| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2012 |
| Session | June |
| Marks | 15 |
| Topic | Polar coordinates |
| Type | Arc length of polar curve |
| Difficulty | Challenging +1.8 This question combines a reduction formula derivation (requiring integration by parts with careful algebraic manipulation) with polar arc length calculation. Part (i) demands technical integration skills and algebraic insight to establish the recurrence relation. Part (ii)(b) requires recognizing that the arc length formula ds = √(r² + (dr/dθ)²)dθ leads to an integral matching the I_n form, then applying the reduction formula recursively—a sophisticated multi-stage problem requiring strong technique and strategic thinking, though the individual components are standard Further Maths topics. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve8.06a Reduction formulae: establish, use, and evaluate recursively |
**(i)** $I_n = \int_0^3 x^{n-1}\!\left(x\sqrt{16+x^2}\right)\mathrm{d}x$ Correct splitting *and* use of parts **M1**
$= \left[x^{n-1}\cdot\frac{(16+x^2)^{3/2}}{3}\right]_0^3 - \int_0^3(n-1)x^{n-2}\frac{(16+x^2)^{3/2}}{3}\,\mathrm{d}x$ **A1**
$= 3^{n-2}\cdot 125 - \left(\frac{n-1}{3}\right)\int_0^3 x^{n-2}(16+x^2)\sqrt{16+x^2}\,\mathrm{d}x$ **M1**
Method to get 2nd integral of correct form
$= 3^{n-2}\cdot 125 - \left(\frac{n-1}{3}\right)\left\{16I_{n-2} + I_n\right\}$ [i.e. reverting to $I$'s in 2nd integral **ft**] **M1**
$\Rightarrow 3I_n = 3^{n-1}\cdot 125 - 16(n-1)I_{n-2} - (n-1)I_n$ Collecting up $I_n$'s **M1**
$(n+2)I_n = 125\times 3^{n-1} - 16(n-1)I_{n-2}$ **AG** **A1**
**[6]**
**(ii)(a)** Spiral (with $r$ increasing) **B1**
From $O$ to just short of $\theta = \pi$ **B1**
**[2]**
**(b)** $r = \frac{1}{4}\theta^4 \Rightarrow \frac{\mathrm{d}r}{\mathrm{d}\theta} = \theta^3$ and $r^2 + \left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^2 = \frac{1}{16}\theta^8 + \theta^6$ **M1 A1**
$L = \int_0^3\frac{1}{4}\theta^3\sqrt{16+\theta^2}\,\mathrm{d}\theta\;\left(= \frac{1}{4}I_3\right)$ **M1 A1**
Now $I_1 = \left[\frac{1}{3}(16+x^2)^{3/2}\right]_0^3 = \frac{61}{3}$ **B1**
and $5I_3 = 125\times 9 - 16\times 2\left(\frac{61}{3}\right) = \frac{1423}{3}$ or $474\frac{1}{3}$ Use of given reduction formula **M1**
so that $L = \frac{1}{20}\times\frac{1423}{3} = \frac{1423}{60}$ or $23\frac{43}{60}$ or awrt 23.7 **ft** only from suitable $kI_3$ **A1**
**NB** The last 3 marks can be earned by integrating in a variety of ways
**[7]**12 (i) Let $I _ { n } = \int _ { 0 } ^ { 3 } x ^ { n } \sqrt { 16 + x ^ { 2 } } \mathrm {~d} x$, for $n \geqslant 0$. Show that, for $n \geqslant 2$,
$$( n + 2 ) I _ { n } = 125 \times 3 ^ { n - 1 } - 16 ( n - 1 ) I _ { n - 2 }$$
(ii) A curve has polar equation $r = \frac { 1 } { 4 } \theta ^ { 4 }$ for $0 \leqslant \theta \leqslant 3$.
\begin{enumerate}[label=(\alph*)]
\item Sketch this curve.
\item Find the exact length of the curve.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2012 Q12 [15]}}