Rates of Change (Edexcel IGCSE Further Pure Maths)

Revision Note

Written by: Roger B

Reviewed by: Dan Finlay

Approximations Using Rates of Change

How can I use rates of change to approximate changes in value?

Remember that a derivative in maths represents a rate of change
e.g. if $y = x^{3}$ then $\frac{d y}{d x} = 3 x^{2}$
- When $x = 2$ , $\frac{d y}{d x} = 3 {(2)}^{2} = 12$
  - That's the rate of change of $y$ with respect to $x$ when $x = 2$
- As soon as $x$ changes away from 2, $\frac{d y}{d x}$ is no longer equal to 12
  - But it will still be close to 12 so long as $x$ is still close to 2
We can use this to approximate the change in one variable based on a change in the other variable:
- $d y \approx \frac{d y}{d x} d x$
- That is, when $x$ changes by a small amount $d x$
  - the change in the value of $y$ , $d y$ ,
  - will be approximately equal to $\frac{d y}{d x}$ times $d x$
This approximation is only valid when the change in $x$ is small
- The smaller the change in $x$ is,
  - the more accurate the approximation will be
You may need to derive rates of change starting from standard geometric formulae
- e.g. the volume of a sphere is $V = \frac{4}{3} π r^{3}$
- Take the derivative with respect to $r$ , $\frac{d V}{d r} = 4 π r^{2}$
  - That's the rate of change of volume with respect to radius
You may also need to use the relation $\frac{d x}{d y} = 1 \div \frac{d y}{d x} = \frac{1}{(\frac{d y}{d x})}$
- e.g. you may need $\frac{d r}{d V}$ to answer a question
  - Then $\frac{d r}{d V} = \frac{1}{(\frac{d V}{d r})} = \frac{1}{4 π r^{2}}$

Examiner Tips and Tricks

Look out for calculus questions asking you to 'estimate' or 'approximate' the change in a quantity
- The $d y \approx \frac{d y}{d x} d x$ approximation is likely to be required
- Remember that's only valid when $d x$ is small

Worked Example

A sphere has a radius of $5 cm$ .

The surface area of the sphere is increased by $15 {cm}^{2}$ .

Using calculus, find an estimate for the increase in the radius of the sphere. Give your answer in $cm$ , correct to 2 significant figures.

'Using calculus' and 'find an estimate for the increase' are hints that we should use $d y \approx \frac{d y}{d x} d x$

Here we know the change in the surface area, $d A$

We want to estimate the change in the radius, $d r$

$d r \approx \frac{d r}{d A} d A$

Write down the formula for the surface area of a sphere from the exam formula sheet

$A = 4 π r^{2}$

Differentiate that with respect to $r$ to find $\frac{d A}{d r}$

$\frac{d A}{d r} = 8 π r$

But we need $\frac{d r}{d A}$ for our approximation formula, so use $\frac{d x}{d y} = \frac{1}{(\frac{d y}{d x})}$

$\frac{d r}{d A} = \frac{1}{(\frac{d A}{d r})} = \frac{1}{8 π r}$

Substitute that into the approximation formula

$d r \approx \frac{1}{8 π r} d A$

We want to know the value of that when $r = 5$ and $d A = 15$

(Note that that is a 'small' change, $d A$ , compared to the total surface area of $100 π = 314.15 . . . {cm}^{2}$ when $r = 5$ )

$d r \approx \frac{1}{8 π (5)} (15) = \frac{3}{8 π} = 0.119366 . . .$

Round to 2 significant figures, as required

$0.12 cm (2 s . f .)$

Connected Rates of Change

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 $V$ of a sphere changes
  - with respect to how its radius $r$ is changing
Context is important when interpreting positive and negative rates of change
- A positive rate of change indicates an increase
  - e.g. the change in volume of water as a bathtub fills
- A negative rate of change indicates a decrease
  - e.g. the change in volume of water in a leaking bucket
- If a question talks about rate of increase or decrease
  - make sure you use the appropriate sign (+/-)

What is meant by connected rates of change?

Connected rates of change are connected by a linking variable or parameter
- They are also called 'related rates of change'
- Often the linking parameter is time, represented by $t$
  - seconds is the standard unit for time
  - but a question may use other units
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

What are the key ideas involved with connected rates of change?

These questions usually involve the chain rule
- $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$

Different letters may be used relative to the context
- e.g. $V$ for volume, $A$ for area, $h$ for height, $r$ for radius
- For time problems you will often use the form $\frac{d y}{d t} = \frac{d y}{d u} \times \frac{d u}{d t}$
  - where $y$ and $u$ represent other quantities in the question
Note that $\frac{d x}{d y} = 1 \div \frac{d y}{d x} = \frac{1}{(\frac{d y}{d x})}$
- Use this if you know a derivative $\frac{d y}{d x}$
  - but you need to know the derivative $\frac{d x}{d y}$ instead
Also note that the chain rule can be extended to more than two terms on the right-hand side
- e.g. $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d v} \times \frac{d v}{d x}$
- This lets you connect more variables using the chain rule
Remember, a derivative is not a fraction
- But if you treat the derivatives on the right-hand side of the chain rule as fractions
  - then their common terms should 'cancel out'
  - to give you the derivative on the left-hand side
- This is a way to check a chain rule formula
  - It can also help you write the formula in the first place

How do I solve problems involving connected rates of change?

Most connected rates of change questions will involve the following steps

STEP 1
Write down the rate of change given and the rate of change required
- Write these down as derivatives
- If unsure of the rates of change involved, use the units given as a clue
  - e.g. $m / s$ (or ${ms}^{- 1}$ , metres per second)
  - This would be the rate of change of length with respect to time
  - The precise 'length of what' would depend on the question
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
  - e.g. volume, surface area, perimeter
- More complicated questions may involve more than three rates of change
  - But those will still be able to be connected by the chain rule
STEP 3
Write down the formula for a related quantity (volume, etc)
- This can then be differentiated
  - which will provide a formula for one or more of your rates of change
STEP 4
Substitute all the known values into the chain rule equation
- Then solve for the value you need to know

Examiner Tips and Tricks

To determine which rate of change to use, look at the units for help
- e.g. A rate of 5 cm³ per second implies volume per time
- so the rate would likely be $\frac{d V}{d t}$

Worked Example

A cuboid has a fixed height of 5 cm, and a square cross-section with side length of $x$ cm in the other two dimensions.

The volume of the cuboid is increasing at a fixed rate of 20 cm³ per second.

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

Write down what is given, and what we need to know

$\frac{d V}{d t} = 20 {cm}^{3} / s$

$What is \frac{d x}{d t} when x = 3 cm ?$

Note that length and volume are both given in terms of centimetres, so we won't need to convert any units

Write down a chain rule equation connecting $\frac{d V}{d t}$ and $\frac{d x}{d t}$

$\frac{d x}{d t} = \frac{d ?}{d ?} \times \frac{d V}{d t}$

The right-hand side should 'cancel out' to be equal with the left-hand side
This means the missing derivative must be $\frac{d x}{d V}$

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

Now we need to find $\frac{d x}{d V}$

Write down the volume of a cuboid formula, $V = height \times length \times width$

Here the height is 5, and the length and width are both $x$

This gives a formula connecting $V$ and $x$

$\begin{array}{rcl} V & = & 5 \times x \times x \\ = & 5 x^{2} \end{array}$

Now differentiate that with respect to $x$

That will give a formula for $\frac{d V}{d x}$ in terms of $x$

$\frac{d V}{d x} = 10 x$

But our chain rule formula needs $\frac{d x}{d V}$

We can find this using $\frac{d x}{d y} = \frac{1}{(\frac{d y}{d x})}$

$\frac{d x}{d V} = \frac{1}{(\frac{d V}{d x})} = \frac{1}{10 x}$

Substitute that into the chain rule formula

$\frac{d x}{d t} = \frac{1}{10 x} \times \frac{d V}{d t}$

We know $\frac{d V}{d t} = 20$ , and we want to know $\frac{d x}{d t}$ when $x = 3$

Substitute those values into the chain rule formula

$When x = 3,$

$\begin{array}{rcl} \frac{d x}{d t} & = & \frac{1}{10 (3)} \times 20 \\ = & \frac{20}{30} \\ = & \frac{2}{3} \end{array}$

Don't forget the units when giving your final answer!

$\frac{d x}{d t} = \frac{2}{3} cm / s$

Last updated: 20 October 2024

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Rates of Change (Edexcel IGCSE Further Pure Maths)

Revision Note

Approximations Using Rates of Change

How can I use rates of change to approximate changes in value?

Connected Rates of Change

What is meant by rates of change?

What is meant by connected rates of change?

What are the key ideas involved with connected rates of change?

How do I solve problems involving connected rates of change?

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

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