| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find second derivative d²y/dx² |
| Difficulty | Standard +0.3 This is a standard Further Maths parametric differentiation question requiring the chain rule formula dy/dx = (dy/dt)/(dx/dt) and the second derivative formula. While it involves trigonometric functions and substitution at a specific parameter value, the techniques are routine for FP1 students with no novel problem-solving required. Slightly above average difficulty due to being Further Maths content and requiring careful algebraic manipulation for the second derivative. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{\dot{y}}{\dot{x}} = \frac{-6\sin 2t}{4\cos 2t} = -\frac{3}{2}\tan 2t\) | M1A1 | Finds first derivative |
| When \(t=\frac{\pi}{3}\), \(\quad \frac{dy}{dx} = \frac{3\sqrt{3}}{2}\) | A1 | Evaluates; Part marks: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right)\times\frac{dt}{dx} = -3\sec^2 2t \times \frac{1}{4}\sec 2t\) | M1A1 | Finds second derivative |
| \(= -\frac{3}{4}\sec^3 2t\) | A1 | |
| When \(t=\frac{\pi}{3}\), \(\quad \frac{d^2y}{dx^2} = \frac{3}{4}\times 8 = 6\) | A1 | Evaluates; Part marks: 4; Total: [7] |
| Alternatively (i): \(\left(\frac{x}{2}\right)^2+\left(\frac{y}{3}\right)^2=1 \Rightarrow y' = -\frac{9x}{4y} = -\frac{9}{4}\times\frac{-2\sqrt{3}}{3} = \frac{3\sqrt{3}}{2}\) | M1A1, A1 | Part marks: 3 |
| Alternatively (ii): \(\frac{1}{2}+\frac{2}{9}\left[(y')^2+yy''\right]=0 \Rightarrow \frac{1}{2}+\frac{3}{2} = \frac{1}{3}y'' \Rightarrow y''=6\) | M1A2, A1 | Part marks: 4; Total: [7] |
## Question 4(i):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{\dot{y}}{\dot{x}} = \frac{-6\sin 2t}{4\cos 2t} = -\frac{3}{2}\tan 2t$ | M1A1 | Finds first derivative |
| When $t=\frac{\pi}{3}$, $\quad \frac{dy}{dx} = \frac{3\sqrt{3}}{2}$ | A1 | Evaluates; Part marks: 3 |
## Question 4(ii):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right)\times\frac{dt}{dx} = -3\sec^2 2t \times \frac{1}{4}\sec 2t$ | M1A1 | Finds second derivative |
| $= -\frac{3}{4}\sec^3 2t$ | A1 | |
| When $t=\frac{\pi}{3}$, $\quad \frac{d^2y}{dx^2} = \frac{3}{4}\times 8 = 6$ | A1 | Evaluates; Part marks: 4; **Total: [7]** |
**Alternatively (i):** $\left(\frac{x}{2}\right)^2+\left(\frac{y}{3}\right)^2=1 \Rightarrow y' = -\frac{9x}{4y} = -\frac{9}{4}\times\frac{-2\sqrt{3}}{3} = \frac{3\sqrt{3}}{2}$ | M1A1, A1 | Part marks: 3
**Alternatively (ii):** $\frac{1}{2}+\frac{2}{9}\left[(y')^2+yy''\right]=0 \Rightarrow \frac{1}{2}+\frac{3}{2} = \frac{1}{3}y'' \Rightarrow y''=6$ | M1A2, A1 | Part marks: 4; **Total: [7]** |
---4 A curve has parametric equations
$$x = 2 \sin 2 t , \quad y = 3 \cos 2 t$$
for $0 < t < \frac { 1 } { 2 } \pi$. For the point on the curve where $t = \frac { 1 } { 3 } \pi$, find the value of\\
(i) $\frac { \mathrm { d } y } { \mathrm {~d} x }$,\\
(ii) $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\hfill \mbox{\textit{CAIE FP1 2011 Q4 [7]}}