Related Rates of Change (DP IB Analysis & Approaches (AA)): Revision Note

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Paul

Last updated

5 June 2025

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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 often time, represented by $t$
- seconds is the standard unit for time but this will depend on context
e.g. Water running into a large hemi-spherical bowl
- both the height and volume of water in the bowl are changing with time
  - time is the linking parameter between the rate of change of height and the rate of change of volume

Use of chain rule and product rule are common in such problems
Be clear about which variables are representing which quantities

STEP 1
Write down any variables and derivatives involved in the problem
e.g. $x, y, t, \frac{d y}{d x}, \frac{d x}{d t}, \frac{d y}{d t}$

STEP 2
Use an appropriate differentiation rule to set up an equation linking ‘rates of change’
e.g. Chain rule: $\frac{d y}{d t} = \frac{d y}{d x} \times \frac{d x}{d t}$

STEP 3
Substitute in known values
e.g. If, when $t = 3$ , $\frac{d x}{d t} = 2$ and $\frac{d y}{d t} = 8$ , then $8 = \frac{d y}{d x} \times 2$

STEP 4
Solve the problem and interpret the answer in context if required
e.g. $\frac{d y}{d x} = \frac{8}{2} = 4$ ‘when $t = 3$ , $y$ changes at a rate of 4, with respect to $x$ ’

Examiner Tips and Tricks

If you struggle to determine which rate to use 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

In a manufacturing process a metal component is heated such that it’s cross-sectional area expands but always retains the shape of a right-angled triangle. At time $t$ seconds the triangle has base $b$ cm and height $h$ cm.

At the time when the component’s cross-sectional area is changing at 4 cm s^-1, the base of the triangle is 3 cm and its height is 6 cm. Also at this time, the rate of change of the height is twice the rate of change of the base.

Find the rate of change of the base at this point of time.

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Related Rates of Change (DP IB Analysis & Approaches (AA)): Revision Note