Rates of Change

This lesson covers connected rates of change, a key topic in A-Level Mathematics. Here we apply the chain rule to link rates at which different quantities change, enabling us to solve problems about expanding shapes, filling containers, and other dynamic real-world situations.

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The Chain Rule for Connected Rates

The chain rule states that if y is a function of u and u is a function of x, then:

dy/dx = (dy/du) × (du/dx)

This principle extends to rates of change with respect to time. If a quantity V depends on a radius r, and both change with time t, then:

dV/dt = (dV/dr) × (dr/dt)

This connects the rate of change of volume with the rate of change of radius.

Exam Tip: The chain rule for connected rates of change is one of the most commonly examined topics. Always identify the rate you want, the rate you know, and the link between the variables.

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Strategy for Connected Rates Problems

1. Identify all the variables involved and which rates are given/required.
2. Write down a formula linking the variables (e.g. V = 4πr³/3 for a sphere).
3. Differentiate the formula with respect to the appropriate variable.
4. Apply the chain rule to connect the known rate to the required rate.
5. Substitute the given values and calculate.

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Example 1: Expanding Sphere

A spherical balloon is being inflated so that its volume increases at a constant rate of 50 cm³ s⁻¹. Find the rate at which the radius is increasing when the radius is 10 cm.

Given: dV/dt = 50 cm³ s⁻¹, r = 10 cm. Find: dr/dt.

Formula: V = (4/3)πr³

Differentiate with respect to r:

dV/dr = 4πr²

Chain rule:

dV/dt = (dV/dr) × (dr/dt)
50 = 4π(10)² × (dr/dt)
50 = 400π × (dr/dt)
dr/dt = 50/(400π)
dr/dt = 1/(8π)
dr/dt ≈ 0.0398 cm s⁻¹

So the radius is increasing at 1/(8π) ≈ 0.0398 cm s⁻¹ when r = 10 cm.

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Example 2: Water Filling a Cone

Water is poured into a conical vessel of semi-vertical angle 30° at a rate of 5 cm³ s⁻¹. Find the rate at which the water level is rising when the depth of water is 6 cm.

Setting up: Let the depth of water be h and the surface radius be r. From the geometry of the cone:

tan 30° = r/h
r = h tan 30° = h/√3

The volume of water (a cone) is:

V = (1/3)πr²h = (1/3)π(h/√3)²h = (1/3)π(h²/3)h = πh³/9

Differentiate with respect to h:

dV/dh = πh²/3

Chain rule:

dV/dt = (dV/dh) × (dh/dt)
5 = (π × 6²/3) × (dh/dt)
5 = 12π × (dh/dt)
dh/dt = 5/(12π) ≈ 0.133 cm s⁻¹

The water level is rising at 5/(12π) ≈ 0.133 cm s⁻¹ when the depth is 6 cm.

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Example 3: Expanding Circle

A circular oil slick has area A cm² and radius r cm. The radius increases at a constant rate of 0.5 cm s⁻¹. Find the rate of increase of the area when the radius is 20 cm.

Formula: A = πr²

Differentiate:

dA/dr = 2πr

Chain rule:

dA/dt = (dA/dr) × (dr/dt)
dA/dt = 2π(20) × 0.5
dA/dt = 20π ≈ 62.8 cm² s⁻¹

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Example 4: Three-Variable Chain

A cube has side length x cm. The side length is increasing at 0.1 cm s⁻¹. Find the rate at which the surface area is increasing when x = 5 cm.

Formula: S = 6x²

Differentiate:

dS/dx = 12x

Chain rule:

dS/dt = (dS/dx) × (dx/dt)
dS/dt = 12(5) × 0.1
dS/dt = 6 cm² s⁻¹

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Using the Chain Rule with Implicit Relationships

Sometimes you need to connect three or more rates through an intermediate variable.

Example 5

A hemisphere of radius r cm is filled with water. The volume of water is V = (2/3)πr³. The surface area of the water is A = πr². Given that the volume increases at 10 cm³ s⁻¹, find the rate of change of the surface area when r = 5 cm.

We want dA/dt. We know dV/dt = 10. We can write:

dA/dt = (dA/dr) × (dr/dt)

We need dr/dt. From the volume:

dV/dt = (dV/dr) × (dr/dt)
10 = 2πr² × (dr/dt)
dr/dt = 10/(2π × 25) = 10/(50π) = 1/(5π)

Now:

dA/dr = 2πr = 10π
dA/dt = 10π × 1/(5π) = 2 cm² s⁻¹

Alternatively, you could compute dA/dV directly:

dA/dt = (dA/dV) × (dV/dt)

Since dA/dV = (dA/dr) ÷ (dV/dr) = 2πr ÷ 2πr² = 1/r:

dA/dt = (1/5) × 10 = 2 cm² s⁻¹ ✓

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Decreasing Rates and Negative Signs

If a quantity is decreasing, its rate of change is negative. For example, if ice is melting and the radius decreases at 0.02 cm s⁻¹, then dr/dt = −0.02.

Example 6

A spherical snowball melts so that its radius decreases at a constant rate of 0.5 cm per minute. Find the rate at which the volume is decreasing when the radius is 8 cm.

V = (4/3)πr³
dV/dr = 4πr²
dV/dt = 4πr² × (dr/dt) = 4π(64) × (−0.5) = −128π ≈ −402 cm³ min⁻¹

The volume is decreasing at 128π ≈ 402 cm³ per minute.

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Common Exam Patterns

1. Sphere: V = (4/3)πr³, dV/dr = 4πr², A = 4πr², dA/dr = 8πr.
2. Cone: V = (1/3)πr²h — use the geometry to eliminate r or h.
3. Cylinder: V = πr²h, S = 2πr² + 2πrh — one variable is usually fixed.
4. Square or cube: A = x², V = x³ — straightforward power-rule differentiation.

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Summary

• Connected rates of change use the chain rule to link the rate of change of one variable to another.
• The key formula is: dA/dt = (dA/dr) × (dr/dt) or more generally dy/dt = (dy/dx) × (dx/dt).
• Always identify: (a) the rate you know, (b) the rate you want, (c) the formula connecting the variables.
• Differentiate the formula, apply the chain rule, substitute the given values.
• Take care with signs: a decreasing quantity has a negative rate.

Exam Tip: Show every step of your chain rule clearly. Write down the formula you are differentiating, state the chain rule connection, substitute, and calculate. The examiners award method marks for each of these stages. Always include correct units (e.g. cm² s⁻¹) in your final answer.

A-Level Deep Dive: Connected Rates of Change

Spec mapping

AQA 7357 specification, Pure section G — Differentiation (Year 2 content) covers differentiation to find rates of change; connected rates of change and related contexts (refer to the official specification document for exact wording). This sub-strand sits in Paper 2 (Pure) but its applications saturate Paper 3 (Mechanics and Statistics). Connected rates of change is the canonical Year 2 topic where the chain rule $\dfrac{dV}{dt} = \dfrac{dV}{dr} \cdot \dfrac{dr}{dt}$dtdV​=drdV​⋅dtdr​ is deployed across two genuinely different physical variables — radius and time — rather than as an algebraic exercise. The AQA formula booklet provides standard derivatives but does not state the chain rule explicitly: candidates are expected to know it.

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Worked example with full mark scheme

Question (8 marks): A spherical balloon is being inflated so that its volume increases at a constant rate of $50\,\text{cm}^3\,\text{s}^{-1}$50cm3s−1. The volume of a sphere is given by $V = \tfrac{4}{3}\pi r^3$V=34​πr3.

(a) Find an expression for $\dfrac{dV}{dr}$drdV​. (1)

(b) Hence find the rate at which the radius is increasing at the instant when $r = 10\,\text{cm}$r=10cm, giving your answer in exact form. (4)

(c) The surface area of the balloon is $S = 4\pi r^2$S=4πr2. Find the rate at which the surface area is increasing when $r = 10\,\text{cm}$r=10cm. (3)

Solution with mark scheme:

(a) Differentiate $V = \tfrac{4}{3}\pi r^3$V=34​πr3 with respect to $r$r:

$\dfrac{dV}{dr} = 4\pi r^2$drdV​=4πr2

B1 — correct derivative. Note this equals the surface area $S$S, a fact worth holding in reserve for part (c).

(b) Step 1 — set up the chain rule.

By the chain rule:

$\dfrac{dV}{dt} = \dfrac{dV}{dr} \cdot \dfrac{dr}{dt}$dtdV​=drdV​⋅dtdr​

M1 — quoting or implicitly using the correct chain-rule relationship between the three rates.

Step 2 — substitute known values.

We have $\dfrac{dV}{dt} = 50$dtdV​=50 and $\dfrac{dV}{dr} = 4\pi r^2 = 4\pi (10)^2 = 400\pi$drdV​=4πr2=4π(10)2=400π at $r = 10$r=10.

M1 — substituting $r = 10$r=10 into $\dfrac{dV}{dr}$drdV​ to obtain a numerical (or exact-$\pi$π) value.

Step 3 — solve for $\dfrac{dr}{dt}$dtdr​.

$50 = 400\pi \cdot \dfrac{dr}{dt} \implies \dfrac{dr}{dt} = \dfrac{50}{400\pi} = \dfrac{1}{8\pi}\,\text{cm}\,\text{s}^{-1}$50=400π⋅dtdr​⟹dtdr​=400π50​=8π1​cms−1

A1 — correct exact value $\dfrac{1}{8\pi}$8π1​.

A1 — units $\text{cm}\,\text{s}^{-1}$cms−1 stated. Omitting the units loses this final accuracy mark on a context-rich question.

(c) Step 1 — chain rule for $S$S.

$\dfrac{dS}{dt} = \dfrac{dS}{dr} \cdot \dfrac{dr}{dt}$dtdS​=drdS​⋅dtdr​

with $\dfrac{dS}{dr} = 8\pi r$drdS​=8πr.

M1 — correct chain-rule setup with the right derivative for $S$S.

Step 2 — evaluate at $r = 10$r=10.

$\dfrac{dS}{dr} = 8\pi(10) = 80\pi$drdS​=8π(10)=80π, and from (b) $\dfrac{dr}{dt} = \dfrac{1}{8\pi}$dtdr​=8π1​:

$\dfrac{dS}{dt} = 80\pi \cdot \dfrac{1}{8\pi} = 10\,\text{cm}^2\,\text{s}^{-1}$dtdS​=80π⋅8π1​=10cm2s−1

A1 — correct value $10$10.

A1 — units $\text{cm}^2\,\text{s}^{-1}$cm2s−1.

Total: 8 marks (B1 M2 A2 + M1 A2).

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Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): A scientist models the spread of an oil slick as a thin circular film on water. The radius $r$r (metres) of the slick at time $t$t (minutes) is observed to grow such that $\dfrac{dr}{dt} = \dfrac{k}{r}$dtdr​=rk​ for some constant $k > 0$k>0. The area $A$A of the slick is $A = \pi r^2$A=πr2.

(a) Show that $\dfrac{dA}{dt}$dtdA​ is independent of $r$r. (3)

(b) Given that the area is observed to be increasing at $0.6\,\text{m}^2\,\text{min}^{-1}$0.6m2min−1, find the value of $k$k to 3 significant figures, and comment on whether the model is plausible. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1b) — applying the chain rule $\dfrac{dA}{dt} = \dfrac{dA}{dr} \cdot \dfrac{dr}{dt} = 2\pi r \cdot \dfrac{k}{r}$dtdA​=drdA​⋅dtdr​=2πr⋅rk​.
• A1 (AO1.1b) — simplifying to $2\pi k$2πk.
• A1 (AO2.4) — concluding "since the result $2\pi k$2πk contains no $r$r, the rate of area growth is independent of $r$r."

(b)

• M1 (AO1.1b) — setting $2\pi k = 0.6$2πk=0.6.
• A1 (AO1.1b) — $k = \dfrac{0.3}{\pi} \approx 0.0955$k=π0.3​≈0.0955.
• B1 (AO3.5b) — modelling comment: e.g. "the model predicts a constant rate of area growth, which is plausible if oil is leaking at a steady rate from a fixed source onto a calm sea, but not if the source is depleting or surface tension changes with film thickness."

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This is a Paper 2 archetype: an AO3-heavy modelling question where the algebra is light but the interpretation — recognising that $r$r has cancelled, and judging whether the model captures the physics — is what separates 4-mark and 6-mark answers.

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Synoptic links

Connects to:

1. Pure section G — Chain rule: connected rates of change is the chain rule applied across a time variable. The same identity $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$dxdy​=dudy​⋅dxdu​ governs both, but in connected-rates problems $x$x is replaced by $t$t and $u$u by a geometric variable like $r$r or $h$h.

2. Mechanics — velocity and acceleration: in kinematics, velocity $v = \dfrac{ds}{dt}$v=dtds​ and acceleration $a = \dfrac{dv}{dt}$a=dtdv​. When motion is described in terms of a position variable $x$x, the relation $a = v\dfrac{dv}{dx}$a=vdxdv​ is exactly a connected-rates identity (chain rule with $\dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot \dfrac{dx}{dt}$dtdv​=dxdv​⋅dtdx​).

3. Pure section O — Numerical methods and modelling: AQA Paper 2 modelling questions frequently embed connected-rates calculations inside word problems where the candidate must first identify the geometric formula and then apply the chain rule.

4. Pure section G — Implicit differentiation: the same chain-rule machinery underlies $\dfrac{d}{dx}f(y) = f'(y)\dfrac{dy}{dx}$dxd​f(y)=f′(y)dxdy​. Mastery of one transfers immediately to the other.

5. Pure section P — Differential equations (Year 2): an equation like $\dfrac{dr}{dt} = \dfrac{k}{r}$dtdr​=rk​ in the specimen question above is itself an ODE; separating variables yields $r^2 = 2kt + C$r2=2kt+C. Connected-rates problems are often the entry point to first-order ODEs.

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Mark-scheme literacy

Connected rates of change questions on AQA 7357 split AO marks distinctively:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Differentiating $V(r)$V(r) correctly, quoting the chain rule, substituting numerical values, computing $\dfrac{dr}{dt}$dtdr​
AO2 (reasoning / interpretation)	20–30%	Recognising which quantity is changing with respect to which, justifying the cancellation pattern, presenting exact answers correctly
AO3 (problem-solving / modelling)	15–25%	Identifying the geometric formula from the context, commenting on model assumptions, evaluating whether the answer is physically reasonable