| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find second derivative d²y/dx² |
| Difficulty | Challenging +1.3 This is a standard Further Maths parametric calculus question requiring arc length integration and second derivative calculation. Part (a) involves computing √((dx/dt)² + (dy/dt)²) which simplifies nicely to |t + 1/t|, leading to a straightforward integration. Part (b) requires the chain rule for d²y/dx² = d/dx(dy/dx) = (d/dt(dy/dx))/(dx/dt), involving routine algebraic manipulation. While requiring multiple techniques and careful algebra, these are well-practiced procedures for Further Maths students with no novel problem-solving insight needed. |
| Spec | 1.07s Parametric and implicit differentiation4.08d Volumes of revolution: about x and y axes |
| Answer | Marks |
|---|---|
| 5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or |
| Answer | Marks |
|---|---|
| 5(a) | x= − −1, y= |
| t t 2 | B1 |
| Answer | Marks |
|---|---|
| x2+y2 =t2−2+t −2+4= t+t −1 2 | M1 A1 |
| Answer | Marks |
|---|---|
| 2 2 | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 5(b) | dy y 2 |
| Answer | Marks |
|---|---|
| dx x t−t−1 | B1 |
| Answer | Marks |
|---|---|
| dt t−t−1 ( t−t−1 )2 | B1 |
| Answer | Marks |
|---|---|
| dx2 dt t−t−1 dx ( t −t−1 )3 | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | x= − −1, y=
t t 2 | B1
( )
x2+y2 =t2−2+t −2+4= t+t −1 2 | M1 A1
2 t+t−1dt = 1t2 +lnt 2 = 15 +2ln2
1 2 1 8
2 2 | M1 A1
5
Question | Answer | Marks
--- 5(b) ---
5(b) | dy y 2
= =
dx x t−t−1 | B1
( )
−2 1+t−2
d 2
=
dt t−t−1 ( t−t−1 )2 | B1
( )
−2 1+t−2
d2y d 2 dt
= × =
dx2 dt t−t−1 dx ( t −t−1 )3 | M1 A1
4
Question | Answer | MarksThe curve $C$ has parametric equations
$$x = \frac{1}{2}t^2 - \ln t, \quad y = 2t + 1, \quad \text{for } \frac{1}{2} \leqslant t \leqslant 2.$$
\begin{enumerate}[label=(\alph*)]
\item Find the exact length of $C$. [5]
\item Find $\frac{d^2y}{dx^2}$ in terms of $t$, simplifying your answer. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q5 [9]}}