| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Topic | Parametric differentiation |
| Type | Show gradient expression then find coordinates |
| Difficulty | Standard +0.3 This is a standard parametric differentiation question requiring the chain rule (dy/dx = (dy/dt)/(dx/dt)) and quotient rule applications, followed by solving dy/dx = 0. The algebra is straightforward with no geometric insight needed, making it slightly easier than average but still requiring competent technique across multiple steps for 7 marks total. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dx}{dt} = \frac{-2t}{(1+t^2)^2}\) | B1 | |
| Attempt use of quotient or product rule for \(\frac{dy}{dt}\) | M1 | |
| \(\frac{(1+t^2)-2t^2}{(1+t^2)^2}\) or \((1+t^2)^{-1}-2t^2(1+t^2)^{-2}\) | A1 | |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) | M1 | |
| Obtain \(\frac{1-t^2}{-2t}\) aef. | A1 | [5] |
| (ii) Put their \(\frac{dy}{dx} = 0\) and attempt to solve | M1 | |
| Obtain \(\left(\frac{1}{2}, \frac{1}{2}\right)\) AND \(\left(\frac{1}{2}, -\frac{1}{2}\right)\) [must obtain both] | A1 | [2] |
(i) $\frac{dx}{dt} = \frac{-2t}{(1+t^2)^2}$ | B1 |
Attempt use of quotient or product rule for $\frac{dy}{dt}$ | M1 |
$\frac{(1+t^2)-2t^2}{(1+t^2)^2}$ or $(1+t^2)^{-1}-2t^2(1+t^2)^{-2}$ | A1 |
Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 |
Obtain $\frac{1-t^2}{-2t}$ aef. | A1 | [5]
(ii) Put their $\frac{dy}{dx} = 0$ and attempt to solve | M1 |
Obtain $\left(\frac{1}{2}, \frac{1}{2}\right)$ AND $\left(\frac{1}{2}, -\frac{1}{2}\right)$ [must obtain both] | A1 | [2]The parametric equations of a curve are $x = \frac{1}{1 + t^2}$ and $y = \frac{t}{1 + t^2}$, $t \in \mathbb{R}$.
\begin{enumerate}[label=(\roman*)]
\item Find $\frac{dy}{dx}$ in terms of $t$. [5]
\item Hence find the coordinates of the stationary points of the curve. [2]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2010 Q5 [7]}}