| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | March |
| Marks | 11 |
| Topic | Parametric curves and Cartesian conversion |
| Type | Properties of specific curves |
| Difficulty | Standard +0.3 Part (i) is a standard parametric differentiation exercise requiring dy/dx = (dy/dt)/(dx/dt) and point-slope form—routine A-level technique. Part (ii) requires substituting point A to find t, then finding point B, followed by triangle area calculation. While multi-step (7 marks), each step uses standard methods with no novel insight required, making this slightly easier than average overall. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dx}{dt}\) where \(\frac{dy}{dt} = 3t^2\), \(\frac{dx}{dt} = 2t\) | *M1 | Correct application of \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dx}{dt}\) with their \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) (with at least one correct) |
| \(\frac{dy}{dx} = \frac{3t^2}{2t} (= \frac{3}{2}t)\) | A1 | Correct derivative (need not be simplified at this stage) |
| \(y - t^3 = \frac{3t^2}{2t}(x - t^2)\) | dep*M1 | Use of \(y - t^3 = m(x - t^2)\) with their \(m\) in terms of \(t\) |
| \(2y = 3tx - t^3\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Substitute \(A\) giving \(t^3 - 3t(\frac{19}{12}) + 2(-\frac{15}{8}) = 0\), and attempt factor theorem with \(f(t) = 4t^3 - 19t - 15\), oe | M1 | Correctly substitute \(A\) into given tangent and attempt to find a factor |
| \(f(-1) = 0 \Rightarrow (t+1)\) is a factor | A1 | |
| \(f(t) = (t+1)(4t^2 - 4t - 15)\) | M1 | Attempt to obtain a quadratic factor; By any correct method |
| \(f(t) = (t+1)(2t-5)(2t+3)\) | A1 | |
| \(t = \frac{5}{2}\) only, as \(t \geq 0\) | A1 | |
| \(y = 0 \Rightarrow B(\frac{25}{12}, 0)\) and area \(= \frac{1}{2} \times \frac{15}{8} \times \frac{25}{12}\) | M1 | Use of their \(t\) to find \(B\) and attempt to find area using their \(B\); Their value of \(t\) must be positive |
| area \(= \frac{125}{64}\) | A1 |
## Part (i):
$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dx}{dt}$ where $\frac{dy}{dt} = 3t^2$, $\frac{dx}{dt} = 2t$ | *M1 | Correct application of $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dx}{dt}$ with their $\frac{dy}{dt}$ and $\frac{dx}{dt}$ (with at least one correct)
$\frac{dy}{dx} = \frac{3t^2}{2t} (= \frac{3}{2}t)$ | A1 | Correct derivative (need not be simplified at this stage)
$y - t^3 = \frac{3t^2}{2t}(x - t^2)$ | dep*M1 | Use of $y - t^3 = m(x - t^2)$ with their $m$ in terms of $t$
$2y = 3tx - t^3$ | A1 | AG
## Part (ii):
Substitute $A$ giving $t^3 - 3t(\frac{19}{12}) + 2(-\frac{15}{8}) = 0$, and attempt factor theorem with $f(t) = 4t^3 - 19t - 15$, oe | M1 | Correctly substitute $A$ into given tangent and attempt to find a factor
$f(-1) = 0 \Rightarrow (t+1)$ is a factor | A1 |
$f(t) = (t+1)(4t^2 - 4t - 15)$ | M1 | Attempt to obtain a quadratic factor; By any correct method
$f(t) = (t+1)(2t-5)(2t+3)$ | A1 |
$t = \frac{5}{2}$ only, as $t \geq 0$ | A1 |
$y = 0 \Rightarrow B(\frac{25}{12}, 0)$ and area $= \frac{1}{2} \times \frac{15}{8} \times \frac{25}{12}$ | M1 | Use of their $t$ to find $B$ and attempt to find area using their $B$; Their value of $t$ must be positive
area $= \frac{125}{64}$ | A1 |A curve is defined, for $t \geqslant 0$, by the parametric equations
$$x = t^2, \quad y = t^3.$$
\begin{enumerate}[label=(\roman*)]
\item Show that the equation of the tangent at the point with parameter $t$ is
$$2y = 3tx - t^3.$$ [4]
\end{enumerate}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item In this question you must show detailed reasoning.
It is given that this tangent passes through the point $A\left(\frac{19}{2}, -\frac{15}{8}\right)$ and it meets the $x$-axis at the point $B$.
Find the area of triangle $OAB$, where $O$ is the origin. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q4 [11]}}