18N.1.AHL.TZ0.H_7

pestleMathematicsAAHLPaper 118N· ahl-5-14-implicit-functions-related-rates-optimisationsource ↗

Consider the curves ${C_1}$ and ${C_2}$ defined as follows

${C_1}\,{\text{:}}\,xy = 4$ , $x > 0$

${C_2}\,{\text{:}}\,{y^2} - {x^2} = 2$ , $x > 0$

Using implicit differentiation, or otherwise, find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ for each curve in terms of $x$ and $y$ .

[4]

Let P( $a$ , $b$ ) be the unique point where the curves ${C_1}$ and ${C_2}$ intersect.

Show that the tangent to ${C_1}$ at P is perpendicular to the tangent to ${C_2}$ at P.

[2]

Markscheme / solution

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

${C_1}\,{\text{:}}\,y + x\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ (M1)

Note: M1 is for use of both product rule and implicit differentiation.

$\Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{y}{x}$ A1

Note: Accept $- \frac{4}{{{x^2}}}$

${C_2}\,{\text{:}}\,2y\frac{{{\text{d}}y}}{{{\text{d}}x}} - 2x = 0$ (M1)

$\Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{x}{y}$ A1

Note: Accept $\pm \frac{x}{{\sqrt {2 + {x^2}} }}$

[4 marks]

substituting $a$ and $b$ for $x$ and $y$ M1

product of gradients at P is $\left( { - \frac{b}{a}} \right)\left( {\frac{a}{b}} \right) = - 1$ or equivalent reasoning R1

Note: The R1 is dependent on the previous M1.

so tangents are perpendicular AG

[2 marks]

Examiners’ report

[N/A]