Year: 2001
Paper: 3
Question Number: 7
Course: LFM Pure
Section: Differential equations
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Sketch the graph of the function $\ln x - {1 \over 2} x^2$.
Show that the differential equation
\[
{\mathrm{d} y \over \mathrm{d} x} = {2xy \over x^2 - 1}
\]
describes a family of parabolas each of which
passes through the points $(1,0)$ and $(-1,0)$ and has its vertex on the $y$-axis. Hence find the equation of the curve that passes through the point $(1,1)$ and intersects each of the above parabolas orthogonally.
Sketch this curve.
[Two curves intersect \textit{orthogonally} if their tangents at the point of
intersection are perpendicular.]
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){ln(#1) - 0.5*(#1)*(#1)};
\def\functiong(#1){ - 0.5*(#1)*(#1)};
\def\xl{-0.5};
\def\xu{7};
\def\yl{-10};
\def\yu{10};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\yrange}
% Define the styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
% Draw a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
\draw[thick, blue, smooth, domain=0.01:\xu, samples=201]
plot ({\x}, {\functionf(\x)});
\draw[thick, red, dashed, domain=0.01:\xu, samples=201]
plot ({\x}, {\functiong(\x)});
\end{scope}
\filldraw (1,-0.5) circle (1.5pt) node[below] {$(1,-\frac12)$};
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
\end{tikzpicture}
\end{center}
\begin{align*}
&& y' &= \frac{2xy}{x^2-1} \\
\Rightarrow && \int \frac{1}{y} \d y &= \int \frac{2x}{x^2-1} \d x \\
\Rightarrow && \ln |y| &= \ln |x^2-1| + C \\
\Rightarrow && y &= A(x^2-1)
\end{align*}
which is a family of parabolas each passing through $(\pm1, 0)$ and with a vertex on the $y$-axis.
The curve we seek must satisfy \begin{align*}
&& y' &= \frac{1-x^2}{2xy} \\
\Rightarrow && \int2 y \d y &= \int \left ( \frac{1}{x} - x \right) \d x \\
\Rightarrow && y^2 &= \ln x - \tfrac12 x^2 + C \\
(1,1): && 1 &= -\tfrac12+C \\
\Rightarrow && C &= \frac32 \\
\Rightarrow && y^2 &= \tfrac32 + \ln x - \tfrac12 x^2
\end{align*}
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){1.5+ln(#1) - 0.5*(#1)*(#1)};
% \def\functiong(#1){ - 0.5*(#1)*(#1)};
\def\xl{-2};
\def\xu{3};
\def\yl{-2};
\def\yu{2};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\yrange}
% Define the styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
% Draw a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
\draw[thick, blue, smooth, domain=0.22907:2.12257, samples=100]
plot ({\x}, {sqrt(\functionf(\x)});
\draw[thick, blue, smooth, domain=0.22907:2.122561, samples=100]
plot ({\x}, {-sqrt(\functionf(\x)});
\foreach \a in {1, ..., 5} {
\draw[thin, red, dashed, domain=\xl:\xu, samples=100]
plot ({\x}, {1/\a*(\x*\x-1)});
\draw[thin, red, dashed, domain=\xl:\xu, samples=100]
plot ({\x}, {-1/\a*(\x*\x-1)});
\draw[thin, red, dashed, domain=\xl:\xu, samples=100]
plot ({\x}, {\a*(\x*\x-1)});
\draw[thin, red, dashed, domain=\xl:\xu, samples=100]
plot ({\x}, {-\a*(\x*\x-1)});
}
\end{scope}
\filldraw (1,0) circle (1.5pt) node[below] {$1$};
\filldraw (-1,0) circle (1.5pt) node[below] {$-1$};
\filldraw (1,1) circle (1.5pt) node[below] {$(1,1)$};
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
\end{tikzpicture}
\end{center}