Pre-U Pre-U 9795/1 2018 June — Question 4 7 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2018
SessionJune
Marks7
TopicPolar coordinates
TypePolar curve with exponential function
DifficultyChallenging +1.2 This is a standard polar arc length question requiring the formula L = ∫√(r² + (dr/dθ)²)dθ. Part (i) involves straightforward differentiation and integration of exponentials with algebraic simplification. Part (ii) requires setting up and solving e^(3β) - 1 = e^(3β), which simplifies to a trivial result. While polar coordinates are Further Maths content, the calculus techniques are routine and the question follows a predictable template, making it moderately above average difficulty but not requiring novel insight.
Spec4.08c Improper integrals: infinite limits or discontinuous integrands4.09c Area enclosed: by polar curve

A curve has polar equation \(r = \frac{3}{10}e^{3\theta}\) for \(\theta \geq 0\). The length of the arc of this curve between \(\theta = 0\) and \(\theta = \alpha\) is denoted by \(L(\alpha)\).
  1. Show that \(L(\alpha) = \frac{1}{3}(e^{3\alpha} - 1)\). [5]
  2. The point \(P\) on the curve corresponding to \(\theta = \beta\) is such that \(L(\beta) = OP\), where \(O\) is the pole. Find the value of \(\beta\). [2]

A curve has polar equation $r = \frac{3}{10}e^{3\theta}$ for $\theta \geq 0$. The length of the arc of this curve between $\theta = 0$ and $\theta = \alpha$ is denoted by $L(\alpha)$.

\begin{enumerate}[label=(\roman*)]
\item Show that $L(\alpha) = \frac{1}{3}(e^{3\alpha} - 1)$. [5]

\item The point $P$ on the curve corresponding to $\theta = \beta$ is such that $L(\beta) = OP$, where $O$ is the pole. Find the value of $\beta$. [2]
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2018 Q4 [7]}}