| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Tangent/normal meets curve again |
| Difficulty | Standard +0.3 This is a standard C4 parametric differentiation question with routine techniques: finding dy/dx using the chain rule, finding a tangent equation, and solving a parametric-algebraic system. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = \frac{(dy/dt)}{(dx/dt)} = \frac{(-1/2)}{2t}\) as unsimplified expression | M1, A1 | (S.R. Award M1 for attempt to change to cartesian eqn & differentiate a1 for dy/dx or dx/dy in terms of x or y) |
| \(= -\frac{1}{2t^3}\) as simplified expression | A1 3 | |
| (ii) \((4,-1/2) \to t = -2\) only | B1, M1, A1 3 | Using \(t = -2\) or \(2\); Satis attempt to find equation of tgt; \(x - 16y = 12\) only AG |
| Satis attempt to find equation of tgt | ||
| \(x - 16y = 12\) only | ||
| (iii) \(t^3 - 12t - 16 = 0\) or \(16y^3 + 12y^2 - 1 = 0\) or \(x^3 - 24x + 144x - 256 = 0\) | B2 4 | S.R. Award B1 for "4 or -2"; S.R. If B0, award M1 for clear indic of method of soln of correct eqn. |
| \(t = 4\) (only) ISW giving cartesian coords | For simplified equiv non-fract cubic; For solving any relevant pair of eqns |
(i) $\frac{dy}{dx} = \frac{(dy/dt)}{(dx/dt)} = \frac{(-1/2)}{2t}$ as unsimplified expression | M1, A1 | **(S.R. Award M1 for attempt to change to cartesian eqn & differentiate a1 for dy/dx or dx/dy in terms of x or y)**
$= -\frac{1}{2t^3}$ as simplified expression | | A1 3 | Not $1/-2t^3$. Not in terms of x &/or y.
(ii) $(4,-1/2) \to t = -2$ only | B1, M1, A1 3 | Using $t = -2$ or $2$; Satis attempt to find equation of tgt; $x - 16y = 12$ only **AG**
Satis attempt to find equation of tgt | |
$x - 16y = 12$ only | |
(iii) $t^3 - 12t - 16 = 0$ or $16y^3 + 12y^2 - 1 = 0$ or $x^3 - 24x + 144x - 256 = 0$ | B2 4 | **S.R.** Award B1 for "4 or -2"; **S.R.** If B0, award M1 for clear indic of method of soln of correct eqn.
$t = 4$ (only) **ISW** giving cartesian coords | | For simplified equiv non-fract cubic; For solving any relevant pair of eqnsA curve is given parametrically by the equations
$$x = t^2, \quad y = \frac{1}{t}.$$
\begin{enumerate}[label=(\roman*)]
\item Find $\frac{dy}{dx}$ in terms of $t$, giving your answer in its simplest form. [3]
\item Show that the equation of the tangent at the point $P\left(4, -\frac{1}{4}\right)$ is $x - 16y = 12$. [3]
\item Find the value of the parameter at the point where the tangent at $P$ meets the curve again. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2005 Q7 [10]}}