| 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 arc length and surface of revolution question requiring straightforward application of formulas. The derivatives are simple polynomials, and the resulting integral √(4t² + (1-t²)²) simplifies cleanly to (1+t²), making both parts routine calculations without requiring problem-solving insight or difficult integration techniques. |
| Spec | 8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(\dot{x} = 2t\), \(\dot{y} = 1 - t^2\) | B1 | |
| \(\dot{x}^2 + \dot{y}^2 = 4t^2 + 1 - 2t^2 + t^4 = (1+t^2)^2\) | M1A1 | ACF |
| \(s = \int_0^1(1+t^2)\,dt = \left[t + \frac{t^3}{3}\right]_0^1 = \frac{4}{3}\) or 1.33 | M1A1 [5] | LR |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(S = 2\pi\int_0^1\left(t - \frac{1}{3}t^3\right)(1+t^2)\,dt = 2\pi\int_0^1\left(t + \frac{2}{3}t^3 - \frac{1}{3}t^5\right)dt\) | M1A1 | LNR |
| \(= 2\pi\left[\frac{1}{2}t^2 + \frac{1}{6}t^4 - \frac{1}{18}t^6\right]_0^1\) | M1A1 | LR |
| \(= \frac{11}{9}\pi\) or 3.84 | A1 [5] |
# Question 8:
## Part (i):
| Answer/Working | Mark | Notes |
|---|---|---|
| $\dot{x} = 2t$, $\dot{y} = 1 - t^2$ | B1 | |
| $\dot{x}^2 + \dot{y}^2 = 4t^2 + 1 - 2t^2 + t^4 = (1+t^2)^2$ | M1A1 | ACF |
| $s = \int_0^1(1+t^2)\,dt = \left[t + \frac{t^3}{3}\right]_0^1 = \frac{4}{3}$ or 1.33 | M1A1 [5] | LR |
## Part (ii):
| Answer/Working | Mark | Notes |
|---|---|---|
| $S = 2\pi\int_0^1\left(t - \frac{1}{3}t^3\right)(1+t^2)\,dt = 2\pi\int_0^1\left(t + \frac{2}{3}t^3 - \frac{1}{3}t^5\right)dt$ | M1A1 | LNR |
| $= 2\pi\left[\frac{1}{2}t^2 + \frac{1}{6}t^4 - \frac{1}{18}t^6\right]_0^1$ | M1A1 | LR |
| $= \frac{11}{9}\pi$ or 3.84 | A1 [5] | |
---8 The curve $C$ has parametric equations
$$x = t ^ { 2 } , \quad y = t - \frac { 1 } { 3 } t ^ { 3 } , \quad \text { for } 0 \leqslant t \leqslant 1 .$$
Find\\
(i) the arc length of $C$,\\
(ii) the surface area generated when $C$ is rotated through $2 \pi$ radians about the $x$-axis.
\hfill \mbox{\textit{CAIE FP1 2014 Q8 [10]}}