Connected rates of change is one of those A Math topics that looks terrifying in Paper 2 and then turns out to be almost mechanical once you know the routine. Most students lose marks not because they cannot differentiate, but because they never wrote down what they were given and what they were asked for. Here is the exact method our tutors used when we sat the 4049 papers ourselves.
What The Question Is Really Asking
Every connected rates question is the chain rule in disguise. You are given one rate with respect to time, and you are asked for another. The link between them is a formula from geometry, mensuration or the question stem itself.
The skeleton is always:
dA/dt = dA/dr × dr/dt
The middle term comes from differentiating your formula. The last term is the rate you were given. That is the whole topic.
The Five Step Routine
- List your variables. Write down every quantity that changes with time, and the letter you are using for it.
- Write the given rate and the required rate in symbols. For example, "given dr/dt = 0.2, find dV/dt when r = 5".
- Find the connecting equation. Volume of a sphere, area of a circle, Pythagoras, or a relationship stated in the question.
- Differentiate the connecting equation with respect to the variable inside it, then build the chain.
- Substitute the instant given in the question, and state units.
Substitute the value of the variable only at the very end. Students who substitute r = 5 before differentiating end up differentiating a constant and getting zero, which is an instant loss of all the method marks.
Worked Example 1: The Expanding Sphere
A spherical balloon is inflated so that its radius increases at 0.3 cm per second. Find the rate of increase of the volume when the radius is 4 cm.
Given dr/dt = 0.3. Required dV/dt when r = 4.
Connecting equation: V = (4/3)πr³, so dV/dr = 4πr².
Chain: dV/dt = dV/dr × dr/dt = 4πr² × 0.3.
At r = 4: dV/dt = 4π(16)(0.3) = 19.2π ≈ 60.3 cm³ per second.
Notice the units. Volume over time gives cm³ s⁻¹. Examiners do award and withhold marks for units in this topic, so never skip them.
Worked Example 2: The Cone With A Ratio
Water is poured into an inverted cone of height 12 cm and base radius 3 cm at a rate of 8 cm³ per second. Find the rate at which the water level rises when the depth is 6 cm.
This is the classic trap. The formula V = (1/3)πr²h has two changing variables, so you must eliminate one first.
By similar triangles, r/h = 3/12, so r = h/4.
Substitute: V = (1/3)π(h/4)²h = πh³/48.
Then dV/dh = πh²/16.
Chain: dV/dt = dV/dh × dh/dt, so 8 = (πh²/16) × dh/dt.
At h = 6: 8 = (36π/16) × dh/dt, giving dh/dt = 128/(36π) ≈ 1.13 cm per second.
Key habit: if your formula has two variables and you are only given one rate, you are missing a similar triangles or given ratio step.
Worked Example 3: The Sliding Ladder
A 5 m ladder leans against a vertical wall. The foot slides away from the wall at 0.4 m per second. Find the rate at which the top slides down when the foot is 3 m from the wall.
Let x be the distance of the foot from the wall and y the height of the top.
Connecting equation: x² + y² = 25.
Differentiate both sides with respect to t: 2x(dx/dt) + 2y(dy/dt) = 0.
When x = 3, y = 4. So 2(3)(0.4) + 2(4)(dy/dt) = 0, giving dy/dt = -0.3.
The negative sign matters. Write "the top slides down at 0.3 m per second" so the examiner sees you understood the sign, not just the arithmetic.
Common Mark Losers
- Substituting the instantaneous value before differentiating.
- Forgetting that a decreasing quantity has a negative rate. Leaking water, cooling liquid and falling height all give negative rates.
- Leaving two variables in the formula when only one rate is given.
- Rounding mid working. Keep exact values such as 19.2π until the final line.
- No units, or the wrong power of the unit.
How To Practise This Efficiently
Do not grind fifty questions. Do ten from different past year papers and force yourself to write the "given" and "required" line first, every single time. When you can set up the chain in under thirty seconds, the topic is done. It is one of the highest return topics in Paper 2 because the marks are structured and predictable.
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