| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| 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 question requiring application of arc length and surface area formulas. While it involves multiple steps and exponential functions, the techniques are routine: differentiate parametrics, substitute into standard formulas, and integrate. The integrals simplify nicely (completing squares for arc length, straightforward for surface area), making this a textbook-style question that's harder than typical A-level but not requiring novel insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian4.08d Volumes of revolution: about x and y axes8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\dot{x} = e^t - 4\), \(\dot{y} = 4e^{\frac{1}{2}t}\) (both) | B1 | |
| \(\dot{x}^2 + \dot{y}^2 = e^{2t} - 8e^t + 16 + 16e^t = (e^t + 4)^2\) | M1A1 | ACF |
| \(s = \int_0^2 (e^t + 4)\,dt = \left[e^t + 4t\right]_0^2 = e^2 + 8 - 1 = e^2 + 7\) | M1A1 | |
| Total | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(S = 2\pi\int_0^2 8e^{\frac{1}{2}t}(e^t + 4)\,dt\) | B1✓ | |
| \(16\pi\int_0^2\left(e^{\frac{3}{2}t} + 4e^{\frac{1}{2}t}\right)dt = 16\pi\left[\frac{2}{3}e^{\frac{3}{2}t} + 8e^{\frac{1}{2}t}\right]_0^2\) | M1A1 | |
| \(= 16\pi\left(\frac{2}{3}e^3 + 8e - \frac{26}{3}\right)\) | M1A1 | ACF |
| Total | [5] | N.B. B1M1A1 can be earned without use of limits |
# Question 6:
## Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\dot{x} = e^t - 4$, $\dot{y} = 4e^{\frac{1}{2}t}$ (both) | B1 | |
| $\dot{x}^2 + \dot{y}^2 = e^{2t} - 8e^t + 16 + 16e^t = (e^t + 4)^2$ | M1A1 | ACF |
| $s = \int_0^2 (e^t + 4)\,dt = \left[e^t + 4t\right]_0^2 = e^2 + 8 - 1 = e^2 + 7$ | M1A1 | |
| Total | [5] | |
## Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $S = 2\pi\int_0^2 8e^{\frac{1}{2}t}(e^t + 4)\,dt$ | B1✓ | |
| $16\pi\int_0^2\left(e^{\frac{3}{2}t} + 4e^{\frac{1}{2}t}\right)dt = 16\pi\left[\frac{2}{3}e^{\frac{3}{2}t} + 8e^{\frac{1}{2}t}\right]_0^2$ | M1A1 | |
| $= 16\pi\left(\frac{2}{3}e^3 + 8e - \frac{26}{3}\right)$ | M1A1 | ACF |
| Total | [5] | N.B. B1M1A1 can be earned without use of limits |
---6 The curve $C$ has parametric equations
$$x = \mathrm { e } ^ { t } - 4 t + 3 , \quad y = 8 \mathrm { e } ^ { \frac { 1 } { 2 } t } , \quad \text { for } 0 \leqslant t \leqslant 2$$
(i) Find, in terms of e , the length of $C$.\\
(ii) 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 FP1 2014 Q6 [10]}}