When two quantities both change over time, their rates are linked by the chain rule: if Q depends on x and x changes with time, then dQ/dt = dQ/dx × dx/dt. The key steps are to write the quantity in terms of one variable, differentiate to get the bridging derivative, and then combine it with the rate you are given.
Connected rates of change
- dQ/dt = dQ/dx × dx/dt
- Circle: A = πr², so dA/dr = 2πr
- Sphere: V = (4/3)πr³, so dV/dr = 4πr²; A = 4πr², so dA/dr = 8πr
- Cone: V = (1/3)πr²h, with r and h linked by similar triangles
A rate of increase is positive; a rate of decrease is negative.
How are two rates connected?
Write down the relationship between the quantities, differentiate it to get the bridging derivative, then join everything with the chain rule. For a circle, A = πr² gives dA/dr = 2πr, so dA/dt = 2πr × dr/dt. Read the question to see which rate you are given and which one you need, then rearrange.
Spheres and other solids
The same idea works for volumes and surface areas. For a sphere, V = (4/3)πr³ gives dV/dr = 4πr², so dV/dt = 4πr² × dr/dt. A neat check is that dV/dr for a sphere equals its surface area.
Cones: reducing to one variable
The classic exam vessel is an inverted cone being filled with water. Its volume, V = (1/3)πr²h, has two variables, so first use similar triangles to link r and h: r/h equals the ratio of the cone's full radius to its full height. Substitute to write V in terms of h alone, then differentiate and use dV/dt = dV/dh × dh/dt.
Vessels with a given volume formula
Some questions give the volume as a function of depth, such as a hemispherical bowl or a cylinder. Differentiate the given formula to get dV/dh, then chain as before. For a cylinder of fixed radius, V = πr²h and dV/dh = πr² is constant.
Rates proportional to a quantity
Many questions say a rate is proportional to something. Translate the words into an equation with a positive constant k, adding a minus sign for a decrease: “the volume decreases at a rate proportional to the surface area” becomes dV/dt = −k × (surface area). Combining this with the chain rule often proves a neat result, for example that a radius changes at a constant rate.
Chaining several rates
Sometimes one rate leads to another. A known volume rate gives the radius rate, which then gives the surface area rate: use dV/dr to find dr/dt, then dA/dr to find dA/dt. If a container fills at one rate and leaks at another, use the net rate.
Worked examples
Example 1. The area of a circle increases at 1.5 cm² s⁻¹. Find the rate at which the radius increases when r = 4 cm.
- A = πr², so dA/dr = 2πr.
- dA/dt = dA/dr × dr/dt, so dr/dt = (dA/dt)/(2πr).
- At r = 4: dr/dt = 1.5/(8π) ≈ 0.0597.
Answer: dr/dt ≈ 0.0597 cm s⁻¹ (3 s.f.).
Marks: M1 for dA/dr, M1 for the chain rule, A1 for 0.0597.
Example 2. The radius of a spherical balloon increases at 0.5 cm s⁻¹. Find dV/dt when r = 3 cm.
- V = (4/3)πr³, so dV/dr = 4πr².
- dV/dt = 4πr² × dr/dt.
- At r = 3: 4π × 9 × 0.5 = 18π.
Answer: dV/dt = 18π cm³ s⁻¹.
Marks: M1 for dV/dr, M1 for the chain rule, A1 for 18π.
Example 3. Water is poured at 8 cm³ s⁻¹ into an inverted cone of height 24 cm and radius 12 cm. Find dh/dt when the depth is 6 cm.
- Similar triangles: r/h = 12/24, so r = h/2.
- V = (1/3)π(h/2)²h = πh³/12.
- dV/dh = πh²/4, which is 9π when h = 6.
- 8 = 9π × dh/dt, so dh/dt = 8/(9π).
Answer: dh/dt = 8/(9π) cm s⁻¹.
Marks: M1 for r = h/2, A1 for V = πh³/12, M1 for the chain rule, A1 for 8/(9π).
Example 4. A hemispherical bowl of radius 10 cm is filled at 4 cm³ s⁻¹. The volume at depth h is V = π(10h² − h³/3). Find dh/dt when h = 2.
- dV/dh = π(20h − h²).
- At h = 2: π(40 − 4) = 36π.
- 4 = 36π × dh/dt, so dh/dt = 1/(9π).
Answer: dh/dt = 1/(9π) cm s⁻¹.
Marks: M1 for dV/dh, A1 for 36π, A1 for 1/(9π).
Example 5. The volume of a sphere increases at 12π cm³ s⁻¹. Find dr/dt and dA/dt when r = 3 cm.
- dV/dt = 4πr² × dr/dt, so 12π = 36π × dr/dt and dr/dt = 1/3.
- A = 4πr², so dA/dr = 8πr.
- dA/dt = 8π × 3 × 1/3 = 8π.
Answer: dr/dt = 1/3 cm s⁻¹ and dA/dt = 8π cm² s⁻¹.
Marks: M1 for dr/dt, M1 for dA/dr, A1 for both rates.
Common mistakes
- Differentiating a formula with two variables. For a cone, use similar triangles to eliminate r before differentiating.
- Losing the sign for a decrease. A decreasing quantity has a negative rate.
- Multiplying the wrong derivatives. Check the chain cancels correctly: dV/dh × dh/dt = dV/dt.
- Substituting too early. Differentiate the general formula first, then substitute the given value.
- Rounding before the end. Keep π and exact values until the final step, unless asked for decimals.
How connected rates of change are examined
Rates of change questions are set on the Pure 4 paper (WMA14) of Edexcel International A-Level, and in the 9MA0 A-Level they come up in second-year Pure. A question usually has two or three parts:
- show that the volume of a vessel can be written in terms of one variable
- find a rate at a given depth or radius, exactly or to a stated accuracy
- prove a result about a rate, such as a radius decreasing at a constant rate
- form a differential equation from a description of a rate
Find full connected rates questions with mark schemes in the Pure 4 differentiation exam questions. The Pure 4 differentiation worksheet and the differentiation revision engine give further practice.
This is an application of the chain rule, and it leads into forming and solving differential equations.
Frequently asked questions
What is a connected rate of change?
A rate that depends on another rate through a formula linking the two quantities. The chain rule connects them.
How do you handle a cone in rates of change questions?
Use similar triangles to write r in terms of h, so the volume depends on h only, then differentiate.
Why is dV/dr equal to the surface area of a sphere?
Differentiating V = (4/3)πr³ gives 4πr², which is the surface area formula. Adding a thin layer of thickness δr adds roughly surface area × δr to the volume.