2001 Paper 3 Q7

Year: 2001
Paper: 3
Question Number: 7

Course: LFM Pure
Section: Differential equations

Difficulty: 1700.0 Banger: 1516.0

Problem

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 orthogonally if their tangents at the point of intersection are perpendicular.]

Solution

TikZ diagram
\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*}
TikZ diagram
Rating Information

Difficulty Rating: 1700.0

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Banger Rating: 1516.0

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Show LaTeX source
Problem source
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.]
Solution source

\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}