| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find second derivative d²y/dx² |
| Difficulty | Standard +0.8 This is a standard Further Maths parametric differentiation question requiring the chain rule for second derivatives. Part (a) involves routine application of dy/dx = (dy/dt)/(dx/dt), while part (b) requires the more challenging formula d²y/dx² = [d/dt(dy/dx)]/(dx/dt), involving product rule on an exponential-polynomial expression. The algebraic manipulation is moderately demanding but follows a well-practiced technique for Further Maths students. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dt} = e^t(t+1)\), \(\frac{dx}{dt} = -t^{-2}\) | B1 | |
| \(\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = -t^2 e^t(t+1) = -e^t(t^3+t^2)\) | M1 A1 | Uses chain rule, AG |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d}{dt}\left(\frac{dy}{dx}\right) = -e^t(3t^2+2t) - e^t(t^3+t^2) = -e^t(t^3+4t^2+2t)\) | M1 A1 | Applies product rule |
| \(\frac{d^2y}{dx^2} = t^3 e^t(t^2+4t+2)\) | M1 A1 | Uses chain rule, multiplies by \(\frac{dt}{dx}\) for M1 |
| Alternative: \(\frac{d^2y}{dx^2} = -e^t\left(3t^2\frac{dt}{dx}+2t\frac{dt}{dx}\right) - e^t\left(t^3+t^2\right)\frac{dt}{dx}\) | (M1 A1) | Differentiates \(-e^t(t^3+t^2)\) implicitly |
| \(\frac{d^2y}{dx^2} = e^t\left(-4t^2-2t-t^3\right)\frac{dt}{dx} = t^3e^t(t^2+4t+2)\) | (M1 A1) | Substitutes \(\frac{dt}{dx} = -t^2\) |
| Total: 4 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dt} = e^t(t+1)$, $\frac{dx}{dt} = -t^{-2}$ | B1 | |
| $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = -t^2 e^t(t+1) = -e^t(t^3+t^2)$ | M1 A1 | Uses chain rule, AG |
| **Total: 3** | | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dt}\left(\frac{dy}{dx}\right) = -e^t(3t^2+2t) - e^t(t^3+t^2) = -e^t(t^3+4t^2+2t)$ | M1 A1 | Applies product rule |
| $\frac{d^2y}{dx^2} = t^3 e^t(t^2+4t+2)$ | M1 A1 | Uses chain rule, multiplies by $\frac{dt}{dx}$ for M1 |
| **Alternative:** $\frac{d^2y}{dx^2} = -e^t\left(3t^2\frac{dt}{dx}+2t\frac{dt}{dx}\right) - e^t\left(t^3+t^2\right)\frac{dt}{dx}$ | (M1 A1) | Differentiates $-e^t(t^3+t^2)$ implicitly |
| $\frac{d^2y}{dx^2} = e^t\left(-4t^2-2t-t^3\right)\frac{dt}{dx} = t^3e^t(t^2+4t+2)$ | (M1 A1) | Substitutes $\frac{dt}{dx} = -t^2$ |
| **Total: 4** | | |2 It is given that
$$x = 1 + \frac { 1 } { t } \quad \text { and } \quad y = t e ^ { t }$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { d y } { d x } = - e ^ { t } \left( t ^ { 3 } + t ^ { 2 } \right)$.
\item Find $\frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dx } ^ { 2 } }$ in terms of $t$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q2 [7]}}