Find the gradient of the tangent to the curve at the point \((\pi ,{\text{ }}\pi )\) .

Hence, show that \(\tan \theta  = \frac{1}{{1 + 2\pi }}\), where \(\theta \) is the acute angle between the tangent to the curve at \((\pi ,{\text{ }}\pi )\) and the line y = x .

attempt to differentiate implicitly     M1

\(2x + \cos y\frac{{{\text{d}}y}}{{{\text{d}}x}} - y - x\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)     A1A1 

Note: A1 for differentiating \({x^2}\) and sin y ; A1 for differentiating xy.

substitute x and y by \(\pi \)     M1

\(2\pi  - \frac{{{\text{d}}y}}{{{\text{d}}x}} - \pi  - \pi \frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{\pi }{{1 + \pi }}\)     M1A1 

Note: M1 for attempt to make dy/dx the subject. This could be seen earlier.

[6 marks]

\(\theta  = \frac{\pi }{4} - \arctan \frac{\pi }{{1 + \pi }}\) (or seen the other way)     M1

\(\tan \theta  = \tan \left( {\frac{\pi }{4} - \arctan \frac{\pi }{{1 + \pi }}} \right) = \frac{{1 - \frac{\pi }{{1 + \pi }}}}{{1 + \frac{\pi }{{1 + \pi }}}}\)     M1A1

\(\tan \theta  = \frac{1}{{1 + 2\pi }}\)     AG

[3 marks]

Part a) proved an easy 6 marks for most candidates, while the majority failed to make any headway with part b), with some attempting to find the equation of their line in the form y = mx + c . Only the best candidates were able to see their way through to the given answer.

Part a) proved an easy 6 marks for most candidates, while the majority failed to make any headway with part b), with some attempting to find the equation of their line in the form y = mx + c . Only the best candidates were able to see their way through to the given answer.