| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find gradient at given parameter |
| Difficulty | Moderate -0.3 This is a straightforward parametric differentiation question requiring standard techniques: finding dy/dx using the chain rule (dy/dt รท dx/dt) and eliminating the parameter to find the cartesian equation. Both parts are routine C4 exercises with clear methods and minimal problem-solving required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| Note: You can mark parts (a) and (b) together. |
| Answer | Marks |
|---|---|
| \(x = 4t + 3\), \(y = 4t + 8 + \frac{5}{2t}\) | |
| \(\frac{dx}{dt} = 4\), \(\frac{dy}{dt} = 4 - \frac{5}{2}t^{-2}\) | B1 |
| Both \(\frac{dx}{dt} = 4\) or \(\frac{dt}{dy}\) and \(\frac{dy}{dt} = 4 - \frac{5}{2}t^{-2}\) | |
| Candidate's \(\frac{dy}{dt}\) divided by a candidate's \(\frac{dx}{dt}\) or \(\frac{dy}{dt}\) multiplied by a candidate's \(\frac{dt}{dx}\) | M1 |
| So, \(\frac{dy}{dx} = \frac{4-\frac{5}{2}t^{-2}}{4} = \left\{1-\frac{5}{8}t^{-2} = 1-\frac{5}{8t^2}\right\}\) | |
| When \(t = 2\), \(\frac{dy}{dx} = \frac{27}{32}\) | A1 |
| [3] |
| Answer | Marks |
|---|---|
| Way 2: Cartesian Method | |
| \(\frac{dy}{dx} = 1 - \frac{10}{(x-3)^2}\) | B1 |
| \(\frac{dy}{dx} = 1 - \frac{10}{(x-3)^2}\), simplified or un-simplified. | |
| \(\frac{dy}{dx} = \pm 2 \pm \frac{\mu}{(x-3)^2}\), \(\lambda \neq 0, \mu \neq 0\) | M1 |
| When \(t = 2, x = 11\): \(\frac{dy}{dx} = \frac{27}{32}\) or \(0.84375\) cao | A1 |
| [3] | |
| Way 3: Cartesian Method | |
| \(\frac{dy}{dx} = \frac{(2x+2)(x-3)-(x^2+2x-5)}{(x-3)^2}\) | B1 |
| Correct expression for \(\frac{dy}{dx}\), simplified or un-simplified. | |
| \(\frac{dy}{dx} = \frac{f'(x)(x-3)-1f(x)}{(x-3)^2}\) | M1 |
| where \(f(x) =\) their "\(x^2 + ax + b\)", \(g(x) = x - 3\) | |
| When \(t = 2, x = 11\): \(\frac{dy}{dx} = \frac{27}{32}\) or \(0.84375\) cao | A1 |
| [3] | |
| \(t = \frac{x - 3}{4} \Rightarrow y = 4\left(\frac{x-3}{4}\right) + 8 + \frac{5}{2\left(\frac{x-3}{4}\right)}\) | M1 |
| Eliminates \(t\) to achieve an equation in only \(x\) and \(y\). | |
| \(y = x - 3 + 8 + \frac{10}{x-3}\) | |
| \(y = \frac{(x-3)(x-3)+8(x-3)+10}{x-3}\) or \(y = \frac{(x+5)(x-3)+10}{x-3}\) or \(y = \frac{(x+5)(x-3)}{x-3} + \frac{10}{x-3}\) | dM1 |
| See notes | |
| or \(y = \frac{(x-3)(x-3)+8(x-3)+10}{x-3}\) or \(y = x + 5 + \frac{10}{x-3}\) | |
| \(\Rightarrow y = \frac{x^2+2x-5}{x-3}\), \(\{a=2\) and \(b = -5\}\) | A1 |
| Correct algebra leading to \(y = \frac{x^2+2x-5}{x-3}\) or \(a = 2\) and \(b = -5\) cso | |
| [3] |
Note: You can mark parts (a) and (b) together. | |
## Part (a)
$x = 4t + 3$, $y = 4t + 8 + \frac{5}{2t}$ | |
$\frac{dx}{dt} = 4$, $\frac{dy}{dt} = 4 - \frac{5}{2}t^{-2}$ | B1 |
Both $\frac{dx}{dt} = 4$ or $\frac{dt}{dy}$ and $\frac{dy}{dt} = 4 - \frac{5}{2}t^{-2}$ | |
Candidate's $\frac{dy}{dt}$ divided by a candidate's $\frac{dx}{dt}$ or $\frac{dy}{dt}$ multiplied by a candidate's $\frac{dt}{dx}$ | M1 |
So, $\frac{dy}{dx} = \frac{4-\frac{5}{2}t^{-2}}{4} = \left\{1-\frac{5}{8}t^{-2} = 1-\frac{5}{8t^2}\right\}$ | |
When $t = 2$, $\frac{dy}{dx} = \frac{27}{32}$ | A1 |
| [3] |
## Part (b)
Way 2: Cartesian Method | |
$\frac{dy}{dx} = 1 - \frac{10}{(x-3)^2}$ | B1 |
$\frac{dy}{dx} = 1 - \frac{10}{(x-3)^2}$, simplified or un-simplified. | |
$\frac{dy}{dx} = \pm 2 \pm \frac{\mu}{(x-3)^2}$, $\lambda \neq 0, \mu \neq 0$ | M1 |
When $t = 2, x = 11$: $\frac{dy}{dx} = \frac{27}{32}$ or $0.84375$ cao | A1 |
| [3] |
Way 3: Cartesian Method | |
$\frac{dy}{dx} = \frac{(2x+2)(x-3)-(x^2+2x-5)}{(x-3)^2}$ | B1 |
Correct expression for $\frac{dy}{dx}$, simplified or un-simplified. | |
$\frac{dy}{dx} = \frac{f'(x)(x-3)-1f(x)}{(x-3)^2}$ | M1 |
where $f(x) =$ their "$x^2 + ax + b$", $g(x) = x - 3$ | |
When $t = 2, x = 11$: $\frac{dy}{dx} = \frac{27}{32}$ or $0.84375$ cao | A1 |
| [3] |
$t = \frac{x - 3}{4} \Rightarrow y = 4\left(\frac{x-3}{4}\right) + 8 + \frac{5}{2\left(\frac{x-3}{4}\right)}$ | M1 |
Eliminates $t$ to achieve an equation in only $x$ and $y$. | |
$y = x - 3 + 8 + \frac{10}{x-3}$ | |
$y = \frac{(x-3)(x-3)+8(x-3)+10}{x-3}$ or $y = \frac{(x+5)(x-3)+10}{x-3}$ or $y = \frac{(x+5)(x-3)}{x-3} + \frac{10}{x-3}$ | dM1 |
See notes | |
or $y = \frac{(x-3)(x-3)+8(x-3)+10}{x-3}$ or $y = x + 5 + \frac{10}{x-3}$ | |
$\Rightarrow y = \frac{x^2+2x-5}{x-3}$, $\{a=2$ and $b = -5\}$ | A1 |
Correct algebra leading to $y = \frac{x^2+2x-5}{x-3}$ or $a = 2$ and $b = -5$ cso | |
| [3] |
---A curve $C$ has parametric equations
$$x = 4t + 3, \quad y = 4t + 8 + \frac{5}{2t}, \quad t \neq 0$$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\frac{dy}{dx}$ at the point on $C$ where $t = 2$, giving your answer as a fraction in its simplest form.
[3]
\item Show that the cartesian equation of the curve $C$ can be written in the form
$$y = \frac{x^2 + ax + b}{x - 3}, \quad x \neq 3$$
where $a$ and $b$ are integers to be determined.
[3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2015 Q5 [6]}}