| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| 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 surface area question requiring the formula S = 2π∫y√((dx/dt)² + (dy/dt)²)dt, differentiation of hyperbolic-type expressions, and a guided substitution. While it involves multiple steps and careful algebra with exponentials, the structure is routine for FP1 and the substitution is provided, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.08h Integration by substitution4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dt} = \frac{-2(e^t - e^{-t})}{(e^t + e^{-t})^2}\) | B1 | |
| \(\frac{dy}{dt} = \frac{(e^t+e^{-t})^2 - (e^t-e^{-t})^2}{(e^t+e^{-t})^2} = \frac{4}{(e^t+e^{-t})^2}\) | B1 | Differentiates and simplifies. Accept \(1 - \left(\frac{e^t-e^{-t}}{e^t+e^{-t}}\right)^2\) |
| \(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \frac{4(e^t-e^{-t})^2+16}{(e^t+e^{-t})^4} = \frac{4(e^t+e^{-t})^2}{(e^t+e^{-t})^4}\) | M1 A1 | Attempt at writing \(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\) as a square |
| \(S = 2\pi\int_0^1\left(\frac{e^t-e^{-t}}{e^t+e^{-t}}\right)\left(\frac{2}{e^t+e^{-t}}\right)dt = 4\pi\int_0^1 \frac{e^t-e^{-t}}{(e^t+e^{-t})^2}\, dt\) | A1 | Uses correct formula, simplifies to AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(S = 4\pi\int_2^{e+e^{-1}} u^{-2}\, du = 4\pi\left[-u^{-1}\right]_2^{e+e^{-1}}\) | M1 A1 | Applies given substitution |
| \(= 4\pi\left(\frac{1}{2} - \frac{1}{e+e^{-1}}\right)\) | A1 | AEF, must be exact |
## Question 5(i):
| $\frac{dx}{dt} = \frac{-2(e^t - e^{-t})}{(e^t + e^{-t})^2}$ | B1 | |
|---|---|---|
| $\frac{dy}{dt} = \frac{(e^t+e^{-t})^2 - (e^t-e^{-t})^2}{(e^t+e^{-t})^2} = \frac{4}{(e^t+e^{-t})^2}$ | B1 | Differentiates and simplifies. Accept $1 - \left(\frac{e^t-e^{-t}}{e^t+e^{-t}}\right)^2$ |
| $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \frac{4(e^t-e^{-t})^2+16}{(e^t+e^{-t})^4} = \frac{4(e^t+e^{-t})^2}{(e^t+e^{-t})^4}$ | M1 A1 | Attempt at writing $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2$ as a square |
| $S = 2\pi\int_0^1\left(\frac{e^t-e^{-t}}{e^t+e^{-t}}\right)\left(\frac{2}{e^t+e^{-t}}\right)dt = 4\pi\int_0^1 \frac{e^t-e^{-t}}{(e^t+e^{-t})^2}\, dt$ | A1 | Uses correct formula, simplifies to AG |
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## Question 5(ii):
| $S = 4\pi\int_2^{e+e^{-1}} u^{-2}\, du = 4\pi\left[-u^{-1}\right]_2^{e+e^{-1}}$ | M1 A1 | Applies given substitution |
|---|---|---|
| $= 4\pi\left(\frac{1}{2} - \frac{1}{e+e^{-1}}\right)$ | A1 | AEF, must be exact |
---5 A curve $C$ is defined parametrically by
$$x = \frac { 2 } { \mathrm { e } ^ { t } + \mathrm { e } ^ { - t } } \quad \text { and } \quad y = \frac { \mathrm { e } ^ { t } - \mathrm { e } ^ { - t } } { \mathrm { e } ^ { t } + \mathrm { e } ^ { - t } }$$
for $0 \leqslant t \leqslant 1$. The area of the surface generated when $C$ is rotated through $2 \pi$ radians about the $x$-axis is denoted by $S$.\\
(i) Show that $S = 4 \pi \int _ { 0 } ^ { 1 } \frac { \mathrm { e } ^ { t } - \mathrm { e } ^ { - t } } { \left( \mathrm { e } ^ { t } + \mathrm { e } ^ { - t } \right) ^ { 2 } } \mathrm {~d} t$.\\
(ii) Using the substitution $u = \mathrm { e } ^ { t } + \mathrm { e } ^ { - t }$, or otherwise, find $S$ in terms of $\pi$ and e .\\
\hfill \mbox{\textit{CAIE FP1 2019 Q5 [8]}}