19M.1.AHL.TZ2.H_6

pestleMathematicsAAHLPaper 119M· sl-5-4-tangents-and-normalsource ↗

The curve $C$ is given by the equation $y = x\,{\text{tan}}\left( {\frac{{\pi xy}}{4}} \right)$ .

At the point (1, 1) , show that $\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2 + \pi }}{{2 - \pi }}$ .

[5]

Hence find the equation of the normal to $C$ at the point (1, 1).

[2]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to differentiate implicitly M1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = x\,{\text{se}}{{\text{c}}^2}\left( {\frac{{\pi xy}}{4}} \right)\left[ {\left( {\frac{\pi }{4}x\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{\pi }{4}y} \right)} \right] + {\text{tan}}\left( {\frac{{\pi xy}}{4}} \right)$ A1A1

Note: Award A1 for each term.

attempt to substitute $x = 1$ , $y = 1$ into their equation for $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ M1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{\pi }{2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{\pi }{2} + 1$

$\frac{{{\text{d}}y}}{{{\text{d}}x}}\left( {1 - \frac{\pi }{2}} \right) = \frac{\pi }{2} + 1$ A1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2 + \pi }}{{2 - \pi }}$ AG

[5 marks]

attempt to use gradient of normal $= \frac{{ - 1}}{{\frac{{{\text{d}}y}}{{{\text{d}}x}}}}$ (M1)

$= \frac{{\pi - 2}}{{\pi + 2}}$

so equation of normal is $y - 1 = \frac{{\pi - 2}}{{\pi + 2}}\left( {x - 1} \right)$ or $y = \frac{{\pi - 2}}{{\pi + 2}}x + \frac{4}{{\pi + 2}}$ A1

[2 marks]

Examiners’ report

[N/A]