2020 H2 Maths Paper 1 Q2 — Worked Solution

Solution

6 marks

We differentiate both sides of $\frac{x^{2}}{1 + x^{2}} + \frac{y^{2}}{1 + y^{2}} = x^{3}y^{5}$ implicitly with respect to $x$ .

For the left side, we apply the quotient rule to each term.

\frac{d}{dx}\left( \frac{x^{2}}{1 + x^{2}} \right)

= \frac{2x\left( 1 + x^{2} \right) - x^{2}(2x)}{\left( 1 + x^{2} \right)^{2}}

= \frac{2x}{\left( 1 + x^{2} \right)^{2}}

\frac{d}{dx}\left( \frac{y^{2}}{1 + y^{2}} \right) = \frac{2y\frac{dy}{dx}\left( 1 + y^{2} \right) - y^{2}\left( 2y\frac{dy}{dx} \right)}{\left( 1 + y^{2} \right)^{2}} = \frac{2y\frac{dy}{dx}}{\left( 1 + y^{2} \right)^{2}}

For the right side, we apply the product rule.

\frac{d}{dx}\left( x^{3}y^{5} \right) = 3x^{2}y^{5} + 5x^{3}y^{4}\frac{dy}{dx}

\frac{2x}{\left( 1 + x^{2} \right)^{2}} + \frac{2y}{\left( 1 + y^{2} \right)^{2}}\frac{dy}{dx} = 3x^{2}y^{5} + 5x^{3}y^{4}\frac{dy}{dx}

To check your answer: substituting

x = 1

y = 1

into the original equation,

\frac{1}{2} + \frac{1}{2} = 1

and

x^{3}y^{5} = 1

, so

(1,1)

does lie on the curve. ✓

Substituting $x = 1$ , $y = 1$ ,

\frac{2}{4} + \frac{2}{4}\frac{dy}{dx} = 3 + 5\frac{dy}{dx}

\Rightarrow \frac{1}{2} + \frac{1}{2}\frac{dy}{dx} = 3 + 5\frac{dy}{dx}

\Rightarrow \frac{1}{2} - 3 = 5\frac{dy}{dx} - \frac{1}{2}\frac{dy}{dx}

\Rightarrow - \frac{5}{2} = \frac{9}{2}\frac{dy}{dx}

\frac{dy}{dx} = - \frac{5}{2} \times \frac{2}{9} = - \frac{5}{9}

The tangent at $(1,1)$ with gradient $- \frac{5}{9}$ is

y - 1 = - \frac{5}{9}(x - 1)

\Rightarrow 9(y - 1) = - 5(x - 1)

\Rightarrow 9y - 9 = - 5x + 5

\Rightarrow 5x + 9y = 14

$\therefore$ the tangent has equation $5x + 9y = 14$ .