Modelling using t-formulae (Edexcel A Level Further Maths: Further Pure 1): Revision Note

You must learn the t-formulae for $\sin \theta$, $\cos \theta$ and $\tan \theta$ as they are not given in the formulae booklet!

The amplitude, $A$ metres, of part of a wave that varies with distance, $x$ metres, is modelled by

$A=1+4\sin \frac{x}{2}-\cos x0<x<2\pi$

(a) Show that

$\frac{\mathrm{d}A}{\mathrm{d}x}=\frac{2\left(1-t\right){\left(1+t\right)}^3}{{\left(1+t^2\right)}^2}$

where $t=\tan \frac{x}{4}$.

Answer:

Differentiate $A=1+4\sin \frac{x}{2}-\cos x$ with respect to $x$ (avoid substituting $t=\tan \frac{x}{4}$ into $A$ first)

$\frac{\mathrm{d}A}{\mathrm{d}x}=2\cos \frac{x}{2}+\sin x$

Now convert $2\cos \frac{x}{2}+\sin x$ into $t$-formulae

If $t=\tan \frac{x}{4}$ then $\sin \frac{x}{2}=\frac{2t}{1+t^2}$ and $\cos \frac{x}{2}=\frac{1-t^2}{1+t^2}$

$\cos \frac{x}{2}=\frac{1-t^2}{1+t^2}$

To find $\sin x$ in terms of $t$, use the double-angle formula $\sin 2A\equiv 2\sin A\cos A$

$\begin{array}{rcl}\sin x& \equiv & 2\sin \frac{x}{2}\cos \frac{x}{2}\\ & =& 2\left(\frac{2t}{1+t^2}\right)\left(\frac{1-t^2}{1+t^2}\right)\\ & =& \frac{4t\left(1-t^2\right)}{{\left(1+t^2\right)}^2}\end{array}$

Substitute $\cos \frac{x}{2}$ and $\begin{array}{rcl}& & \sin x\end{array}$ into $\frac{\mathrm{d}A}{\mathrm{d}x}$

$\frac{\mathrm{d}A}{\mathrm{d}x}=2\left(\frac{1-t^2}{1+t^2}\right)+\begin{array}{rcl}& & \frac{4t\left(1-t^2\right)}{{\left(1+t^2\right)}^2}\end{array}$

Add the algebraic fractions using a common denominator of ${\left(1+t^2\right)}^2$

$\frac{\mathrm{d}A}{\mathrm{d}x}=\frac{2\left(1-t^2\right)\left(1+t^2\right)+4t\left(1-t^2\right)}{{\left(1+t^2\right)}^2}$

Factorise out $2\left(1-t^2\right)$ from the numerator

$\frac{\mathrm{d}A}{\mathrm{d}x}=\frac{2\left(1-t^2\right)\left[\left(1+t^2\right)+2t\right]}{{\left(1+t^2\right)}^2}$

Rearrange $\left(1+t^2\right)+2t$ to $1+2t+t^2$, which factorises to ${\left(1+t\right)}^2$

$\frac{\mathrm{d}A}{\mathrm{d}x}=\frac{2\left(1-t^2\right){\left(1+t\right)}^2}{{\left(1+t^2\right)}^2}$

To make this look like the answer in the question, use the difference of two squares to write $1-t^2$ as $\left(1+t\right)\left(1-t\right)$

$\frac{\mathrm{d}A}{\mathrm{d}x}=\frac{2\left(1+t\right)\left(1-t\right){\left(1+t\right)}^2}{{\left(1+t^2\right)}^2}$

Now combine the $\left(1+t\right)$ and ${\left(1+t\right)}^2$ in the numerator into one single power

$\frac{\mathrm{d}A}{\mathrm{d}x}=\frac{2\left(1-t\right){\left(1+t\right)}^3}{{\left(1+t^2\right)}^2}$

(b) Given that

$\frac{{\mathrm{d}}^{2}A}{\mathrm{d}{x}^2}=\frac{{\left(1+t\right)}^2\left(t^2-4t+1\right)}{{\left(1+t^2\right)}^2}$

determine whether the model represents a peak (crest) or a trough (dip) of the wave.

Answer:

A peak or trough means a maximum or a minimum point, which both occur when $\frac{\mathrm{d}A}{\mathrm{d}x}=0$

Find the value of $t$ that makes $\frac{\mathrm{d}A}{\mathrm{d}x}=0$ using the answer in part (a)

$\frac{2\left(1-t\right){\left(1+t\right)}^3}{{\left(1+t^2\right)}^2}=0$

Setting the numerator equal to zero gives

$t=1\mathrm{or}t=-1$

Check to see if these solutions are within the range $0<x<2\pi$ given in the question

For $t=\tan \frac{x}{4}=1$ there is one possible solution in the range

$\begin{array}{rcl}\frac{x}{4}& =& \mathrm{artan}\left(1\right)+n\pi \\ \frac{x}{4}& =& \frac{\pi }{4}+n\pi \\ x& =& \pi +4n\pi \\ x& =& ...-3\pi ,\pi ,5\pi ,...\end{array}$

$\begin{array}{rcl}x& =& \pi \end{array}$

For $t=\tan \frac{x}{4}=-1$ there are no possible solutions in the range

$\begin{array}{rcl}\frac{x}{4}& =& \mathrm{artan}\left(-1\right)+n\pi \\ \frac{x}{4}& =& -\frac{\pi }{4}+n\pi \\ x& =& -\pi +4n\pi \\ x& =& ...,-\pi ,3\pi ,...\end{array}$

This means the maximum or minimum is at $x=\pi$ when $t=1$

To find out the nature of the stationary point, substitute $t=1$ into $\frac{{\mathrm{d}}^{2}A}{\mathrm{d}{x}^2}$

$\begin{array}{rcl}\frac{{\mathrm{d}}^{2}A}{\mathrm{d}{x}^2}& =& \frac{{\left(1+1\right)}^2\left(1^2-4\times 1+1\right)}{{\left(1+1^2\right)}^2}\\ & =& \frac{2^2\times \left(-2\right)}{2^2}<0\end{array}$

The second derivative is negative, so $x=\pi$ is a maximum point

The model represents a peak (crest) of the wave