G4PureStretch

Connected Rates of Change

When several quantities change with time and are linked by an equation, the chain rule connects their rates. This is the basis of many real-world “how fast is …” problems.

30 min Differentiation

The big picture

For a quantity V determined by a changing radius r, $dV/dt=(dV/dr)(dr/dt)$ . Rearrange only after the two derivatives have been identified. If the radius changes, its rate need not be constant even when the volume rate is constant. Use the geometry to link the variables. Units reveal a wrong chain: cubic centimetres per second divided by square centimetres gives centimetres per second. State whether the radius is increasing or decreasing.

Edexcel A level Maths: 9.10 Connected Rates of ChangeCurated by Kepler, not produced by us. Watch it, then work through the notes below.

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What you'll be able to do

Link rates with the chain rule
Set up dV/dt = dV/dr · dr/dt type relations
Identify the known and required rates
Solve connected-rate problems

Connecting rates

If quantities are related, their rates of change are connected by the chain rule. For example, $\dfrac{dV}{dt} = \dfrac{dV}{dr}\cdot\dfrac{dr}{dt}$ .

\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt}

Chain rule linking rates.

Example 1A sphere’s radius grows at

2

cm/s. Find

\dfrac{dV}{dt}

when

r = 3

. (

V = \tfrac43\pi r^3

\tfrac{dV}{dr} = 4\pi r^2

\tfrac{dV}{dt} = 4\pi r^2 \cdot \tfrac{dr}{dt} = 4\pi(9)(2)

Answer

72\pi

cm³/s

Tip — Write down each rate you know and the one you want, then bridge them with the chain rule.

Choose the intermediate variable and check the units

Use the geometry to link the variables. Units reveal a wrong chain: cubic centimetres per second divided by square centimetres gives centimetres per second. State whether the radius is increasing or decreasing.

Example 1Water fills a spherical balloon at

36\pi\text{ cm}^3\text{/s}

. Find the rate of increase of radius when r = 3 cm.

V=\tfrac43\pi r^3

, so

dV/dr=4\pi r^2

36\pi=4\pi(3)^2\,dr/dt

dr/dt=36\pi/(36\pi)=1

cm/s.

4The radius rate decreases at larger radius because the same volume increase is spread over more surface area.

AnswerThe radius is increasing at 1 cm/s at that instant.

Think like an examiner

Write the defining equations and show the substitution before evaluating.
Check the domain, units and interpretation of the final result.

Common misconceptions

Adding the rates instead of using the chain rule.

Multiply the linking derivatives.

Substituting the value of r before differentiating.

Differentiate first, then substitute.

Key results to remember

\frac{dA}{dt} = \frac{dA}{dx}\cdot\frac{dx}{dt}

General connected rates.

V = \tfrac43\pi r^3 \Rightarrow \frac{dV}{dr} = 4\pi r^2

Sphere example.

Key takeaways

Connected rates link via the chain rule.
dV/dt = dV/dr · dr/dt.
List known/required rates, then bridge them; substitute last.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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