| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2019 |
| Session | Specimen |
| Marks | 5 |
| Topic | Parametric differentiation |
| Type | Find tangent equation at parameter |
| Difficulty | Moderate -0.3 This is a straightforward parametric differentiation question requiring the standard formula dy/dx = (dy/dt)/(dx/dt), followed by finding a tangent equation at a specific parameter value. The exponential and linear terms differentiate cleanly, and substituting t=0 gives simple values. Slightly easier than average due to the routine nature and clean arithmetic, though it does require correct application of the chain rule for parametric equations. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
**(a)**
Either $\frac{dy}{dt} = 2e^{2t} - 3$ or $\frac{dx}{dt} = 2e^{2t} - 5$ — **B1**
$\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ used — **M1**
$= \frac{2e^{2t}-3}{2e^{2t}-5}$ — **A1** [3]
**(b)**
Substitute $t = 0$ to obtain gradient $= \frac{-1}{-3}$ or equivalent — **B1**
Obtain $x = 1$ — **B1**
Obtain $y = 1$ — **B1**
Form equation of a straight line — **M1**
Obtain $3y - x = 2$ — **A1** [5]8 The parametric equations of a curve are
$$x = \mathrm { e } ^ { 2 t } - 5 t , \quad y = \mathrm { e } ^ { 2 t } - 3 t .$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.
\item Find the equation of the tangent to the curve at the point when $t = 0$, giving your answer in the form $a y + b x + c = 0$ where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2019 Q8 [5]}}