| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find second derivative d²y/dx² |
| Difficulty | Challenging +1.2 This is a Further Maths parametric differentiation question requiring the chain rule and product rule with inverse trig functions. Part (a) is scaffolded (show that), and part (b) requires applying d²y/dx² = (d/dt(dy/dx))/(dx/dt), which is a standard technique. The inverse trig derivatives are bookwork, making this moderately above average but routine for Further Maths students. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dt}=\cos^{-1}t - \frac{t}{\sqrt{1-t^2}}, \quad \frac{dx}{dt}=\frac{1}{\sqrt{1-t^2}}\) | B1 | |
| \(\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}=-t+\sqrt{1-t^2}\cos^{-1}t\) | M1 A1 | Uses chain rule, AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d}{dt}\!\left(\frac{dy}{dx}\right)=-1-\frac{\sqrt{1-t^2}}{\sqrt{1-t^2}}-\frac{t\cos^{-1}t}{\sqrt{1-t^2}}=-2-\frac{t\cos^{-1}t}{\sqrt{1-t^2}}\) | M1 A1 | Applies product rule |
| \(\frac{d^2y}{dx^2}=-2\sqrt{1-t^2}-t\cos^{-1}t\) | M1 A1 | Uses chain rule |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dt}=\cos^{-1}t - \frac{t}{\sqrt{1-t^2}}, \quad \frac{dx}{dt}=\frac{1}{\sqrt{1-t^2}}$ | B1 | |
| $\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}=-t+\sqrt{1-t^2}\cos^{-1}t$ | M1 A1 | Uses chain rule, AG |
---
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dt}\!\left(\frac{dy}{dx}\right)=-1-\frac{\sqrt{1-t^2}}{\sqrt{1-t^2}}-\frac{t\cos^{-1}t}{\sqrt{1-t^2}}=-2-\frac{t\cos^{-1}t}{\sqrt{1-t^2}}$ | M1 A1 | Applies product rule |
| $\frac{d^2y}{dx^2}=-2\sqrt{1-t^2}-t\cos^{-1}t$ | M1 A1 | Uses chain rule |
---3 It is given that
$$\mathrm { x } = \sin ^ { - 1 } \mathrm { t } \quad \text { and } \quad \mathrm { y } = \mathrm { tcos } ^ { - 1 } \mathrm { t } , \quad \text { for } 0 \leqslant t < 1 .$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { d y } { d x } = - t + \sqrt { 1 - t ^ { 2 } } \cos ^ { - 1 } t$.
\item Find $\frac { d ^ { 2 } y } { d x ^ { 2 } }$ in terms of $t$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q3 [7]}}