain rule)
Solve for $\frac{dr}{dt}$

Since $r > 0$ ⁠, $4\pi r^2 \ne 0$ and we may divide:
$\frac{dr}{dt} = \frac{1}{4\pi r^2}\,\frac{dV}{dt} = \frac{900}{4\pi r^2} = \frac{225}{\pi r^2} .$
This time the given rate is $\frac{dV}{dt}$ and the unknown is $\frac{dr}{dt}$ ⁠, so we rearrange the chain-rule equation. Doing this for a general $r$ first shows that the radius grows more and more slowly as the balloon gets bigger.

Dividing by an expression is only allowed when it is non-zero; here $r > 0$ guarantees it. Lesson: rate of change of quantities
Substitute the instant $r = 15$
$\left.\frac{dr}{dt}\right|_{r=15} = \frac{225}{\pi(15)^2} = \frac{225}{225\pi} = \frac{1}{\pi} .$
So when the radius is 15 cm, the radius is increasing at $\frac{1}{\pi}$ cm/s, about 0.318 cm/s.

The instant goes in last. The answer is positive, so the radius increases, as it must while gas is pumped in. Units: the radius is a length in cm and time is in seconds, so cm/s.

At $r = 15$ the surface area is $4\pi(15)^2 = 900\pi$ cm $^2$ ⁠, and $\frac{900}{900\pi} = \frac{1}{\pi}$ ⁠: the volume rate divided by the surface area. Basics: area and volume formulas

The balloon at the instant $r = 15$ cm. Gas arrives at 900 cm $^3$ /s and spreads as a thin layer over the whole surface, $900\pi$ cm $^2$ ⁠, so the layer thickens at $\frac{900}{900\pi} = \frac{1}{\pi}$ cm per second. The layer's thickness is exaggerated.

Result

$\dfrac{dr}{dt} = \dfrac{225}{\pi r^2}$ ⁠, so when the radius is 15 cm it is increasing at

\frac{dr}{dt} = \frac{1}{\pi}\ \text{cm/s} \approx 0.318\ \text{cm/s}.

(Textbook answer: $\frac{1}{\pi}$ cm/s.)

Check it with numbers

At $r = 15$ the volume is $\frac{4}{3}\pi(15)^3 = 4500\pi \approx 14137.17$ cm $^3$ ⁠. A hundredth of a second later 9 cm $^3$ more gas has entered, and a hundredth of a second earlier there was 9 cm $^3$ less. Solving $r = \left(\frac{3V}{4\pi}\right)^{1/3}$ for these two volumes gives radii of about 15.0031824 cm and 14.9968162 cm, so

\frac{15.0031824 - 14.9968162}{0.02} \approx 0.318310 ,

which matches $\frac{1}{\pi} \approx 0.318310$ ⁠.

Common mistakes

Dividing by the volume. $\frac{900}{4500\pi}$ is not the answer (its units are 1/s, not cm/s). The factor linking the rates is $\frac{dV}{dr} = 4\pi r^2$ ⁠, not $V$ itself.
Forgetting to solve. The chain rule gives $900 = 4\pi r^2\,\frac{dr}{dt}$ ⁠; the answer is $\frac{dr}{dt}$ ⁠, not $4\pi r^2 = 900\pi$ ⁠.
Using the circle area $\pi r^2$ instead of the sphere volume $\frac{4}{3}\pi r^3$ ⁠.
Substituting $r = 15$ into $V$ first, which makes $V$ the constant $4500\pi$ ⁠.

What to take away

In a related-rates problem the unknown rate can sit on either side of the chain rule; differentiate the relation, then solve for what is asked.
For a sphere, $\frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt}$ ⁠: volume rate = surface area $\times$ rate of the radius.
A steady inflow makes the radius grow ever more slowly: $\frac{dr}{dt} = \frac{225}{\pi r^2}$ falls as $r$ grows.

Every fact used, and where it comes from

Fact	Used in	Source
$V = \frac{4}{3}\pi r^3$ ⁠; surface area $4\pi r^2$	Steps 1, 4	Mensuration (Basics: area and volume formulas)
Chain rule for rates with respect to time	Step 2	Section 6.2; Theorem 4, Section 5.3.1 (Basics: related rates)
$\frac{d}{dr}(r^3) = 3r^2$ ⁠; constant multiple rule	Step 2	Table 5.3 and algebra of derivatives, Section 5.3
Positive rate means increasing	Step 4	Note after Example 3, Section 6.2 (Lesson: rate of change of quantities)
Solving the chain-rule equation for the unknown rate	Step 3	Method of Example 2, equation (1), Section 6.2

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Exercise 6.1 · Q8: Balloon filled at 900 cm³/s: rate of the radius at r=15r = 15r=15 cm

Basics you need

Solution

Name the quantities and write the relation

Differentiate with respect to time (chain rule)

Solve for drdt\frac{dr}{dt}dtdr​

Substitute the instant r=15r = 15r=15

Every fact used, and where it comes from

Exercise 6.1 · Q8: Balloon filled at 900 cm³/s: rate of the radius at $r = 15$ cm

Solve for $\frac{dr}{dt}$

Substitute the instant $r = 15$