| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find normal equation at parameter |
| Difficulty | Moderate -0.3 This is a straightforward parametric differentiation question requiring standard application of dy/dx = (dy/dt)/(dx/dt) and finding a normal line equation. The logarithmic differentiation adds minor complexity, but both parts follow routine procedures with no problem-solving insight needed. Slightly easier than average due to its mechanical nature. |
| 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 |
|---|---|
| 4(i) | dy 2 |
| Answer | Marks |
|---|---|
| dt 2t−1 | B1 |
| Answer | Marks |
|---|---|
| dx dt dt | M1 |
| Answer | Marks |
|---|---|
| 4t −2t | A1 |
| Total: | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(ii) | Use correct method to find the gradient of the normal at t = 1 | M1 |
| Use a correct method to form an equation for the normal at t = 1 | M1 | |
| Obtain final answerx+3y−14=0, or horizontal equivalent | A1 | |
| Total: | 3 | |
| Question | Answer | Marks |
Question 4:
--- 4(i) ---
4(i) | dy 2
State =4+
dt 2t−1 | B1
dy dy dx
Use = ÷
dx dt dt | M1
dy 8t−2 2 2
Obtain answer = , or equivalent e.g. +
dx 2t(2t−1) t 2
4t −2t | A1
Total: | 3
--- 4(ii) ---
4(ii) | Use correct method to find the gradient of the normal at t = 1 | M1
Use a correct method to form an equation for the normal at t = 1 | M1
Obtain final answerx+3y−14=0, or horizontal equivalent | A1
Total: | 3
Question | Answer | MarksThe parametric equations of a curve are
$$x = t^2 + 1, \quad y = 4t + \ln(2t - 1).$$
\begin{enumerate}[label=(\roman*)]
\item Express $\frac{dy}{dx}$ in terms of $t$. [3]
\item Find the equation of the normal to the curve at the point where $t = 1$. Give your answer in the form $ax + by + c = 0$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2017 Q4 [6]}}