Related Rates of Change (DP IB Applications & Interpretation (AI)): Revision Note

Written by: Paul

Reviewed by: Dan Finlay

Updated on 15 July 2025

What is meant by rates of change?

A rate of change is a measure of how a quantity is changing with respect to another quantity
Mathematically rates of change are derivatives
- $\frac{d V}{d r}$ could be the rate at which the volume of a sphere changes relative to how its radius is changing
Context is important when interpreting positive and negative rates of change
- A positive rate of change would indicate an increase
  - e.g. the change in volume of water as a bathtub fills
- A negative rate of change would indicate a decrease
  - e.g. the change in volume of water in a leaking bucket

Related rates of change are connected by a linking variable or parameter
- This is usually time, represented by $t$
  - Seconds is the standard unit for time but this will depend on context
E.g. water running into a large bowl
- Both the height and volume of water in the bowl change with time
- Time is the linking parameter

Use the chain rule

$y = g (u), where u = f (x) \Rightarrow \frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$

The chain rule is given in the formula booklet in the form above
- Different letters may be used relative to the context of a question
  - e.g. $V$ for volume, $S$ or $A$ for surface area, $h$ for height, $r$ for radius
Problems often involve one quantity being constant
- This means another quantity can be expressed in terms of a single variable
  - which makes finding a derivative a lot easier

Examiner Tips and Tricks

For time problems, it can be more convenient to use the chain rule in a form like

$\frac{d y}{d t} = \frac{d x}{d t} \times \frac{d y}{d x}$

If it is more convenient to find $\frac{d x}{d y}$ than $\frac{d y}{d x}$ then this can be rewritten as

$\frac{d y}{d t} = \frac{d x}{d t} \div \frac{d x}{d y}$

STEP 1
Write down the rate of change given and the rate of change required
STEP 2
Use the chain rule to form an equation connecting these rates of change with a third rate
- E.g. if you know rate of change of radius, $\frac{d r}{d t}$ , and want to find out the rate of change of volume, $\frac{d V}{d t}$
  - You can write a chain rule equation with 'blanks'
    - $\frac{d V}{d t} = \frac{d r}{d t} \times \frac{▯}{▯}$
  - then fill in the blanks to make the equation valid
    - $\frac{d V}{d t} = \frac{d r}{d t} \times \frac{d V}{d r}$

Examiner Tips and Tricks

You can 'cancel' as with normal fractions to check that your chain rule equation is valid

$\frac{d r}{d t} \times \frac{d V}{d r} = \frac{d r}{d t} \times \frac{d V}{d r} = \frac{d V}{d t} ✓$

STEP 3
Write down the formula for the related quantity, and differentiate to find the missing derivative
- E.g. the volume of a sphere is $V = \frac{4}{3} π r^{3}$
  - So differentiating with respect to $r$ gives $\frac{d V}{d r} = 4 π r^{2}$
STEP 4
Substitute the derivative and known rate of change into the equation and solve

Examiner Tips and Tricks

If you unsure which rate to use in an exam then you can look at the units to help

E.g. A rate of 5 cm³ per second implies volume per time, so the rate would be $\frac{d V}{d t}$

Worked Example

A cuboid has a square cross-sectional area with side length $x$ cm, and a fixed height of 5 cm.

The volume of the cuboid is increasing at a rate of 20 cm³ s^-1.

Find the rate at which the side length is increasing at the point when its side length is 3 cm.

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Related Rates of Change (DP IB Applications & Interpretation (AI)): Revision Note