Rates of Change (Edexcel IGCSE Further Maths)

Approximations Using Rates of Change

Spec point — 9G

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 $\frac{dy}{dx}=3x^{2}$

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

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

  • As soon as $x$ changes away from 2, $\frac{dy}{dx}$ 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:

  • $dy\approx\frac{dy}{dx}dx$

  • That is, when $x$ changes by a small amount $dx$

    • the change in the value of $y$, $dy$,
    • will be approximately equal to $\frac{dy}{dx}$ times $dx$

• 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 $V=\frac{4}{3}πr^{3}$

  • Take the derivative with respect to $r$, $\frac{dV}{dr}=4πr^{2}$

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

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

  • e.g. you may need $\frac{dr}{dV}$ to answer a question

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

Exam Hint

• Look out for calculus questions asking you to 'estimate' or 'approximate' the change in a quantity

  • The $dy\approx\frac{dy}{dx}dx$ approximation is likely to be required
  • Remember that's only valid when $dx$ is small

Worked Example A sphere has a radius of $5\mathrm{cm}$.

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

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

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

Here we know the change in the surface area, $dA$

We want to estimate the change in the radius, $dr$

$dr\approx\frac{dr}{dA}dA$

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

$A=4πr^{2}$

Differentiate that with respect to $r$ to find $\frac{dA}{dr}$* *

$\frac{dA}{dr}=8πr$

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

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

Substitute that into the approximation formula

$dr\approx\frac{1}{8πr}dA$

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

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

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

Round to 2 significant figures, as required

$0.12\mathrm{cm}(2s.f.)$

Connected Rates of Change

Spec point — 9G

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

  • $\frac{dV}{dr}$ 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

  • $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$

• 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 $\frac{dy}{dt}=\frac{dy}{du}\times\frac{du}{dt}$

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

• Note that $\frac{ \mathrm{d} x}{ \mathrm{d} y} = 1 \div \frac{ \mathrm{d} y}{ \mathrm{d} x} = \frac{1}{( \frac{ \mathrm{d} y}{ \mathrm{d} x} )}$

  • Use this if you know a derivative $\frac{dy}{dx}$

    • but you need to know the derivative $\frac{dx}{dy}$ instead

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

  • e.g. $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dv}\times\frac{dv}{dx}$
  • 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 $\mathrm{ms}^{-1}$, 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

Exam Hint

• 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 $\frac{dV}{dt}$

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

$\frac{dV}{dt}=20\mathrm{cm}^{3}/s$

$\mathrm{What} \mathrm{is}\frac{dx}{dt}\mathrm{when}x=3\mathrm{cm}?$

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 $\frac{dV}{dt}$ and $\frac{dx}{dt}$

$\frac{dx}{dt}=\frac{d?}{d?}\times\frac{dV}{dt}$

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

$\frac{dx}{dt}=\frac{dx}{dV}\times\frac{dV}{dt}$

Now we need to find $\frac{dx}{dV}$

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

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

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

$\begin{gathered} V = 5 \times x \times x \\ = 5 x^{2} \end{gathered}$

Now differentiate that with respect to $x$

That will give a formula for $\frac{dV}{dx}$ in terms of $x$

$\frac{dV}{dx}=10x$

But our chain rule formula needs $\frac{dx}{dV}$

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

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

Substitute that into the chain rule formula

$\frac{dx}{dt}=\frac{1}{10x}\times\frac{dV}{dt}$

We know $\frac{dV}{dt}=20$, and we want to know $\frac{dx}{dt}$ when $x=3$

Substitute those values into the chain rule formula

$\mathrm{When}x=3,$

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

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

$\frac{dx}{dt}=\frac{2}{3}\mathrm{cm}/s$