Modelling with Differentiation (Cambridge International IGCSE Further Maths)

Modelling with differentiation

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 $A=4πr^{2}$ then $\frac{dA}{dr}=8πr$
  • $\frac{dA}{dr}$ 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

  • $\frac{d}{dt}$

• 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

Exam Hint

• 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 $V=\frac{4}{3}πr^{3}.$

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

Differentiate the formula given for the volume of a sphere.

$\frac{dV}{dr}=3\times\frac{4}{3}π\times r^{2}$

$\frac{dV}{dr}=4πr^{2}$

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

Exam Hint

• 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 $4x$ cm, width $x$ cm, and height $( \frac{3}{x} - 5 )$ cm.

(a)$\begin{gathered} \\ \\ \end{gathered}$ Show that the volume, $V$ cm³ is given by $V=12x-20x^{2}$.

The volume of a cuboid is "$V=length\times width\times height$"

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

Expand and simplify

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

$\therefore V=12x-20x^{2}$* *

(b) Find the maximum volume of the cuboid. Differentiate V with respect to x

$\frac{dV}{dx}=12-40x$

At the maximum volume, $\frac{dV}{dx}=0$

$\therefore12-40x=0$

*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

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

Final answer: 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

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

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

Final answer: Therefore V = 1.8 cm³is a maximum volume