IB Maths AI HL
Further Differentiation
Paper 1 & 2
~7 min read
Related Rates of Change
A balloon inflates: its radius grows, and so does its volume — the two rates are linked. Related rates of change connect how two quantities change over time by routing them through a shared variable (usually t). The engine is the chain rule: you build a chain of derivatives that “cancel” to connect the rate you know with the rate you want.
📘 What you need to know
- A rate of change is a derivative — e.g. dVdt is how fast volume changes with time.
- Sign = direction: positive means increasing (filling), negative means decreasing (leaking).
- Related rates share a linking variable, almost always time t.
- The tool is the chain rule: dydx = dydu × dudx (in the booklet).
- Build the chain so the derivatives “cancel” to give the rate you want.
- Units help: “cm³ per second” tells you the rate is dVdt.
Rates as derivatives
Every “how fast is … changing” phrase is a derivative with respect to time. Reading the units of a given rate tells you exactly which derivative it is.
positive rate
increasing
e.g. a bathtub filling — volume rises, dVdt > 0.
negative rate
decreasing
e.g. a leaking bucket — volume falls, dVdt < 0.
The chain rule (the engine)
dydx = dydu × dudx
✓ given in the formula booklet
🧠 “The d’s cancel”
Treat the chain like fractions: dVdt = drdt × dVdr — the dr‘s “cancel” to leave dVdt. If your chain doesn’t cancel cleanly, it’s wrong.
Solving a related-rates problem
🧭 Recipe — related rates
- Write down the rates: the one you’re given and the one you need.
- Form a chain linking them through a third rate, with “blanks” to fill: e.g. dVdt = drdt × dVdr.
- Differentiate the linking formula (e.g. V in terms of r) to fill the missing derivative.
- Substitute and solve at the required moment.
Cancel-check: you can write the chain in whatever order makes it valid — for time problems, forms like dydt = dxdt × dydx are common. If it’s easier to find dxdy than dydx, divide instead: dydt = dxdt ÷ dxdy.
🤔 Why does one quantity often stay constant?
Many shapes have two variables (say radius and height). If one is fixed by the problem, the other quantity can be written in terms of a single variable — which makes the linking derivative simple to find. That’s why questions specify things like “fixed height of 5 cm”: it collapses the formula down to one variable so the chain rule has something clean to bite on.
Worked examples
A cuboid has a square cross-section of side x cm and a fixed height of 5 cm. Its volume increases at 20 cm³ s⁻¹.
WE 1Write down the rates given and required
Read the units to identify each derivative.
Given: dVdt = 20 (cm³ per second → volume per time)
Required: dxdt
WE 2Form the chain-rule equation linking them
Connect dVdt and dxdt through a third rate.
fill the blank so the d’s cancel:
dVdt = dxdt × dVdx
WE 3Find the linking derivative dVdx
Volume of the cuboid (square cross-section × fixed height) in terms of x, then differentiate.
V = x² × 5 = 5x²
dVdx = 10x
WE 4Find the rate the side length increases when x = 3
Substitute into the chain and solve for dxdt.
dVdt = dxdt × 10x → 20 = dxdt × 10(3)
20 = 30 · dxdt
dxdt = 23 cm s⁻¹
WE 5A sphere’s radius grows at 0.2 cm s⁻¹. How fast is its volume growing when r = 5?
Same method with V = 43πr3.
Given drdt = 0.2, want dVdt
dVdr = 4πr² = 4π(25) = 100π
dVdt = drdt × dVdr = 0.2 × 100π
dVdt = 20π cm³ s⁻¹
💡 Top tips
- Read the units to pin down each rate — “cm³ per second” is dVdt.
- Write the chain with blanks first, then fill it so the d’s cancel.
- Cancel-check your chain like fractions before substituting.
- Use a fixed quantity to reduce the linking formula to one variable.
- Substitute the moment last — differentiate the general formula, then plug in the given value.
- Keep π exact unless asked for a decimal.
⚠ Common mistakes
- Substituting too early — differentiate the general formula first, then put in the value.
- An invalid chain — if the derivatives don’t cancel to the rate you want, the equation is wrong.
- Mixing up the rate’s direction — a leak or shrink gives a negative rate.
- Wrong linking formula — make sure the volume/area formula matches the shape and any fixed dimensions.
- Ignoring units — they’re the quickest check that you’ve identified the right derivative.
Next up — Second Order Derivatives. So far every derivative has been a first derivative — a rate of change. The next topic differentiates again to get f′′(x), the rate of change of the gradient itself. That second derivative is the key to testing the nature of stationary points and the concavity of a curve in the topics that follow.
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