Modelling with Differentiation

Did this video help you?

How is differentiation used in modelling questions?

Derivatives can be calculated for any variables – not just y and x
The derivative is a formula giving the rate of change of one variable with respect to the other variable
- For example if $A = 4 π r^{2}$ then $\frac{d A}{d r} = 8 π r$
- $\frac{d A}{d r}$ is the rate of change of $A$ with respect to $r$
The phrase 'increasing at a rate of' means the rate of change of one variable with respect to time
- $\frac{d}{d t}$

Differentiation can be used to find maximum and minimum points of a function
- In modelling, this is called optimisation
- Second derivative tests help to determine is the point is a maximum or minimum

Examiner Tip

Read the question carefully to determine which variables you will need to use
- The question may give you a formula to help you

Worked example

The volume, $V$ , of a sphere of radius $r$ is given by $V = \frac{4}{3} π r^{3} .$

Find the rate of change of the volume with respect to the radius.

Differentiate the formula given for the volume of a sphere.

$\frac{d V}{d r} = 3 \times \frac{4}{3} π \times r^{2}$

$\frac{d V}{d r} = 4 π r^{2}$

Optimisation

What is optimisation?

In general, optimisation is finding the best way to do something
In mathematics, optimisation is finding the maximum or minimum output of a function
- For example, finding the maximum possible profit or minimum costs

Differentiation can be used to solve optimisation problems in modelling questions
- For example you may want to
  - Maximise the volume of a container
  - Minimise the amount of fuel used

Example of using differentiation to maximise a function

Examiner Tip

Exam questions on this topic will often be divided into two parts:
- First a 'Show that...' part where you derive a given formula from the information in the question
- And then a 'Find...' part where you use differentiation to answer a question about the formula
Even if you can't answer the first part you can still use the formula to answer the second part

Worked example

A cuboid has length $4 x$ cm, width $x$ cm, and height $(\frac{3}{x} - 5)$ cm.

(a)

\begin{matrix}  \end{matrix}

Show that the volume, $V$ cm³ is given by $V = 12 x - 20 x^{2}$ .

The volume of a cuboid is " $V = l e n g t h \times w i d t h \times h e i g h t$ "

$∴ V = 4 x \times x \times (\frac{3}{x} - 5)$

Expand and simplify

$\begin{array}{rcl} V & = & 4 x^{2} (\frac{3}{x} - 5) \\ V & = & \frac{12 x^{2}}{x} - 20 x^{2} \end{array}$

$∴ V = 12 x - 20 x^{2}$

(b)

Find the maximum volume of the cuboid.

Differentiate V with respect to x

$\frac{d V}{d x} = 12 - 40 x$

At the maximum volume, $\frac{d V}{d x} = 0$

$∴ 12 - 40 x = 0$

Solve for x

$\begin{array}{rcl} 40 x & = & 12 \\ x & = & \frac{12}{40} = 0.3 \end{array}$

So the value of x, at the maximum volume is 0.3
Find the maximum volume by substituting x = 0.3 in to the formula for V

$V = 12 (0.3) - 20 {(0.3)}^{2} = 1.8$

The maximum volume of the cuboid is 1.8 cm³

(c)

Prove that your answer is a maximum value.

Using the second derivative is usually the easiest way to find the nature of a stationary point

$\frac{d^{2} y}{d x^{2}} = - 40 (< 0)$

The value of the second derivative (at $x = 0.3$ ) is negative

Therefore V = 1.8 cm³ is a maximum volume

Modelling with Differentiation (CIE IGCSE Additional Maths)

Revision Note