| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find tangent equation at parameter |
| Difficulty | Moderate -0.3 This is a standard parametric differentiation question requiring the chain rule (dy/dx = (dy/dt)/(dx/dt)), substitution into point-slope form, and solving simultaneous linear equations. All techniques are routine for C4 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt}\) | M1 | Used, not just quoted |
| \(\frac{1}{t}\) or \(t^{-1}\) | A1 | 2 Not \(\frac{2}{2t}\) as final answer |
| SR: M1 for Cart conv, finding \(\frac{dy}{dx}\) & ans involv \(t\) + A1 | M1 is attempt only, accuracy not involved |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Finding equation of tangent (using \(p\) or \(t\)) | M1 | |
| \(py = x + p^2\) working | A1 | 2 AG; \(p\) essential; at least 1 line inter |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((25,-10) \Rightarrow p=-5\) or \(-5y=x+25\) seen | B1 | \(5y=x+25\) seen \(\Rightarrow\) B0 |
| Substitution of their values of \(p\) into given tgt eqn | M1 | Producing 2 equations |
| Solving the 2 equations simultaneously | M1 | |
| \((-15,-2)\) \(x=-15\), \(y=-2\) | A1 | 4 Common wrong ans \((15,8)\Rightarrow\) B0, M2, A0 |
# Question 5:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt}$ | M1 | Used, not just quoted |
| $\frac{1}{t}$ or $t^{-1}$ | A1 | **2** Not $\frac{2}{2t}$ as final answer |
| SR: M1 for Cart conv, finding $\frac{dy}{dx}$ & ans involv $t$ + A1 | | M1 is attempt only, accuracy not involved |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Finding equation of tangent (using $p$ or $t$) | M1 | |
| $py = x + p^2$ working | A1 | **2** AG; $p$ essential; at least 1 line inter |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(25,-10) \Rightarrow p=-5$ or $-5y=x+25$ seen | B1 | $5y=x+25$ seen $\Rightarrow$ B0 |
| Substitution of their values of $p$ into given tgt eqn | M1 | Producing 2 equations |
| Solving the 2 equations simultaneously | M1 | |
| $(-15,-2)$ $x=-15$, $y=-2$ | A1 | **4** Common wrong ans $(15,8)\Rightarrow$ B0, M2, A0 |5 A curve is given parametrically by the equations $x = t ^ { 2 } , y = 2 t$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$, giving your answer in its simplest form.\\
(ii) Show that the equation of the tangent to the curve at $\left( p ^ { 2 } , 2 p \right)$ is
$$p y = x + p ^ { 2 } .$$
(iii) Find the coordinates of the point where the tangent at $( 9,6 )$ meets the tangent at $( 25 , - 10 )$.
\hfill \mbox{\textit{OCR C4 2006 Q5 [8]}}