C3 June 2014 Q3

Notes

B1* Either sub \(y=\dfrac{\pi}{8}\) into \(x=8y\tan(2y)\Rightarrow x=8\times\dfrac{\pi}{8}\tan\left(2\times\dfrac{\pi}{8}\right)=\pi\)
Or sub \(x=\pi,\ y=\dfrac{\pi}{8}\) into \(x=8y\tan(2y)\Rightarrow\pi=8\times\dfrac{\pi}{8}\tan\left(2\times\dfrac{\pi}{8}\right)=\pi\times 1=\pi\)
This is a proof and therefore an expectation that at least one intermediate line must be seen, including a term in tangent.
Accept as a minimum \(y=\dfrac{\pi}{8}\Rightarrow x=\pi\tan\left(\dfrac{\pi}{4}\right)=\pi\)
Or \(\pi=\pi\times\tan\left(\dfrac{\pi}{4}\right)=\pi\) ✓
This is a given answer however, and as such there can be no errors.

Notes

M1 Applies the product rule to \(8y\tan 2y\) achieving \(A\tan 2y+By\sec^2(2y)\)

A1 One term correct. Either \(8\tan 2y\) or \(+16y\sec^2(2y)\). There is no requirement for \(\dfrac{\mathrm{d}x}{\mathrm{d}y}=\)

A1 Both lhs and rhs correct. \(\dfrac{\mathrm{d}x}{\mathrm{d}y}=8\tan 2y+16y\sec^2(2y)\)
It is an intermediate line and the expression does not need to be simplified.
Accept \(\dfrac{\mathrm{d}x}{\mathrm{d}y}=\tan 2y\times 8+8y\times 2\sec^2(2y)\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{\tan 2y\times 8+8y\times 2\sec^2(2y)}\) or using implicit differentiation \(1=\tan 2y\times 8\dfrac{\mathrm{d}y}{\mathrm{d}x}+8y\times 2\sec^2(2y)\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

M1 For fully substituting \(y=\dfrac{\pi}{8}\) into their \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to find a 'numerical' value
Accept \(\dfrac{\mathrm{d}x}{\mathrm{d}y}=\text{awrt }20.6\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\text{awrt }0.05\) as evidence

M1 For a correct attempt at an equation of the tangent at the point \(\left(\pi,\dfrac{\pi}{8}\right)\).
The gradient must be an inverted numerical value of their \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\)
Look for \(\dfrac{y-\frac{\pi}{8}}{x-\pi}=\dfrac{1}{\text{numerical }\frac{\mathrm{d}x}{\mathrm{d}y}}\),
Watch for negative reciprocals which is M0
If the form \(y=mx+c\) is used it must be a full method to find a ‘numerical’ value to \(c\).

A1 A correct equation of the tangent.
Accept \(\dfrac{y-\frac{\pi}{8}}{x-\pi}=\dfrac{1}{8+4\pi}\) or if \(y=mx+c\) is used accept \(m=\dfrac{1}{8+4\pi}\) and \(c=\dfrac{\pi}{8}-\dfrac{\pi}{8+4\pi}\)
Watch for answers like this which are correct \(x-\pi=(8+4\pi)\left(y-\dfrac{\pi}{8}\right)\)
Accept the decimal answers awrt 2sf \(y=0.049x+0.24\), awrt 2sf \(21y=x+4.9\), \(\dfrac{y-0.39}{x-3.1}=0.049\)
Accept a mixture of decimals and \(\pi's\) for example \(20.6\left(y-\dfrac{\pi}{8}\right)=x-\pi\)

A1 Correct answer and solution only. \((8+4\pi)y=x+\dfrac{\pi^2}{2}\)
Accept exact alternatives such as \(4(2+\pi)y=x+0.5\pi^2\) and because the question does not ask for \(a\) and \(b\) to be simplified in the form \(ay=x+b\), accept versions like
\((8+4\pi)y=x+\dfrac{\pi}{8}(8+4\pi)-\pi\) and \((8+4\pi)y=x+(8+4\pi)\left(\dfrac{\pi}{8}-\dfrac{\pi}{8+4\pi}\right)\)