Rates of change and connected rates of change
Edexcel International GCSE Further Pure Mathematics· 9G· 25 min read
1. 1. Basic Rates of Change★★☆☆☆⏱ 5 min
The rate of change of a quantity (y) with respect to a second quantity (x) is given by the first derivative (\frac{dy}{dx}). When the independent variable is time (t), this represents how fast (y) increases or decreases over time, with units matching (y) per unit time.
Rate of Change
The derivative of a dependent variable with respect to an independent variable, measuring the speed of change of the dependent variable. A positive value indicates an increasing quantity, negative indicates decreasing.
Example:
If (V) is the volume of a balloon in cm³ and (t) is time in seconds, (\frac{dV}{dt}) has units cm³/s, measuring how fast the balloon inflates or deflates.
The surface area of a cube is given by (A = 6x^2), where (x) is the side length in cm. Find the rate of change of surface area with respect to side length when (x = 3) cm.
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Step 1: Differentiate the area function with respect to (x)
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Step 2: Substitute (x = 3) cm into the derivative
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Final answer: The rate of change of surface area is 36 cm² per cm of side length increase.
Exam tip:
Always include units in your final answer for rates questions; marks are explicitly awarded for correct units as well as correct numerical values.
2. 2. Connected Rates of Change★★★☆☆⏱ 8 min
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When two quantities are related by a known formula, their rates of change can be linked using the chain rule. This is the core of connected rates problems, where you will be given one rate and asked to find a second related rate.
Connected Rates of Change
Rates of change of two or more related quantities, linked via the chain rule to allow calculation of an unknown rate from a known rate.
Example:
For an expanding sphere, the rate of change of volume (\frac{dV}{dt}) can be linked to the rate of change of radius (\frac{dr}{dt}) via the chain rule: (\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}).
A spherical balloon is inflated so that its radius increases at a constant rate of 2 cm/s. Find the rate of increase of the volume of the balloon when the radius is 5 cm. (Volume of a sphere: (V = \frac{4}{3}\pi r^3), given on the formula sheet.)
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Step 1: List known and unknown values
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Step 2: Write the chain rule relationship linking the two rates
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Step 3: Differentiate the volume formula with respect to (r)
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Step 4: Substitute all known values into the chain rule
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Final answer: The rate of increase of volume is (200\pi) cm³/s, or 628 cm³/s to 3 significant figures if a numerical value is requested.
3. 3. Small Change Approximation★★★☆☆⏱ 7 min
For very small changes in an independent variable (x), we can approximate the corresponding change in the dependent variable (y) using the gradient of the function at the point of interest. This approximation is widely used in estimation problems on the 4PM1 exam.
Small Change Approximation
An approximate formula for the change in (y) ((\delta y)) resulting from a small change in (x) ((\delta x)). The approximation becomes more accurate as (\delta x) gets smaller.
Example:
If (y = x^2), a small increase of (\delta x = 0.01) at (x = 10) gives (\delta y \approx 2(10)(0.01) = 0.2), very close to the exact change of (10.01^2 - 10^2 = 0.2001).
The volume of a cube is given by (V = x^3), where (x) is the side length. Use the small change approximation to estimate the change in volume when the side length increases from 8 cm to 8.02 cm.
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Step 1: Identify values for (x) and (\delta x)
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Step 2: Differentiate the volume function with respect to (x)
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Step 3: Substitute into the small change approximation formula
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Final answer: The approximate increase in volume is 3.84 cm³.
When is the small change approximation most accurate?
When (\delta x) is very large
When (\delta x) is very small
When (\frac{dy}{dx}) is zero
When (y) is a linear function
Reveal answer
1 —Correct! The approximation relies on the curve being almost straight over very small intervals of (x).
4. 4. Solving Mixed Exam-Style Rates Problems★★★★☆⏱ 5 min
Most exam questions will combine context, unit conversion, and either connected rates or small change approximation. Follow a consistent step-by-step approach to avoid mistakes.
Read the question carefully to identify known values, unknown values, and the relationship between quantities.
Write down any relevant formulae (use the formula sheet for mensuration values where allowed).
Set up the chain rule (for connected rates) or small change approximation formula explicitly.
Substitute values, paying close attention to units and the sign of rates (negative for decreasing quantities).
Check your final answer has the correct units and makes sense in the context of the problem.
5. Common Pitfalls
Wrong move:
Forgetting to include the chain rule link in connected rates problems, e.g. writing (\frac{dV}{dt} = 4\pi r^2) without multiplying by (\frac{dr}{dt}).
Why:
You are directly differentiating with respect to time instead of linking via the intermediate variable.
Correct move:
Always write out the full chain rule relationship explicitly before substituting values.
Wrong move:
Using a positive rate for a decreasing quantity (e.g. water leaking from a tank).
Why:
The sign of the rate indicates direction of change, so incorrect signs lead to wrong final answers.
Correct move:
Assign negative values to rates of decreasing quantities, and check your final answer sign matches the context.
Wrong move:
Calculating the exact change instead of using the small change approximation when instructed.
Why:
The question specifically tests your knowledge of the approximation formula, so exact calculations get zero marks.
Correct move:
Use only the (\delta y \approx \frac{dy}{dx}\delta x) formula when the question explicitly asks for the small change approximation.
Wrong move:
Forgetting to include units in the final answer.
Why:
Exam marks are explicitly awarded for correct units for rates questions.
Correct move:
Always check units match the quantity being measured (e.g. cm³/s for volume rate of change).
Wrong move:
Mixing up the order of variables in the chain rule, e.g. writing (\frac{dV}{dt} = \frac{dr}{dV} \times \frac{dr}{dt}).
Why:
The chain rule requires the intermediate variable to cancel out (e.g. (dr) in numerator and denominator).
Correct move:
Write the chain rule so the intermediate variable appears once in the numerator and once in the denominator to cancel correctly.
6. Quick Reference Cheatsheet
Concept | Formula | Key Notes |
|---|---|---|
Basic Rate of Change | (\frac{dy}{dx}) | Units = (units of y) / (units of x) |
Connected Rates (Chain Rule) | (\frac{dy}{dt} = \frac{dy}{du} \times \frac{du}{dt}) | Intermediate variable (u) cancels out |
Small Change Approximation | (\delta y \approx \frac{dy}{dx} \delta x) | Only valid for very small (\delta x) |
Sign Convention | Negative = decreasing, Positive = increasing | Check sign matches problem context |
7. Frequently Asked
Do I need to show the chain rule step explicitly in connected rates questions?
Yes, you must write out the chain rule relationship (e.g. (\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt})) before substituting values to get full marks. Partial marks are awarded for correct chain rule setup even if your final calculation is wrong.
Going deeper
What's Next
Now that you have mastered rates of change and connected rates for Edexcel IGCSE Further Pure Maths 4PM1, you are ready to move to more advanced calculus topics. These skills are foundational for all subsequent calculus work, including integration applications and kinematics problems. Make sure to practice past paper questions to get comfortable with the context and problem structures commonly tested, and always double-check your chain rule setup and units to avoid easy mark losses.