Connected Rates of Change

This lesson covers connected rates of change — using the chain rule to relate rates of change of different variables — as required by the Edexcel A-Level Mathematics specification (9MA0). These problems are among the most challenging at A-Level and require careful application of the chain rule.

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The Core Idea

If two variables are related, their rates of change are connected through the chain rule:

dA/dt = (dA/dr) × (dr/dt)

More generally, if we know the rate of change of one variable and the relationship between variables, we can find the rate of change of another.

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The Chain Rule for Connected Rates

Given a chain of relationships:

Want to find	Know	Chain Rule
dV/dt	dV/dr and dr/dt	dV/dt = (dV/dr) × (dr/dt)
dA/dt	dA/dr and dr/dt	dA/dt = (dA/dr) × (dr/dt)
dV/dt	dV/dh, dh/dr, dr/dt	dV/dt = (dV/dh) × (dh/dr) × (dr/dt)

The key is to construct a chain that links what you want to what you know.

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Worked Example 1: Expanding Circle

A circular oil spill has radius r cm at time t seconds. The radius increases at a rate of 2 cm/s. Find the rate of increase of the area when the radius is 10 cm.

Given: dr/dt = 2 cm/s. Find: dA/dt when r = 10.

Solution: A = πr² dA/dr = 2πr

Chain rule: dA/dt = (dA/dr) × (dr/dt) = 2πr × 2 = 4πr

When r = 10: dA/dt = 4π(10) = 40π ≈ 125.7 cm²/s

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Worked Example 2: Expanding Sphere

A spherical balloon is being inflated. Its volume increases at a constant rate of 50 cm³/s. Find the rate of increase of the radius when the radius is 5 cm.

Given: dV/dt = 50 cm³/s. Find: dr/dt when r = 5.

Solution: V = (4/3)πr³ dV/dr = 4πr²

Chain rule: dV/dt = (dV/dr) × (dr/dt) 50 = 4πr² × (dr/dt) dr/dt = 50 / (4πr²)

When r = 5: dr/dt = 50 / (4π × 25) = 50 / (100π) = 1/(2π) ≈ 0.159 cm/s

Follow-up: Rate of Change of Surface Area

Surface area: S = 4πr² dS/dr = 8πr

dS/dt = (dS/dr) × (dr/dt) = 8πr × 1/(2π) = 4r = 4 × 5 = 20 cm²/s (when r = 5)

Exam Tip: Write out the chain rule relationship BEFORE substituting numbers. This makes your method clear and helps you identify which derivatives you need.

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Worked Example 3: Filling a Cone

Water flows into an inverted cone of height 12 cm and base radius 4 cm at a rate of 2 cm³/s. Find the rate at which the water level is rising when the depth is 6 cm.

Given: dV/dt = 2 cm³/s. Find: dh/dt when h = 6.

Setting up the relationship: The cone has height 12 and radius 4, so at any depth h, the water surface radius r satisfies: r/h = 4/12 = 1/3, so r = h/3

Volume of water (cone): V = (1/3)πr²h = (1/3)π(h/3)²h = (1/3)π(h²/9)h = πh³/27

Differentiating: dV/dh = 3πh²/27 = πh²/9

Chain rule: dV/dt = (dV/dh) × (dh/dt) 2 = (πh²/9) × (dh/dt) dh/dt = 18/(πh²)

When h = 6: dh/dt = 18/(π × 36) = 18/(36π) = 1/(2π) ≈ 0.159 cm/s

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Worked Example 4: Related Rates with Multiple Variables

The volume of a cylinder is V = πr²h. The radius increases at 0.5 cm/s and the height decreases at 0.3 cm/s. Find the rate of change of volume when r = 4 cm and h = 10 cm.

Given: dr/dt = 0.5, dh/dt = -0.3. Find: dV/dt.

Solution: V = πr²h

This involves two changing variables, so we need partial differentiation (or equivalently, differentiate the product):

dV/dt = π(2rh × dr/dt + r² × dh/dt)

This uses the product rule: d/dt(r²h) = 2r(dr/dt)h + r²(dh/dt).

When r = 4, h = 10: dV/dt = π(2 × 4 × 10 × 0.5 + 16 × (-0.3)) = π(40 - 4.8) = 35.2π ≈ 110.6 cm³/s

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Worked Example 5: Using Inverse Derivatives

Sometimes it is easier to find dx/dy rather than dy/dx. Remember:

dy/dx = 1 / (dx/dy)

Worked Example

The area of a square is increasing at 10 cm²/s. Find the rate of increase of the side length when the area is 100 cm².

Given: dA/dt = 10. Find: dx/dt when A = 100 (x is side length).

A = x², so x = √A. When A = 100: x = 10.

dA/dx = 2x

Chain rule: dA/dt = (dA/dx) × (dx/dt) 10 = 2x × (dx/dt) dx/dt = 10/(2x) = 10/(2 × 10) = 0.5 cm/s

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Worked Example 6: Modelling Problem

A population of bacteria grows according to N = 2000e^(0.15t). At what rate is the population increasing when N = 5000?

Solution: dN/dt = 2000 × 0.15 × e^(0.15t) = 300e^(0.15t)

We need to find t when N = 5000: 5000 = 2000e^(0.15t) 2.5 = e^(0.15t) 0.15t = ln 2.5 t = ln 2.5 / 0.15

dN/dt = 300e^(0.15t) = 300 × 2.5 = 750 bacteria per unit time

Alternatively: since dN/dt = 0.15N, when N = 5000: dN/dt = 0.15 × 5000 = 750

Exam Tip: For exponential models N = N₀eᵏᵗ, the rate of change is always dN/dt = kN. So if you know N, you can find dN/dt immediately without finding t first.

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Worked Example 7: Triangle Problem

Two sides of a triangle are a = 10 cm and b = 15 cm. The included angle θ is increasing at 0.1 radians per second. Find the rate of change of the area when θ = π/3.

Solution: Area = (1/2)ab sin θ = (1/2)(10)(15) sin θ = 75 sin θ

dA/dθ = 75 cos θ

Chain rule: dA/dt = (dA/dθ) × (dθ/dt) = 75 cos θ × 0.1

When θ = π/3: dA/dt = 75 × cos(π/3) × 0.1 = 75 × 0.5 × 0.1 = 3.75 cm²/s

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Worked Example 8: Sliding Ladder

A ladder 5 m long leans against a vertical wall. The foot slides away from the wall at 0.3 m/s. Find the rate at which the top of the ladder slides down when the foot is 3 m from the wall.

Solution: Let x = distance of foot from wall, y = height of top on wall. x² + y² = 25 (Pythagoras)

Differentiate implicitly with respect to t: 2x(dx/dt) + 2y(dy/dt) = 0

When x = 3: y = √(25 - 9) = 4.

Substitute: 2(3)(0.3) + 2(4)(dy/dt) = 0 1.8 + 8(dy/dt) = 0 dy/dt = -1.8/8 = -0.225 m/s

The negative sign indicates the top is sliding down at 0.225 m/s.

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Summary of Problem Types

Type	Given Rate	Find Rate	Key Relationship
Circle	dr/dt	dA/dt	A = πr²
Sphere	dV/dt	dr/dt	V = (4/3)πr³
Cone filling	dV/dt	dh/dt	V = (1/3)πr²h (with r/h = const)
Sliding ladder	dx/dt	dy/dt	x² + y² = L²
Population	N known	dN/dt	N = N₀eᵏᵗ, dN/dt = kN

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Common Mistakes

Mistake	Correction
Forgetting to use the chain rule	Always write: "what I want = chain of derivatives"
Not expressing everything in one variable	Use the constraint to eliminate extra variables BEFORE differentiating
Getting the chain rule direction wrong	Check: the "dt" must cancel in the chain. dV/dt = (dV/dr)(dr/dt) ✓
Forgetting negative signs	If a quantity is decreasing, its rate of change is negative
Substituting values too early	Differentiate first, THEN substitute numerical values

Exam Tip: The most important step in connected rates problems is setting up the chain rule correctly. Write it out symbolically before doing any calculation.

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Summary

• Connected rates of change use the chain rule to relate rates of different quantities.
• The key formula: dA/dt = (dA/dr) × (dr/dt) (or more links in the chain).
• Eliminate extra variables using the given constraint BEFORE differentiating.
• Differentiate with respect to t (or the relevant variable), then substitute known values.
• Negative rates indicate decreasing quantities.
• For exponential models, dN/dt = kN gives the rate directly if N is known.

A-Level Deep Dive: Connected Rates of Change

Spec mapping

Edexcel 9MA0-02 specification section 9 — Differentiation, sub-strand 9.7 (Year 2 Pure) covers differentiation to connected rates of change problems, including the use of the chain rule (refer to the official specification document for exact wording). Examined predominantly on Paper 2 — Pure Mathematics, but the technique surfaces in Paper 3 — Statistics & Mechanics whenever a kinematic quantity (volume of fuel burned, displacement of a piston) varies with another (time, temperature). The Edexcel formula booklet provides the chain rule $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$dxdy​=dudy​⋅dxdu​ but does not give geometric formulas — $V = \tfrac{4}{3}\pi r^3$V=34​πr3 for a sphere or $V = \tfrac{1}{3}\pi r^2 h$V=31​πr2h for a cone must be memorised.

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Worked example with full mark scheme

Question (8 marks): A spherical balloon is being inflated with helium so that its volume increases at a constant rate of $50\,\text{cm}^3\,\text{s}^{-1}$50cm3s−1.

(a) Find, in terms of $\pi$π, the rate at which the radius of the balloon is increasing at the instant when the radius is $10\,\text{cm}$10cm. (5)

(b) Find the rate at which the surface area of the balloon is increasing at the same instant. (3)

Solution with mark scheme:

(a) Step 1 — write down the volume formula and differentiate.

The volume of a sphere of radius $r$r is $V = \tfrac{4}{3}\pi r^3$V=34​πr3. Differentiating with respect to $r$r:

$\dfrac{dV}{dr} = 4\pi r^2$drdV​=4πr2

M1 — correct expression for $dV/dr$dV/dr. Common slip: differentiating $\tfrac{4}{3}\pi r^3$34​πr3 as $\tfrac{4}{3}\pi \cdot 3r^2 = 4\pi r^2$34​π⋅3r2=4πr2 is right, but candidates often forget the $\tfrac{4}{3}$34​ entirely and write $3\pi r^2$3πr2.

Step 2 — set up the chain rule.

Volume and radius both depend on time, so:

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

M1 — correct chain-rule statement linking the three rates. Candidates who write $\dfrac{dr}{dt} = \dfrac{dV}{dt} \cdot \dfrac{dr}{dV}$dtdr​=dtdV​⋅dVdr​ are equivalently correct provided $dr/dV = 1/(dV/dr)$dr/dV=1/(dV/dr) is then used.

Step 3 — substitute known values.

At the instant in question, $dV/dt = 50$dV/dt=50 and $r = 10$r=10, so $dV/dr = 4\pi(10)^2 = 400\pi$dV/dr=4π(10)2=400π.

$50 = 400\pi \cdot \dfrac{dr}{dt}$50=400π⋅dtdr​

M1 — correct substitution of all three numerical values at the right instant.

Step 4 — solve for $dr/dt$dr/dt.

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

A1 — correct exact value in terms of $\pi$π.

A1 — units stated correctly: $\text{cm}\,\text{s}^{-1}$cms−1 (or equivalent). Edexcel routinely awards a final A mark for correct units on rate-of-change questions; omitting them is the most common mark-loss point.

(b) Step 1 — write down the surface-area formula and differentiate.

$S = 4\pi r^2$S=4πr2, so $\dfrac{dS}{dr} = 8\pi r$drdS​=8πr.

M1 — correct $dS/dr$dS/dr.

Step 2 — chain rule.

$\dfrac{dS}{dt} = \dfrac{dS}{dr} \cdot \dfrac{dr}{dt} = 8\pi(10) \cdot \dfrac{1}{8\pi} = 10$dtdS​=drdS​⋅dtdr​=8π(10)⋅8π1​=10

M1 — chain rule correctly applied using $dr/dt$dr/dt from part (a).

A1 — $dS/dt = 10\,\text{cm}^2\,\text{s}^{-1}$dS/dt=10cm2s−1 with units.

Total: 8 marks (M5 A3). Notice how part (a)'s answer feeds directly into part (b) — the "Hence" structure is implicit. A candidate who botches part (a) but applies a sensible (wrong) value through part (b) can still earn the (b) method marks via follow-through, but loses both A1 marks.

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): A cylindrical tank of radius $2\,\text{m}$2m is being filled with water. At time $t$t seconds the depth of water is $h\,\text{m}$hm and the volume is $V\,\text{m}^3$Vm3. Water enters the tank at a rate proportional to $\sqrt{h}$h​, so that $\dfrac{dV}{dt} = k\sqrt{h}$dtdV​=kh​ for some constant $k > 0$k>0.

(a) Show that $\dfrac{dh}{dt} = \dfrac{k\sqrt{h}}{4\pi}$dtdh​=4πkh​​. (3)

(b) Given that $h = 1$h=1 when $t = 0$t=0 and $h = 4$h=4 when $t = 60$t=60, find the value of $k$k to 3 significant figures. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO3.1a) — translating the cylinder geometry into $V = \pi(2)^2 h = 4\pi h$V=π(2)2h=4πh, recognising that radius is constant so $V$V depends only on $h$h.
• M1 (AO1.1b) — chain rule: $\dfrac{dV}{dt} = \dfrac{dV}{dh} \cdot \dfrac{dh}{dt} = 4\pi \cdot \dfrac{dh}{dt}$dtdV​=dhdV​⋅dtdh​=4π⋅dtdh​.
• A1 (AO2.1) — equating $4\pi \dfrac{dh}{dt} = k\sqrt{h}$4πdtdh​=kh​ and rearranging to the printed form.

(b)

• M1 (AO3.4) — separating variables: $\dfrac{4\pi}{\sqrt{h}}\,dh = k\,dt$h​4π​dh=kdt.
• M1 (AO1.1b) — integrating both sides: $8\pi\sqrt{h} = kt + C$8πh​=kt+C, applying boundary condition $h = 1, t = 0$h=1,t=0 to get $C = 8\pi$C=8π.
• A1 (AO3.5b) — substituting $h = 4, t = 60$h=4,t=60: $16\pi = 60k + 8\pi$16π=60k+8π, so $k = \dfrac{8\pi}{60} = \dfrac{2\pi}{15} \approx 0.419$k=608π​=152π​≈0.419.