Mathematics 9709 · Pure Mathematics 1

Rates of change

A rate of change is how fast something changes over time. Think of a balloon being blown up, an oil patch spreading, or a pile of ore growing.
Use the chain rule from §7.2 to link a rate you are given to a rate you want.

Connected rates link through a shared letter (usually time $t$ ): $\dfrac{dV}{dt}=\dfrac{dV}{dr}\times\dfrac{dr}{dt}$ . Differentiate the model to get the middle part, then multiply.

If you need the rate the other way, divide: $\dfrac{dr}{dt}=\dfrac{dV}{dt}\div\dfrac{dV}{dr}$ . You never subtract rates.

The chain rule, linking rates

The question usually gives you one rate (“the radius increases at $2$ cm/s” means $\tfrac{dr}{dt}=2$ ).
It also gives a model that links the quantities ( $V=\tfrac43\pi r^3$ ). Differentiate it to get $\tfrac{dV}{dr}$ .
Multiply the two to get the rate you want.

$\dfrac{dV}{dt}=\dfrac{dV}{dr}\times\dfrac{dr}{dt}$

Worked example

An oil patch's radius grows at 2 m/hr. Find $\dfrac{dA}{dt}$ when $r = 25$ m

Model: $A=\pi r^2$ , so $\dfrac{dA}{dr}=2\pi r$ . At $r=25$ that is $50\pi$ .
Given $\dfrac{dr}{dt}=2$ , chain them: $\dfrac{dA}{dt}=\dfrac{dA}{dr}\times\dfrac{dr}{dt}=50\pi\times 2$ .

Answer

\dfrac{dA}{dt}=100\pi\ \text{m}^2\text{/hr}

What the exam asks

The common one is a sphere being blown up, with $V=\tfrac43\pi r^3$ and a constant $\tfrac{dV}{dt}$ .
You may first find the radius at some time, then how fast the radius is growing. The second part needs the divide form $\tfrac{dr}{dt}=\tfrac{dV}{dt}\div\tfrac{dV}{dr}$ . (9709/12 May/June 2020 Q3; 9709/12 Oct/Nov 2021 Q9)

Worked example

Real question: balloon volume grows at 600 cm³/s. Find the radius and $\dfrac{dr}{dt}$ after 30 s

9709/12 May/June 2020 Q3. Part (a) is just putting a number in. Part (b) uses the divide form.

After 30 s the volume is $600\times30=18000$ cm³. From $\tfrac43\pi r^3=18000$ : $r^3=\dfrac{13500}{\pi}\Rightarrow r=16.3$ cm (3 sf).
Differentiate the model: $\dfrac{dV}{dr}=4\pi r^2$ .
Divide the rates: $\dfrac{dr}{dt}=\dfrac{dV}{dt}\div\dfrac{dV}{dr}=\dfrac{600}{4\pi (16.3)^2}$ .

Answer

r\approx16.3\ \text{cm},\quad \dfrac{dr}{dt}\approx0.181\ \text{cm/s}

Where the marks go

Common mistake

Differentiate the model. Do not just multiply by

r

. For a sphere you need

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

, not

\tfrac43\pi r^3

itself. (9709/12 May/June 2020 Q3)

Common mistake

Be clear which rate is given and which is wanted. Then put them together:

\dfrac{dr}{dt}=\dfrac{dV}{dt}\div\dfrac{dV}{dr}

. You divide the rates, never subtract them. Keep the division the right way up.

Now you try

A pile of ore has radius $r$ m and volume $V$ m³, where $V=\tfrac32\big(r-\tfrac12\big)^3-1$ for $r\ge 2$ . Ore is added at a constant $1.5$ m³/s. Find how fast the radius is increasing when $r=5.5$ m. (9709/12 Oct/Nov 2021 Q9)

Your turn— tap to reveal the worked answer (9709/12 Oct/Nov 2021 Q9(a))

Differentiate the model (chain rule): $\dfrac{dV}{dr}=\tfrac92\big(r-\tfrac12\big)^2$ .

At $r=5.5$ : $\dfrac{dV}{dr}=\tfrac92(5)^2=\tfrac{225}{2}$ .

Then $\dfrac{dr}{dt}=\dfrac{dV}{dt}\div\dfrac{dV}{dr}=1.5\div\tfrac{225}{2}=\tfrac{1}{75}$ .

Answer

\dfrac{dr}{dt}=\tfrac{1}{75}\approx0.0133\ \text{m/s}

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Rates of change

A rate of change is how fast something changes over time. Think of a balloon being blown up, an oil patch spreading, or a pile of ore growing.
Use the chain rule from §7.2 to link a rate you are given to a rate you want.

Connected rates link through a shared letter (usually time $t$ ): $\dfrac{dV}{dt}=\dfrac{dV}{dr}\times\dfrac{dr}{dt}$ . Differentiate the model to get the middle part, then multiply.

If you need the rate the other way, divide: $\dfrac{dr}{dt}=\dfrac{dV}{dt}\div\dfrac{dV}{dr}$ . You never subtract rates.

The chain rule, linking rates

The question usually gives you one rate (“the radius increases at $2$ cm/s” means $\tfrac{dr}{dt}=2$ ).
It also gives a model that links the quantities ( $V=\tfrac43\pi r^3$ ). Differentiate it to get $\tfrac{dV}{dr}$ .
Multiply the two to get the rate you want.

$\dfrac{dV}{dt}=\dfrac{dV}{dr}\times\dfrac{dr}{dt}$

Worked example

An oil patch's radius grows at 2 m/hr. Find $\dfrac{dA}{dt}$ when $r = 25$ m

Model: $A=\pi r^2$ , so $\dfrac{dA}{dr}=2\pi r$ . At $r=25$ that is $50\pi$ .
Given $\dfrac{dr}{dt}=2$ , chain them: $\dfrac{dA}{dt}=\dfrac{dA}{dr}\times\dfrac{dr}{dt}=50\pi\times 2$ .

Answer

\dfrac{dA}{dt}=100\pi\ \text{m}^2\text{/hr}

What the exam asks

The common one is a sphere being blown up, with $V=\tfrac43\pi r^3$ and a constant $\tfrac{dV}{dt}$ .
You may first find the radius at some time, then how fast the radius is growing. The second part needs the divide form $\tfrac{dr}{dt}=\tfrac{dV}{dt}\div\tfrac{dV}{dr}$ . (9709/12 May/June 2020 Q3; 9709/12 Oct/Nov 2021 Q9)

Worked example

Real question: balloon volume grows at 600 cm³/s. Find the radius and $\dfrac{dr}{dt}$ after 30 s

9709/12 May/June 2020 Q3. Part (a) is just putting a number in. Part (b) uses the divide form.

After 30 s the volume is $600\times30=18000$ cm³. From $\tfrac43\pi r^3=18000$ : $r^3=\dfrac{13500}{\pi}\Rightarrow r=16.3$ cm (3 sf).
Differentiate the model: $\dfrac{dV}{dr}=4\pi r^2$ .
Divide the rates: $\dfrac{dr}{dt}=\dfrac{dV}{dt}\div\dfrac{dV}{dr}=\dfrac{600}{4\pi (16.3)^2}$ .

Answer

r\approx16.3\ \text{cm},\quad \dfrac{dr}{dt}\approx0.181\ \text{cm/s}

Where the marks go

Common mistake

Differentiate the model. Do not just multiply by

r

. For a sphere you need

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

, not

\tfrac43\pi r^3

itself. (9709/12 May/June 2020 Q3)

Common mistake

Be clear which rate is given and which is wanted. Then put them together:

\dfrac{dr}{dt}=\dfrac{dV}{dt}\div\dfrac{dV}{dr}

. You divide the rates, never subtract them. Keep the division the right way up.

Now you try

Your turn— tap to reveal the worked answer (9709/12 Oct/Nov 2021 Q9(a))

Differentiate the model (chain rule): $\dfrac{dV}{dr}=\tfrac92\big(r-\tfrac12\big)^2$ .

At $r=5.5$ : $\dfrac{dV}{dr}=\tfrac92(5)^2=\tfrac{225}{2}$ .

Then $\dfrac{dr}{dt}=\dfrac{dV}{dt}\div\dfrac{dV}{dr}=1.5\div\tfrac{225}{2}=\tfrac{1}{75}$ .

Answer

\dfrac{dr}{dt}=\tfrac{1}{75}\approx0.0133\ \text{m/s}