Year: 2016
Paper: 3
Question Number: 6
Course: UFM Pure
Section: Hyperbolic functions
A substantially larger number of candidates took the paper this year: 14% more than in 2015. However, the mean score was virtually identical to that in 2015. Five questions were very popular, with two being attempted by in excess of 90% of the candidates, but once again, all questions were attempted by significant numbers, with only one dipping under 10% attempting it, and every question was answered perfectly by at least one candidate. Most candidates kept to six sensible attempts, although some did several more scoring weakly overall, except in six outstanding cases that earned very high marks.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
Show, by finding $R$ and $\gamma$, that $A \sinh x + B\cosh x $ can be written in the form $R\cosh (x+\gamma)$ if $B>A>0$. Determine the corresponding forms in the other cases that arise, for $A>0$, according to the value of $B$.
Two curves have equations $y = \textrm{sech} x$ and $y = a\tanh x + b\,$, where $a>0$.
\begin{questionparts}
\item In the case $b>a$, show that if the curves intersect then the $x$-coordinates of the points of intersection can be written in the form
\[
\pm\textrm{arcosh} \left( \frac 1 {\sqrt{b^2-a^2}}\right) - {\rm artanh \,} \frac a b
.\]
\item Find the corresponding result in the case $a>b>0\,$.
\item Find necessary and sufficient conditions on $a$ and $b$ for the curves to intersect at two distinct points.
\item Find necessary and sufficient conditions on $a$ and $b$ for the curves to touch and, given that they touch, express the $y$-coordinate of the point of contact in terms of $a$.
\end{questionparts}\begin{align*}
&& R\cosh(x + \gamma) &=R \cosh x \cosh \gamma + R \sinh x \sinh \gamma \\
\Rightarrow && R \cosh \gamma &= B \\
&& R \sinh \gamma &= A \\
\Rightarrow && R^2 &= B^2 - A^2 \\
\Rightarrow && \tanh \gamma &= \frac{A}{B} \\
\end{align*}
Therefore it is possible, by writing $R = \sqrt{B^2-A^2}$ and $\gamma = \textrm{artanh} \left ( \frac{A}{B} \right)$. This works as long as $|B| > A > 0$.
Supposing $A >|B| $, try $S \sinh (x + \delta) = S \sinh x \cosh \delta +S \cosh x \sinh \delta$
\begin{align*}
&& S \cosh \delta &= A \\
&& -S \sinh \delta &= B \\
\Rightarrow && S^2 &= A^2 - B^2 \\
\Rightarrow && \tanh \delta &= \frac{B}{A} \\
\end{align*}
Therefore in this case we can write $\sqrt{A^2-B^2} \sinh \left (x + \tanh^{-1} \left ( \frac{B}{A} \right) \right)$
If $A = \pm B > 0$ we can we have $A \sinh x + B \cosh x = \pm Ae^{\pm x}$
\begin{questionparts}
\item Suppose $y \cosh x = 1$ and $y \cosh x = a \sinh x +b \cosh x$ so \begin{align*}
&& 1 & = a \sinh x + b \cosh x \\
&&&= \sqrt{b^2-a^2} \cosh(x + \textrm{artanh} \frac{a}{b} ) \\
\Rightarrow && x + \textrm{artanh} \frac{a}{b} &= \pm \textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) \\
\Rightarrow && x &= \pm \textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) -\textrm{artanh} \frac{a}{b}
\end{align*}
\item If $ a > b > 0$ we have
\begin{align*}
&& 1 & = \sqrt{a^2-b^2} \sinh \left ( x - \textrm{artanh} \frac{b}{a} \right) \\
\Rightarrow && x &= \textrm{arsinh} \left ( \frac{1}{\sqrt{a^2-b^2}} \right) + \textrm{artanh} \left ( \frac{b}{a} \right)
\end{align*}
\item To intersect at distinct points we must have $b > a$ and $\textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) \neq 0$ which is always true.
\item For the curves to touch, we need them to intersect and have matching derivatives, ie
\begin{align*}
&& -\tanh x \cdot \textrm{sech}x &= a\textrm{sech}^2 x \\
\Rightarrow && 0 &= \textrm{sech}^2 x (a + \sinh x) \\
\Rightarrow && x &= -\textrm{arsinh} \, a\\
\Rightarrow && \sinh x &= - a\\
\Rightarrow &&\cosh x &= \sqrt{1 + a^2} \\
\end{align*}
So if the curves touch, we must have $1 = -a^2+b\sqrt{1+a^2} \Rightarrow b = \sqrt{1+a^2}$ and since this does work it is a necessary and sufficient condition.
We will also have the $y$ coordinate is $\frac{1}{\sqrt{1+a^2}}$
\end{questionparts}
Just under 40% attempted this question, and did so without great success, scoring about one-third marks. The majority knew the formula for cosh of a sum, or if not, could use definitions and compare coefficients to obtain the first result in the stem. However, they were very weak on the n = 0 case. Explanation for part (i) was poor, and the plus or minus was frequently not properly understood; as a consequence, a specious plus or minus often appeared in part (ii). Arguing necessary and sufficient conditions in (iii) and (iv) was weak. However, a small number of good candidates did complete this question.