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?

• Remember that a derivative in maths represents a rate of change

• e.g. if  $y=x^3$  then  $\frac{\mathrm{d}y}{\mathrm{d}x}=3x^2$

  • When  $x=2$,  $\frac{\mathrm{d}y}{\mathrm{d}x}=3{\left(2\right)}^2=12$

    • That's the rate of change of $y$ with respect to $x$ when $x=2$

  • As soon as $x$ changes away from 2, $\frac{\mathrm{d}y}{\mathrm{d}x}$ 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:

  • $\mathrm{d}y\approx \frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x$

  • That is, when $x$ changes by a small amount $\mathrm{d}x$

    • the change in the value of $y$,  $\mathrm{d}y$,

    • will be approximately equal to $\frac{\mathrm{d}y}{\mathrm{d}x}$ times $\mathrm{d}x$

• 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  $V=\frac{4}{3}\pi r^3$

  • Take the derivative with respect to $r$,  $\frac{\mathrm{d}V}{\mathrm{d}r}=4\pi r^2$

    • That's the rate of change of volume with respect to radius

• You may also need to use the relation  $\frac{\mathrm{d}x}{\mathrm{d}y}=1÷\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\left({\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}}\right)}$

  • e.g. you may need $\frac{\mathrm{d}r}{\mathrm{d}V}$ to answer a question

    • Then  $\frac{\mathrm{d}r}{\mathrm{d}V}=\frac{1}{\left({\displaystyle \frac{\mathrm{d}V}{\mathrm{d}r}}\right)}=\frac{1}{4\pi r^2}$

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

  • $\frac{\mathrm{d}V}{\mathrm{d}r}$ 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 

  • $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\times \frac{\mathrm{d}u}{\mathrm{d}x}$

• 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  $\frac{\mathrm{d}y}{\mathrm{d}t}=\frac{\mathrm{d}y}{\mathrm{d}u}\times \frac{\mathrm{d}u}{\mathrm{d}t}$

    • where $y$ and $u$ represent other quantities in the question

• Note that  $\frac{\mathrm{d}x}{\mathrm{d}y}=1÷\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\left({\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}}\right)}$

  • Use this if you know a derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$

    • but you need to know the derivative $\frac{\mathrm{d}x}{\mathrm{d}y}$ instead 

• Also note that the chain rule can be extended to more than two terms on the right-hand side

  • e.g.  $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\times \frac{\mathrm{d}u}{\mathrm{d}v}\times \frac{\mathrm{d}v}{\mathrm{d}x}$

  • 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.  $\mathrm{m}/\mathrm{s}$ (or ${\mathrm{ms}}^{-1}$, 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