2007 Paper 3 Q4

Year: 2007
Paper: 3
Question Number: 4

Course: LFM Pure and Mechanics
Section: Parametric equations

Difficulty: 1700.0 Banger: 1484.0

Problem

A curve is given parametrically by \begin{align*} x&= a\big( \cos t +\ln \tan \tfrac12 t\big)\,,\\ y&= a\sin t\,, \end{align*} where \(0 < t < \frac12 \pi\) and \(a\) is a positive constant. Show that \(\ds \frac{\d y}{\d x} = \tan t\) and sketch the curve. Let \(P\) be the point with parameter \(t\) and let \(Q\) be the point where the tangent to the curve at \(P\) meets the \(x\)-axis. Show that \(PQ=a\). The {\sl radius of curvature}, \(\rho\), at \(P\) is defined by \[ \rho= \frac {\big(\dot x ^2+\dot y^2\big)^{\frac32}} {\vert \dot x \ddot y - \dot y \ddot x\vert \ \ } \,, \] where the dots denote differentiation with respect to \(t\). Show that \(\rho =a\cot t\). The point \(C\) lies on the normal to the curve at \(P\), a distance \(\rho\) from \(P\) and above the curve. Show that \(CQ\) is parallel to the \(y\)-axis.

No solution available for this problem.

Examiner's report
— 2007 STEP 3, Question 4
Average

This question was quite popular. A lot of attempts involved rambling trigonometrical manipulations, and few spotted the standard differential of ln tan t/2. The curve sketch was often omitted or incorrect, and there was a lot of complicated working using e.g. the equation of the normal etc. to find the centre of curvature.

Source: Cambridge STEP 2007 Examiner's Report · 2007-full.pdf
Rating Information

Difficulty Rating: 1700.0

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Banger Rating: 1484.0

Banger Comparisons: 1

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Problem source
A curve is given parametrically by
\begin{align*}
x&= a\big( \cos t +\ln \tan \tfrac12 t\big)\,,\\
y&= a\sin t\,,
\end{align*}
where $0 < t < \frac12 \pi$ and $a$ is a positive constant. Show that 
$\ds \frac{\d y}{\d x} = \tan t$ and sketch the curve.
Let  $P$ be the  point with parameter $t$ and let $Q$ be the 
point where  the tangent 
to the curve at $P$ meets the $x$-axis.
Show that $PQ=a$.
The {\sl radius of curvature}, $\rho$, at $P$ 
 is defined by
\[
\rho=
\frac 
{\big(\dot x ^2+\dot y^2\big)^{\frac32}}
{\vert \dot x \ddot y - \dot y \ddot x\vert   \ \ }
\,,
\]
where the dots denote differentiation with respect to $t$. Show that
$\rho =a\cot t$.
The point  $C$ 
lies on the normal to the curve at $P$, a distance $\rho$ from $P$
and  above the curve.
Show that $CQ$ is parallel to the $y$-axis.