KBAT: Related Rates of Change

Q: Why does the radius slow down if air goes in at a constant rate?

Because the rate works out as drdt=3r2\frac{dr}{dt}=\frac{3}{r^{2}}dtdr=r23, which is inversely proportional to r2r^{2}r2. A larger sphere needs far more volume for each extra centimetre of radius, that is what dVdr=4πr2\frac{dV}{dr}=4\pi r^{2}drdV=4πr2 says, so the same steady inflow raises the radius less and less. Doubling the radius from 333 to 666 cm makes it grow four times more slowly, from 13\frac{1}{3}31 to 112\frac{1}{12}121 cm s−1^{-1}−1.

Q: What is the most common mistake?

Substituting the value of rrr before differentiating, so the derivative becomes a constant and the relationship disappears, or confusing ddt\frac{d}{dt}dtd with ddr\frac{d}{dr}drd. Dropping the π\piπ is another. Add Math Paper 2 is 2 hours 30 minutes and 100 marks with analytic marking, so a correctly stated chain rule earns method marks even if the arithmetic later slips.

Q: How did you find \frac{dA}{dt} without a fresh calculation of \frac{dr}{dt}?

By chaining again: dAdt=dAdr⋅drdt=8πr⋅3r2=24πr\frac{dA}{dt}=\frac{dA}{dr}\cdot\frac{dr}{dt}=8\pi r\cdot\frac{3}{r^{2}}=\frac{24\pi}{r}dtdA=drdA⋅dtdr=8πr⋅r23=r24π. Because drdt\frac{dr}{dt}dtdr was already found in general form, the area rate follows in one substitution, giving 8π8\pi8π cm2^{2}2 s−1^{-1}−1 at r=3r=3r=3. Reusing the general rate is quicker and keeps the working tidy.

What makes this a KBAT question

A routine differentiation question hands you $y$ in terms of $x$ and asks for $\frac{dy}{dx}$ . A related-rates KBAT question gives you a rate you can measure, air pumped into a balloon, and asks for a rate you cannot measure directly, like how fast the radius grows, then wants you to explain the result.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: the chain rule itself is ordinary Form 5 work, but deciding which quantities are linked, building the bridge $\frac{dV}{dr}$ between them, and reading why the radius slows as the balloon grows are left to you. In Add Math this rewards students who see differentiation as a way to connect changing quantities in the real world, not just a rule for turning $r^{3}$ into $3r^{2}$ .

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Try it yourself before reading the solution.

Q1[9 marks]

A spherical balloon is inflated so that air is pumped in at a constant rate of $12\pi$ cm $^3$ s $^{-1}$ . The balloon keeps its spherical shape, with volume $V=\frac{4}{3}\pi r^{3}$ and surface area $A=4\pi r^{2}$ , where $r$ is the radius in cm.

(a) Find the rate at which the radius is increasing when $r = 3$ cm. (b) Find the rate at which the surface area is increasing at that instant.

(c) Show that $\frac{dr}{dt}=\frac{3}{r^{2}}$ , and use this to explain why the radius grows more and more slowly as the balloon gets bigger. Find $\frac{dr}{dt}$ when $r = 6$ cm and compare it with your answer in (a).

Show worked solution

Understand. The volume $V$ depends on the radius $r$ , and $r$ changes with time $t$ .

The chain rule links their rates: $\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}$ . We are given $\frac{dV}{dt}=12\pi$ and want $\frac{dr}{dt}$ and $\frac{dA}{dt}$ .

Plan. Differentiate $V$ with respect to $r$ to build the bridge $\frac{dV}{dr}$ ; rearrange the chain rule for $\frac{dr}{dt}$ .

For the area, use $\frac{dA}{dt}=\frac{dA}{dr}\cdot\frac{dr}{dt}$ . Keep $r$ as a symbol until the last line of each part.

Execute and check. (a) Differentiate the volume:

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

Rearrange the chain rule for the rate we want:

\frac{dr}{dt}=\frac{dV}{dt}\div\frac{dV}{dr}=\frac{12\pi}{4\pi r^{2}}=\frac{3}{r^{2}}

At $r = 3$ :

\frac{dr}{dt}=\frac{3}{3^{2}}=\frac{3}{9}=\frac{1}{3}\ \text{cm s}^{-1}

(b) For the surface area, $A=4\pi r^{2}$ gives $\frac{dA}{dr}=8\pi r$ , so:

\frac{dA}{dt}=\frac{dA}{dr}\cdot\frac{dr}{dt}=8\pi r\cdot\frac{3}{r^{2}}=\frac{24\pi}{r}

\text{At }r=3:\quad\frac{dA}{dt}=\frac{24\pi}{3}=8\pi\ \text{cm}^{2}\,\text{s}^{-1}\;(\approx 25.13\ \text{cm}^{2}\,\text{s}^{-1})

(c) From part (a) the general rate is $\frac{dr}{dt}=\frac{3}{r^{2}}$ , which is inversely proportional to $r^{2}$ . So as $r$ grows, $\frac{dr}{dt}$ falls quickly.

At $r = 6$ :

\frac{dr}{dt}=\frac{3}{6^{2}}=\frac{3}{36}=\frac{1}{12}\ \text{cm s}^{-1}

That is a quarter of the $\frac{1}{3}$ cm s $^{-1}$ found at $r = 3$ : doubling the radius makes it grow four times more slowly. The reason is that $\frac{dV}{dr}=4\pi r^{2}$ grows with $r^{2}$ , so a larger balloon swallows far more volume for each extra centimetre of radius; the same steady inflow then lifts the radius only slightly.

Check. The units are right: $\frac{dr}{dt}$ in cm s $^{-1}$ and $\frac{dA}{dt}$ in cm $^{2}$ s $^{-1}$ , both positive because the balloon is growing.

As a strong check, put the general rate back into the chain rule: $\frac{dV}{dt}=4\pi r^{2}\cdot\frac{3}{r^{2}}=12\pi$ , recovering the given inflow at every radius.

Finding a sensible first step

The dependable first move on any related-rates question is to write the chain that links the rates before differentiating: $\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}$ . Note what you are given, $\frac{dV}{dt}=12\pi$ , and what you want, $\frac{dr}{dt}$ ; the two are joined only through $r$ by the bridge $\frac{dV}{dr}$ .

So the true first step is to write the volume formula and differentiate it with respect to $r$ , giving $\frac{dV}{dr}=4\pi r^{2}$ . Resist substituting $r = 3$ too early: keep $r$ as a symbol so the relationship $\frac{dr}{dt}=\frac{3}{r^{2}}$ stays visible, which is exactly what part (c) needs.

Substitute the value of $r$ only at the end of each part. Naming the chain first stops the most common error, differentiating with respect to $t$ when you mean $r$ .

What markers reward

Marking is analytic, so method marks are awarded line by line. On a related-rates question a marker looks for:

The chain rule written down: $\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}$ .
Correct differentiation of the linking formula: $\frac{dV}{dr}=4\pi r^{2}$ from $V=\frac{4}{3}\pi r^{3}$ .
The rate rearranged as $\frac{dr}{dt}=\frac{12\pi}{4\pi r^{2}}=\frac{3}{r^{2}}$ , then $\frac{1}{3}$ cm s $^{-1}$ at $r=3$ .
The surface-area chain $\frac{dA}{dt}=8\pi r\cdot\frac{dr}{dt}=\frac{24\pi}{r}$ , giving $8\pi$ cm $^{2}$ s $^{-1}$ .
Exact answers kept in terms of $\pi$ , with units on every rate.
The general $\frac{dr}{dt}=\frac{3}{r^{2}}$ used to reach $\frac{1}{12}$ cm s $^{-1}$ at $r=6$ .
A sentence of reasoning linking the $r^{2}$ growth of $\frac{dV}{dr}$ to the slowing radius.

How a teacher helps

Related-rates questions reward students who set up the chain cleanly and then explain in words, and both habits grow fastest with feedback. In a one-to-one lesson our teachers ask you to write the chain rule before differentiating anything, to name which variable is the bridge, and to keep the radius symbolic so the general rate stays in view.

We treat the explanation in part (c) as part of the answer, because that sentence about the $r^{2}$ term earns the reasoning mark. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

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Frequently asked questions

Why does the radius slow down if air goes in at a constant rate?

Because the rate works out as $\frac{dr}{dt}=\frac{3}{r^{2}}$ , which is inversely proportional to $r^{2}$ . A larger sphere needs far more volume for each extra centimetre of radius, that is what $\frac{dV}{dr}=4\pi r^{2}$ says, so the same steady inflow raises the radius less and less.

Doubling the radius from $3$ to $6$ cm makes it grow four times more slowly, from $\frac{1}{3}$ to $\frac{1}{12}$ cm s $^{-1}$ .

How do I set up a related-rates problem?

Write the chain rule that links the given rate to the wanted rate, identify the variable that bridges them, and differentiate the connecting formula with respect to that variable. Here $\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}$ , the bridge is $\frac{dV}{dr}=4\pi r^{2}$ , and rearranging gives the rate you want.

Keep the variable symbolic and substitute its value only at the end.

What is the most common mistake?

Substituting the value of $r$ before differentiating, so the derivative becomes a constant and the relationship disappears, or confusing $\frac{d}{dt}$ with $\frac{d}{dr}$ . Dropping the $\pi$ is another.

Add Math Paper 2 is 2 hours 30 minutes and 100 marks with analytic marking, so a correctly stated chain rule earns method marks even if the arithmetic later slips.

How did you find

\frac{dA}{dt}

without a fresh calculation of

\frac{dr}{dt}

By chaining again: $\frac{dA}{dt}=\frac{dA}{dr}\cdot\frac{dr}{dt}=8\pi r\cdot\frac{3}{r^{2}}=\frac{24\pi}{r}$ . Because $\frac{dr}{dt}$ was already found in general form, the area rate follows in one substitution, giving $8\pi$ cm $^{2}$ s $^{-1}$ at $r=3$ .

Reusing the general rate is quicker and keeps the working tidy.

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026