| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2016 |
| Session | Specimen |
| Marks | 8 |
| 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 given parameter value. The exponential and linear terms differentiate cleanly, and substituting t=0 gives simple values. Slightly easier than average due to being a routine two-part application of standard techniques with no complications. |
| Spec | 1.07s Parametric and implicit differentiation |
(i) 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]
(ii) 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]8 The parametric equations of a curve are
$$x = \mathrm { e } ^ { 2 t } - 5 t , \quad y = \mathrm { e } ^ { 2 t } - 3 t .$$
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.\\
(ii) 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.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2016 Q8 [8]}}