| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2014 |
| Session | June |
| Marks | 4 |
| 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 ÷ dx/dt) followed by solving a simple equation involving exponentials. The algebra is routine and the techniques are standard A-level material, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
**(i)** Attempt $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \times \frac{\mathrm{d}t}{\mathrm{d}x}$ — M1
Obtain $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^t - 5}{e^t - 2}$ — A1 **[2]**
**(ii)** Equate their derivative to 3 and attempt to solve — M1
Obtain $e^t = 0.5$ — A1
Attempt ln on both sides and use power law — M1
Obtain $t = -\ln 2$ AG — A1 **[4]**
OR
Substitute $t = -\ln 2$ into $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^t-5}{e^t-2}$ — M1
Use power log law to show or imply $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^{\ln\frac{1}{2}}-5}{e^{\ln\frac{1}{2}}-2}$ — M1
Obtain $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{1}{2}-5}{\frac{1}{2}-2}$ — A1
Obtain 3 — A18 The parametric equations of a curve are given by
$$x = \mathrm { e } ^ { t } - 2 t , \quad y = \mathrm { e } ^ { t } - 5 t$$
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.\\
(ii) Show that $t = - \ln 2$ at the point on the curve where the gradient is 3 .
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2014 Q8 [4]}}