| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Show dy/dx equals expression |
| Difficulty | Standard +0.3 This is a standard C4 parametric differentiation question requiring routine application of dy/dx = (dy/dt)/(dx/dt), finding a normal line, and eliminating the parameter. All techniques are textbook exercises with straightforward algebra, making it slightly easier than average for A-level but typical for C4 content. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dx}{dt} = \frac{1 \times (2-t) - tx(-1)}{(2-t)^2} = \frac{2}{(2-t)^2}\) | M1 B1 | |
| \(\frac{dy}{dt} = -(1+t)^{-2}\) | ||
| \(\frac{dy}{dx} = -\frac{1}{(1+t)^2} + \frac{2}{(2-t)^2} = -\frac{(2-t)^2}{2(1+t)^2} = -\frac{1}{2}\left(\frac{2-t}{1+t}\right)^2\) | M1 A1 | |
| (b) at \(t = 1, x = 1, y = \frac{1}{2}\), grad \(= -\frac{1}{8}\) | B1 | |
| grad of normal \(= 8\) | ||
| \(\therefore y - \frac{1}{2} = 8(x - 1)\) | M1 A1 | [\(y = 8x - \frac{15}{2}\)] |
| (c) \(x(2-t) = t\) | M1 | |
| \(2x = t(1 + x), \quad t = \frac{2x}{1+x}\) | A1 | |
| \(y = \frac{1}{1 + \frac{2x}{1+x}} = \frac{1+x}{(1+x) + 2x} = \frac{1+x}{1+3x}\) | M1 A1 | (11 marks total) |
**(a)** $\frac{dx}{dt} = \frac{1 \times (2-t) - tx(-1)}{(2-t)^2} = \frac{2}{(2-t)^2}$ | M1 B1 |
$\frac{dy}{dt} = -(1+t)^{-2}$ | |
$\frac{dy}{dx} = -\frac{1}{(1+t)^2} + \frac{2}{(2-t)^2} = -\frac{(2-t)^2}{2(1+t)^2} = -\frac{1}{2}\left(\frac{2-t}{1+t}\right)^2$ | M1 A1 |
**(b)** at $t = 1, x = 1, y = \frac{1}{2}$, grad $= -\frac{1}{8}$ | B1 |
grad of normal $= 8$ | |
$\therefore y - \frac{1}{2} = 8(x - 1)$ | M1 A1 | [$y = 8x - \frac{15}{2}$] |
**(c)** $x(2-t) = t$ | M1 |
$2x = t(1 + x), \quad t = \frac{2x}{1+x}$ | A1 |
$y = \frac{1}{1 + \frac{2x}{1+x}} = \frac{1+x}{(1+x) + 2x} = \frac{1+x}{1+3x}$ | M1 A1 | (11 marks total) |A curve has parametric equations
$$x = \frac{t}{2-t}, \quad y = \frac{1}{1+t}, \quad -1 < t < 2.$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{dy}{dx} = -\frac{1}{2}\left(\frac{2-t}{1+t}\right)^2$. [4]
\item Find an equation for the normal to the curve at the point where $t = 1$. [3]
\item Show that the cartesian equation of the curve can be written in the form
$$y = \frac{1+x}{1+3x}.$$ [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q5 [11]}}