Pre-U Pre-U 9795/1 2020 Specimen — Question 12 6 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2020
SessionSpecimen
Marks6
TopicPolar coordinates
TypeArc length of polar curve
DifficultyChallenging +1.8 This is a challenging multi-part question requiring: (a) integration by parts to establish a reduction formula involving a square root integrand, and (b) applying the polar arc length formula L = ∫√(r² + (dr/dθ)²)dθ, which leads to the integral from part (a). The connection between parts requires insight, and the reduction formula derivation is non-trivial. However, it's a structured question with clear guidance, making it more accessible than truly open-ended problems.
Spec4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve8.06a Reduction formulae: establish, use, and evaluate recursively
12
  1. Let \(I _ { \mathrm { n } } = \int _ { 0 } ^ { 3 } x ^ { n } \sqrt { 16 + x ^ { 2 } } \mathrm {~d} x\), for \(n \geqslant 0\). Show that, for \(n \geqslant 2\), $$( n + 2 ) I _ { n } = 125 \times 3 ^ { n - 1 } - 16 ( n - 1 ) I _ { n - 2 }$$
  2. A curve has polar equation \(r = \frac { 1 } { 4 } \theta ^ { 4 }\) for \(0 \leqslant \theta \leqslant 3\).
    1. Sketch this curve.
    2. Find the exact length of the curve.
(a)
\(I_n = \int_0^3 x^{n-1}(x\sqrt{16+x^2})\,\text{d}x\) Correct splitting *and* use of parts [M1]
\(= \left[x^{n-1} \cdot \dfrac{(16+x^2)^{\frac{3}{2}}}{3}\right]_0^3 - \int_0^3(n-1)x^{n-2}\dfrac{(16+x^2)^{\frac{3}{2}}}{3}\,\text{d}x\) [A1]
\(= 3^{n-2} \cdot 125 - \left(\dfrac{n-1}{3}\right)\int_0^3 x^{n-2}(16+x^2)\sqrt{16+x^2}\,\text{d}x\) Method to get 2nd integral of correct form [M1]
\(= 3^{n-2} \cdot 125 - \left(\dfrac{n-1}{3}\right)\{16I_{n-2} + I_n\}\) [i.e. reverting to \(I\)'s in 2nd integral] [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]
Total: 6 marks
(b)(i)
Spiral (with \(r\) increasing) [B1]
From \(O\) to just short of \(\theta = \pi\) [B1]
Total: 2 marks
(b)(ii)
\(r = \frac{1}{4}\theta^4 \Rightarrow \dfrac{\text{d}r}{\text{d}\theta} = \theta^3\) and \(r^2 + \left(\dfrac{\text{d}r}{\text{d}\theta}\right)^2 = \frac{1}{16}\theta^8 + \theta^6\) [M1A1]
\(L = \int_0^3 \frac{1}{4}\theta^3\sqrt{16+\theta^2}\,\text{d}\theta \left(= \frac{1}{4}I_3\right)\) [M1A1]
Now \(I_1 = \left[\dfrac{1}{3}(16+x^2)^{\frac{3}{2}}\right]_0^3 = \dfrac{61}{3}\) [B1]
and \(5I_3 = 125 \times 9 - 16 \times 2\left(\dfrac{61}{3}\right) = \dfrac{1423}{3}\) or \(474\frac{1}{3}\) Use of given reduction formula [M1]
so that \(L = \dfrac{1}{20} \times \dfrac{1423}{3} = \dfrac{1423}{60}\) or \(23\dfrac{43}{60}\) ft only from suitable \(k\,I_3\) [A1, A1ft]
Total: 7 marks
**(a)**
$I_n = \int_0^3 x^{n-1}(x\sqrt{16+x^2})\,\text{d}x$ Correct splitting *and* use of parts [M1]

$= \left[x^{n-1} \cdot \dfrac{(16+x^2)^{\frac{3}{2}}}{3}\right]_0^3 - \int_0^3(n-1)x^{n-2}\dfrac{(16+x^2)^{\frac{3}{2}}}{3}\,\text{d}x$ [A1]

$= 3^{n-2} \cdot 125 - \left(\dfrac{n-1}{3}\right)\int_0^3 x^{n-2}(16+x^2)\sqrt{16+x^2}\,\text{d}x$ Method to get 2nd integral of correct form [M1]

$= 3^{n-2} \cdot 125 - \left(\dfrac{n-1}{3}\right)\{16I_{n-2} + I_n\}$ [i.e. reverting to $I$'s in 2nd integral] [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]

**Total: 6 marks**

**(b)(i)**
Spiral (with $r$ increasing) [B1]

From $O$ to just short of $\theta = \pi$ [B1]

**Total: 2 marks**

**(b)(ii)**
$r = \frac{1}{4}\theta^4 \Rightarrow \dfrac{\text{d}r}{\text{d}\theta} = \theta^3$ and $r^2 + \left(\dfrac{\text{d}r}{\text{d}\theta}\right)^2 = \frac{1}{16}\theta^8 + \theta^6$ [M1A1]

$L = \int_0^3 \frac{1}{4}\theta^3\sqrt{16+\theta^2}\,\text{d}\theta \left(= \frac{1}{4}I_3\right)$ [M1A1]

Now $I_1 = \left[\dfrac{1}{3}(16+x^2)^{\frac{3}{2}}\right]_0^3 = \dfrac{61}{3}$ [B1]

and $5I_3 = 125 \times 9 - 16 \times 2\left(\dfrac{61}{3}\right) = \dfrac{1423}{3}$ or $474\frac{1}{3}$ Use of given reduction formula [M1]

so that $L = \dfrac{1}{20} \times \dfrac{1423}{3} = \dfrac{1423}{60}$ or $23\dfrac{43}{60}$ **ft** only from suitable $k\,I_3$ [A1, A1ft]

**Total: 7 marks**
12
\begin{enumerate}[label=(\alph*)]
\item Let $I _ { \mathrm { n } } = \int _ { 0 } ^ { 3 } x ^ { n } \sqrt { 16 + x ^ { 2 } } \mathrm {~d} x$, for $n \geqslant 0$. Show that, for $n \geqslant 2$,

$$( n + 2 ) I _ { n } = 125 \times 3 ^ { n - 1 } - 16 ( n - 1 ) I _ { n - 2 }$$
\item A curve has polar equation $r = \frac { 1 } { 4 } \theta ^ { 4 }$ for $0 \leqslant \theta \leqslant 3$.
\begin{enumerate}[label=(\roman*)]
\item Sketch this curve.
\item Find the exact length of the curve.

\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2020 Q12 [6]}}
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