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

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Paul

Last updated

10 December 2024

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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 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 of chain rule

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

Chain rule is given in the formula booklet in the format above
- Different letters may be used relative to the context
  - e.g. $V$ for volume, $S$ for surface area, $h$ for height, $r$ for radius
Problems often involve one quantity being constant
- so another quantity can be expressed in terms of a single variable
- this makes finding a derivative a lot easier
For time problems at least, it is more convenient to use

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

and if it is more convenient to find $\frac{d x}{d y}$ than $\frac{d y}{d x}$ then use chain rule in the form

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

Neither of these alternative versions of chain rule are in the formula booklet

STEP 1

Write down the rate of change given and the rate of change required

(If unsure of the rates of change involved, use the units given as a clue

e.g. $m s^{- 1}$ (metres per second) would be the rate of change of length, per time, $\frac{d l}{d t}$ )

STEP 2

Use chain rule to form an equation connecting these rates of change with a third rate
The third rate of change will come from a related quantity such as volume, surface area, perimeter

STEP 3

Write down the formula for the related quantity (volume, etc) accounting for any fixed quantities

Find the third rate of change of the related quantity (derivative) using differentiation

STEP 4

Substitute the derivative and known rate of change into the equation and solve it

Examiner Tips and Tricks

If you struggle to determine 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 of 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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Previous:Techniques of DifferentiationNext:Second Order Derivatives

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