2013 H2 Mathematics Paper 1 Question 11

Differentiation I: Tangents and Normals, Parametric Curves

Answers

{y = tx-t^3}

{y=p^2q+pq^2}

{M \left( \frac{3}{2}, \frac{1}{\sqrt{2}} \right)}

{\frac{4}{15\sqrt{2}} \textrm{ units}^2}

\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} &= 6 t \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= 6 t^2 \\ \end{align*}

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t} \\ &=6 t^2 \div 6 t \\ &=t \\ \end{align*}

Tangent at the point with parameter

{t:}

\begin{gather*} y - 2 t^3 = t \left( x - 3 t^2 \right) \\ y = tx - 3t^3 + 2t^3 \\ \end{gather*}

Hence equation of tangent is

y = tx - t^3 \; \blacksquare

Tangents at points

{P}

and

{Q:}

\begin{align} && \quad y = px - p^3 \\ && \quad y = qx - q^3 \\ \end{align}

{R,}

\begin{align*} px-p^3 &= qx-q^3 \\ (p-q)x &= p^3 - q^3 \\ &= (p-q)(p^2+pq+q^2) \\ x &= p^2 + pq + q^2 \; \blacksquare \end{align*}

\begin{align*} y &= p(p^2+pq+q^2) - p^3 \\ &= p^2q + pq^2 \; \blacksquare \end{align*}

When

{pq=-1,}

\begin{equation} x = p^2 - 1 + q^2 \end{equation}

\begin{align*} y &= -p - q \\ y^2 &= (-p-q)^2 \\ &= p^2 + 2pq + q^2 \\ &= p^2 -2 + q^2 \end{align*}

\begin{equation} y^2 + 1 = p^2 - 1 + q^2 \end{equation}

Equations

{(3)}

and

{(4)}

show that

{R}

lies on the curve with equation

{x=y^2+1 \; \blacksquare}

\begin{align} && \quad x &= 3t^2 \\ && \quad y &= 2t^3 \\ && \quad x &= y^2 + 1 \end{align}

Substituting equations

{(5)}

and

{(6)}

into

{(7),}

\begin{gather*} 3t^2 = (2t^3)^2 + 1 \\ 4t^6 - 3t^2 + 1 = 0 \; \blacksquare \end{gather*}

Let

{u=t^2}

\begin{gather*} 4u^3 - 3u + 1 = 0 \\ (2u-1)^2(u+1) = 0 \\ u = \frac{1}{2} \; \textrm{ or } u = -1 \end{gather*}

t^2 = \frac{1}{2} \textrm{ or } t^2 = -1 \textrm{ (NA)}

Since

{y\geq 0,}

\begin{align*} t &=\frac{1}{\sqrt{2}} \\ x &= 3t^2 \\ &= \frac{3}{2} \\ y &= 2t^3 \\ &= \frac{1}{\sqrt{2}} \end{align*}

Exact coordinates of

{M:}

\left( \frac{3}{2}, \frac{1}{\sqrt{2}} \right) \; \blacksquare

\begin{align*} & \textrm{Area under curve } C \textrm{ from } O \textrm{ to } M \\ & = \int_0^{\frac{3}{2}} y \; \mathrm{d} x \\ & = \int_0^{\frac{1}{\sqrt{2}}} 2t^3 (6t) \; \mathrm{d} t \\ & = \int_0^{\frac{1}{\sqrt{2}}} 12t^4 \; \mathrm{d} t \\ & = \left[ \frac{12t^5}{5} \right]_0^{\frac{1}{\sqrt{2}}} \\ & = \frac{12}{5} \left( \frac{1}{\sqrt{2}} \right)^4 \left( \frac{1}{\sqrt{2}} \right) \\ & = \frac{3}{5\sqrt{2}} \end{align*}

The curve

{L}

cuts the

{x\textrm{-}}

axis at

{x=1}

\begin{align*} & \textrm{Area under curve } L \textrm{ from } x=1 \textrm{ to } M \\ & = \int_1^{\frac{3}{2}} y \; \mathrm{d} x \\ & = \int_1^{\frac{3}{2}} \sqrt{x-1} \; \mathrm{d} x \\ & = \left[ \frac{2}{3} (x-1)^{\frac{3}{2}} \right]_1^{\frac{3}{2}} \\ & = \frac{2}{3} \left( \frac{1}{2} \right)^{\frac{3}{2}} \\ & = \frac{2}{3} \left( \frac{1}{2} \right) \sqrt{\frac{1}{2}} \\ & = \frac{1}{3\sqrt{2}} \\ \end{align*}

\begin{align*} & \textrm{Area of shaded region} \\ & = \frac{3}{5\sqrt{2}} - \frac{1}{3\sqrt{2}} \\ & = \frac{4}{15\sqrt{2}} \textrm{ units}^2 \; \blacksquare \end{align*}