C3 January 2012 Q4

Notes

M1 For differentiation of \(2\tan\left(\boldsymbol{y} + \frac{\pi}{12}\right) \to 2\sec^2\left(\boldsymbol{y} + \frac{\pi}{12}\right)\). There is no need to identify this with \(\frac{dx}{dy}\)

A1 For correctly writing \(\dfrac{dx}{dy} = 2\sec^2\left(y + \frac{\pi}{12}\right)\) or \(\dfrac{dy}{dx} = \dfrac{1}{2\sec^2\left(y + \frac{\pi}{12}\right)}\)

M1 Substitute \(y = \frac{\pi}{4}\) into their \(\frac{dx}{dy}\). Accept if \(\frac{dx}{dy}\) is inverted and \(y = \frac{\pi}{4}\) substituted into \(\frac{dy}{dx}\).

A1 \(\dfrac{dx}{dy} = 8\) or \(\dfrac{dy}{dx} = \dfrac{1}{8}\) oe

B1 Obtains the value of \(x = 2\sqrt{3}\) corresponding to \(y = \frac{\pi}{4}\). Accept awrt 3.46

M1 This mark requires all of the necessary elements for finding a numerical equation of the normal.
Either Invert their value of \(\frac{dx}{dy}\), to find \(\frac{dy}{dx}\), then use \(m_1 \times m_2 = -1\) to find the numerical gradient of the normal
Or use their numerical value of \(-\frac{dx}{dy}\)
Having done this then use \(\left(y - \frac{\pi}{4}\right) = \text{their } m\left(x - \text{their } 2\sqrt{3}\right)\)
The \(2\sqrt{3}\) could appear as awrt 3.46, the \(\frac{\pi}{4}\) as awrt 0.79,
This cannot be awarded for finding the equation of a tangent.
Watch for candidates who correctly use \(\left(x - \text{their } 2\sqrt{3}\right) = -\text{their numerical } \frac{dy}{dx}\left(y - \frac{\pi}{4}\right)\)
If they use ‘y=mx+c’ it must be a full method to find c.

A1 Any correct form of the answer. It does not need to be simplified and the question does not ask for an exact answer.
\(\left(y - \frac{\pi}{4}\right) = -8\left(x - 2\sqrt{3}\right)\), \(\dfrac{y - \frac{\pi}{4}}{x - 2\sqrt{3}} = -8\), \(y = -8x + \frac{\pi}{4} + 16\sqrt{3}\), y=-8x+ (awrt) 28.5

Alternatives using arctan (first 3 marks)

M1 Differentiates \(y = \arctan\left(\frac{x}{2}\right) - \frac{\pi}{12}\) to get \(\dfrac{1}{1 + \left(\frac{x}{2}\right)^2} \times \text{constant}\). Don’t worry about the lhs

A1 Achieves \(\dfrac{dy}{dx} = \dfrac{1}{1 + \left(\frac{x}{2}\right)^2} \times \dfrac{1}{2}\)

M1 This method mark requires \(x\) to be found, which then needs to be substituted into \(\frac{dy}{dx}\)

The rest of the marks are then the same.

Or implicitly (first 2 marks)

M1 Differentiates implicitly to get \(1 = 2\sec^2\left(y + \frac{\pi}{12}\right) \times \frac{dy}{dx}\)

A1 Rearranges to get \(\frac{dy}{dx}\) or \(\frac{dx}{dy}\) in terms of y

The rest of the marks are the same

Or by compound angle identities

\(x = 2\tan\left(y + \frac{\pi}{12}\right) = \dfrac{2\tan y + 2\tan\left(\frac{\pi}{12}\right)}{1 - \tan y\tan\frac{\pi}{12}}\) oe

M1 Differentiates using quotient rule-see question 1 in applying this. Additionally the \(\tan y\) must have been differentiated to \(\sec^2 y\). There is no need to assign to \(\frac{dx}{dy}\)

A1 The correct answer for \(\dfrac{dx}{dy} = \dfrac{\left(1 - \tan y\tan\frac{\pi}{12}\right) \times 2\sec^2 y - \left(2\tan y + 2\tan\left(\frac{\pi}{12}\right)\right) \times -\sec^2 y\tan\frac{\pi}{12}}{\left(1 - \tan y\tan\frac{\pi}{12}\right)^2}\)

The rest of the marks are as the main scheme