Modelling with Differentiation (Cambridge (CIE) O Level Additional Maths): Revision Note

• Read the question carefully to determine which variables you will need to use

  • The question may give you a formula to help you 

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

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

Differentiate the formula given for the volume of a sphere. 

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

$\frac{\mathbf{d}\mathit{V}}{\mathbf{d}\mathit{r}}\mathbf{=}\mathbf{4}\mathbf{\pi }{\mathit{r}}^{\mathbf{2}}$

• 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

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

(a)$\begin{array}{c}\\ \\ \end{array}$ Show that the volume, $V$ cm³ is given by $V=12x-20x^2$.
 
The volume of a cuboid is "$V=length\times width\times height$"

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

Expand and simplify

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

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

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

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

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

$\therefore 12-40x=0$

Solve for x

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

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

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

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 

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

The value of the second derivative (at $\mathit{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}$) is negative

Therefore V = 1.8 cm³ is a maximum volume