| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Parametric surface area of revolution |
| Difficulty | Challenging +1.2 This is a standard Further Maths parametric arc length and surface of revolution question requiring computation of √((dx/dt)² + (dy/dt)²) and integration. The derivatives are straightforward (e^t - 4 and 4e^(t/2)), and the resulting integrand simplifies to a perfect square (e^t - 2e^(t/2))², making the integration routine. While it requires multiple steps and careful algebra, it follows a well-practiced template without requiring novel insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dot{x} = e^t - 4;\ \dot{y} = 4e^{\frac{1}{2}t}\) | B1 | find \(dx/dt\) and \(dy/dt\) |
| \(ds = \sqrt{e^{2t} + 16 + 8e^t} = e^t + 4\) | M1A1 | |
| Arc length \(= \int_0^2 (e^t + 4)dt = [e^t + 4t]_0^2 = e^2 + 7\) | M1A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S = 2\pi\int_0^2 8e^{\frac{1}{2}t}(e^t + 4)dt\) | B1FT | FT; limits not required |
| \(= 16\pi\int_0^2 (e^{1.5t} + 4e^{0.5t})dt\) | M1 | limits not required |
| \(16\pi\left[\frac{2}{3}e^{1.5t} + 8e^{0.5t}\right]_0^2\) | M1A1 | correct integration |
| \(= 16\pi\left(\frac{2}{3}e^3 + 8e - \frac{26}{3}\right)\) | A1 | CAO |
| Total: 5 |
# Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dot{x} = e^t - 4;\ \dot{y} = 4e^{\frac{1}{2}t}$ | B1 | find $dx/dt$ and $dy/dt$ |
| $ds = \sqrt{e^{2t} + 16 + 8e^t} = e^t + 4$ | M1A1 | |
| Arc length $= \int_0^2 (e^t + 4)dt = [e^t + 4t]_0^2 = e^2 + 7$ | M1A1 | |
| **Total: 5** | | |
# Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S = 2\pi\int_0^2 8e^{\frac{1}{2}t}(e^t + 4)dt$ | B1FT | FT; limits not required |
| $= 16\pi\int_0^2 (e^{1.5t} + 4e^{0.5t})dt$ | M1 | limits not required |
| $16\pi\left[\frac{2}{3}e^{1.5t} + 8e^{0.5t}\right]_0^2$ | M1A1 | correct integration |
| $= 16\pi\left(\frac{2}{3}e^3 + 8e - \frac{26}{3}\right)$ | A1 | CAO |
| **Total: 5** | | |5 The curve $C$ has parametric equations
$$x = \mathrm { e } ^ { t } - 4 t + 3 , \quad y = 8 \mathrm { e } ^ { \frac { 1 } { 2 } t } , \quad \text { for } 0 \leqslant t \leqslant 2 .$$
(a) Find, in terms of e , the length of $C$.\\
(b) Find, in terms of $\pi$ and e , the area of the surface generated when $C$ is rotated through $2 \pi$ radians about the $x$-axis.\\
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q5 [10]}}