2023 H2 Mathematics Paper 2 Question 4

Differentiation I: Tangents and Normals, Parametric Curves

Definite Integrals: Areas and Volumes

Answers

\frac{12152}{75}.

( $5, 4).$

\left(- \frac{16}{5}, - \frac{25}{4}\right).

When $x = 21,$

\begin{align*} 2 t^2 + 3 &= 21 \\ 2 t^2 &= 18 \\ t^2 &= 9 \end{align*}

Since $t \geq \frac{1}{5}, t = 3$

When $y=0, t=\frac{1}{5}.$

\begin{align*} x &= 2 t^2 + 3 \\ \frac{\operatorname{d}\!x}{\operatorname{d}\!t} &= 4 t \end{align*}

\begin{align*} & \text{Area} \\ &= \int_a^{21} y \, \mathrm{d}x \\ &= \int_{\frac{1}{5}}^{3} \left(5 t - 1\right) \left(4 t\right) \, \mathrm{d}t \\ &= \left[ \frac{20}{3} t^3 - 2 t^2 \right]_{\frac{1}{5}}^{3} \\ &= \frac{12152}{75} \; \blacksquare \end{align*}

Equating equations of the two curves,

\begin{align} 2 t^2 + 3 &= 5u \\ 5 t - 1 &= \frac{5}{u} \end{align}

Making $u$ the subject of (1) and (2) and equating them,

\begin{align*} \frac{2 t^2 + 3}{5} &= \frac{4}{5 t - 1} \\ (2 t^2 + 3)(5 t - 1) &= 25 \\ 10 t^3 - 2 t^2 + 15 t - 3 &= 20 \\ 10 t^3 - 2 t^2 + 15 t - 23 &= 0 \end{align*}

\begin{equation} (t - 1)(10 t^2 + 8 t + 23) = 0 \end{equation}

\begin{align*} &\text{Discriminant for } 10 t^2 + 8 t + 23 \\ &= 10 t^2 + 8 t + 23 \\ &= b^2 - 4ac \\ &= 8^2 - 4(10)(23) \\ &= - 856 \\ &< 0 \end{align*}

Hence $10 t^2 + 8 t + 23=0$ has no real roots.

Hence equation (3) only has one real root $t=1$ which corresponds to $A$ and there are no other points of intersection $\; \blacksquare$

When $t=1,$

\begin{align*} x &= 5 \\ y &= 4 \end{align*}

{A = (5, 4) \; \blacksquare}

\begin{align*} \frac{\operatorname{d}\!y}{\operatorname{d}\!x} &= \frac{\operatorname{d}\!y}{\operatorname{d}\!t} \div \frac{\operatorname{d}\!x}{\operatorname{d}\!t} \\ &= \frac{5}{4 t} \end{align*}

When $t=1,$

{ \frac{\operatorname{d}\!y}{\operatorname{d}\!x} = \frac{5}{4}}

Equation of tangent at $A:$

\begin{align*} y - 4 &= \frac{5}{4} \left( x - 5 \right) \\ y &= \frac{5}{4} x - \frac{9}{4} \end{align*}

Substituting the parametric equations of curve $D,$

\begin{gather*} \frac{4}{u} = \frac{25}{4} u - \frac{9}{4} \\ \frac{25}{4} u^2 - \frac{9}{4} u = 4 \\ 25 u^2 - 9 u = 16 \\ 25 u^2 - 9 u - 16 = 0 \\ \left( 25 u + 16 \right) \left( u - 1 \right) = 0 \\ u = - \frac{16}{25} \quad \text{or} \quad u = 1 \end{gather*}

We note that $u=1$ corresponds to $A$

Hence, at the point where the tangent to curve $C$ at point $A$ meets curve $D$ for a second time

\begin{align*} u &= - \frac{16}{25} \\ x &= - \frac{16}{5} \\ y &= - \frac{25}{4} \end{align*}

Coordinates of point:

{ \left(- \frac{16}{5}, - \frac{25}{4}\right) \; \blacksquare}

Knowing how to work with curves with equation in parametric form for both integration and differentiation is important in solving this question.

The use of the discriminant to justify the number of points of intersection is also tested in part (b).