| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2012 |
| Session | Specimen |
| Marks | 8 |
| Topic | Parametric differentiation |
| Type | Find parameter value given gradient condition |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = dy/dt รท dx/dt) followed by solving a simple exponential equation. Both parts are standard textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
**(i)**
- Either $\frac{\mathrm{d}y}{\mathrm{d}t} = 2e^{2t} - 2$ or $\frac{\mathrm{d}x}{\mathrm{d}t} = 2e^{2t} - 5$ [B1]
- $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t}$ used [M1]
- $= \frac{2e^{2t}-2}{2e^{2t}-5}$ [A1]
**(ii)**
- Set $\frac{\mathrm{d}y}{\mathrm{d}x} = 2$ [M1]
- Correctly rearrange to $e^{2t} = k$ [M1]
- Obtain $e^{2t} = 4$ [A1]
- Correct use of logs [M1]
- $t = \ln 2$ (allow $\frac{1}{2}\ln 4$) [A1]
**Total: 8 marks**9 The parametric equations of a curve are
$$x = \mathrm { e } ^ { 2 t } - 5 t , \quad y = \mathrm { e } ^ { 2 t } - 2 t$$
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.\\
(ii) Find the exact value of $t$ at the point on the curve where the gradient is 2 .
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2012 Q9 [8]}}