| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find second derivative d²y/dx² |
| Difficulty | Standard +0.8 This is a Further Maths parametric differentiation question requiring the product rule, quotient rule, and the chain rule formula for d²y/dx². Part (i) is routine verification, but part (ii) requires careful application of d²y/dx² = (d/dt(dy/dx))/(dx/dt) with algebraic manipulation of a complex expression involving exponentials and rational functions—more demanding than standard A-level but typical for FP1. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dx}{dt} = (2t - 2t^3) e^{-t^2}\), \(\frac{dy}{dt} = (1 - 2t^2) e^{-t^2}\) (both, AEF) | B1 | |
| Obtains displayed result (AG) | M1A1 | [3] |
| (ii) Any correct result for \(\frac{d(dy/dx)}{dt}\) in terms of \(t\) (*) | M1A1A1 | |
| M1 – Quotient Rule A1A1 for terms in numerator \(\frac{d^2y}{dx^2} = (*) \times \frac{dt}{dx}\) expressed in terms of \(t\) | M1 | |
| Simplify to, e.g., \(\frac{(-1 + t^2 - 2t^4)e^{t^2}}{4t^3(1-t^2)^2}\) | A1 | |
| Two terms in numerator must be combined. | [5] |
**(i)** $\frac{dx}{dt} = (2t - 2t^3) e^{-t^2}$, $\frac{dy}{dt} = (1 - 2t^2) e^{-t^2}$ (both, AEF) | B1 |
Obtains displayed result (AG) | M1A1 | [3] |
**(ii)** Any correct result for $\frac{d(dy/dx)}{dt}$ in terms of $t$ (*) | M1A1A1 |
M1 – Quotient Rule A1A1 for terms in numerator $\frac{d^2y}{dx^2} = (*) \times \frac{dt}{dx}$ expressed in terms of $t$ | M1 |
Simplify to, e.g., $\frac{(-1 + t^2 - 2t^4)e^{t^2}}{4t^3(1-t^2)^2}$ | A1 |
Two terms in numerator must be combined. | [5] |
---7 It is given that
$$x = t ^ { 2 } \mathrm { e } ^ { - t ^ { 2 } } \quad \text { and } \quad y = t \mathrm { e } ^ { - t ^ { 2 } }$$
(i) Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 - 2 t ^ { 2 } } { 2 t - 2 t ^ { 3 } }$$
(ii) Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ in terms of $t$.
\hfill \mbox{\textit{CAIE FP1 2010 Q7 [8]}}