2019 Paper 3 Q7
Year: 2019
Paper: 3
Question Number: 7
Course: LFM Stats And Pure
Section: Curve Sketching
Difficulty: 1500.0
Banger: 1500.0
curve sketching
implicit curve
quartic
tangent-normal
quadratic in disguise
symmetry
asymptotic behaviour
Devil's Curve
Problem
The Devil's Curve is given by
$$y^2(y^2 - b^2) = x^2(x^2 - a^2),$$
where \(a\) and \(b\) are positive constants.
- In the case \(a = b\), sketch the Devil's Curve.
- Now consider the case \(a = 2\) and \(b = \sqrt{5}\), and \(x \geq 0\), \(y \geq 0\).
- Show by considering a quadratic equation in \(x^2\) that either \(0 \leq y \leq 1\) or \(y \geq 2\).
- Describe the curve very close to and very far from the origin.
- Find the points at which the tangent to the curve is parallel to the \(x\)-axis and the point at which the tangent to the curve is parallel to the \(y\)-axis.
- Sketch the Devil's Curve in the case \(a = 2\) and \(b = \sqrt{5}\) again, but with \(-\infty < x < \infty\) and \(-\infty < y < \infty\).
Solution
- Suppose \(a=b\), ie
\begin{align*}
&& y^2(y^2-a^2) &= x^2(x^2-a^2) \\
\Rightarrow && 0 &= x^4-y^4-a^2(x^2-y^2) \\
&&&= (x^2-y^2)(x^2+y^2-a^2)
\end{align*}
Therefore we have the lines \(y = \pm x\) and a circle radius \(a\).
- Since \(x^4 - 4x^2 - y^2(y^2-5)= 0\), we must have \(0 \leq \Delta = 16 + 4y^2(y^2-5) \Rightarrow y^4-5y^2+4 = (y^2-4)(y^2-1) \geq 0\), therefore \(0 \leq y \leq 1\) or \(y \geq 2\) (since we are only considering positive values of \(y\)).
- When \((x, y) \approx 0\) the equation is more like \(4x^2 \approx 5y^2\) or \(y \approx \frac{2}{\sqrt{5}}x\) If \(|x|, |y|\) are very large, it is more like \(x^4 \approx y^4\), ie \(y \approx x\)
- \(\,\) \begin{align*} && (2y(y^2-5)+y^2(2y))y' &= 2x(x^2-4)+2x^3 \\ \Rightarrow && (4y^3-10y)y' &= 4x^3-8x \end{align*} Therefore the gradient is parallel to the \(x\)-axis when \(x = 0, x = \sqrt{2}\). We need \(x = 0, y \neq 0\), ie \(y = \sqrt{5}\), so \((0, \sqrt{5})\) and \((\sqrt{2}, 0)\) It is parallel to the \(y\)-axis when \(y = 0\) or \(y = \sqrt{\frac52}\), ie \((2, 0)\)
- \(\,\)
Examiner's report
— 2019 STEP 3, Question 7
Mean: 6.7 / 20
67% attempted
'about one third marks' → 6.7/20; 'two thirds' → 67%
From the paper introduction
There was a significant rise in the total entry this year with an increase of nearly 8.5% on 2018. One question was attempted by over 90%, two others were very popular, and three further questions were attempted by 60% or more. No question was generally avoided and even the least popular attracted more than 10% of the candidates. 88% restricted themselves to attempting no more than 7 questions, and only a handful, but not the very best, scored strongly attempting more than 7 questions.
Source: Cambridge STEP 2019 Examiner's Report
· 2019-p3.pdf
Rating Information
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Show LaTeX source
Problem source
The \textit{Devil's Curve} is given by
$$y^2(y^2 - b^2) = x^2(x^2 - a^2),$$
where $a$ and $b$ are positive constants.
\begin{questionparts}
\item In the case $a = b$, sketch the Devil's Curve.
\item Now consider the case $a = 2$ and $b = \sqrt{5}$, and $x \geq 0$, $y \geq 0$.
\begin{enumerate}
\item Show by considering a quadratic equation in $x^2$ that either $0 \leq y \leq 1$ or $y \geq 2$.
\item Describe the curve very close to and very far from the origin.
\item Find the points at which the tangent to the curve is parallel to the $x$-axis and the point at which the tangent to the curve is parallel to the $y$-axis.
\end{enumerate}
Sketch the Devil's Curve in this case.
\item Sketch the Devil's Curve in the case $a = 2$ and $b = \sqrt{5}$ again, but with $-\infty < x < \infty$ and $-\infty < y < \infty$.
\end{questionparts}Solution source
\begin{questionparts}
\item Suppose $a=b$, ie
\begin{align*}
&& y^2(y^2-a^2) &= x^2(x^2-a^2) \\
\Rightarrow && 0 &= x^4-y^4-a^2(x^2-y^2) \\
&&&= (x^2-y^2)(x^2+y^2-a^2)
\end{align*}
Therefore we have the lines $y = \pm x$ and a circle radius $a$.
\begin{center}
\begin{tikzpicture}
\def\a{2};
\def\functionf(#1){abs((#1)-\c};
\def\xl{-5};
\def\xu{5};
\def\yl{-5};
\def\yu{5};
% 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] (\xl, {\xl}) -- (\xu, \xu)
node[pos=0.1, sloped, below] {\tiny $y=x$};
\draw[thick, blue] (\xl, {-\xl}) -- (\xu, {-\xu})
node[pos=0.1, sloped, below] {\tiny $y=-x$};
\draw[thick, blue] (\a,0) arc (0:360:\a)
node[pos=0.15, sloped, above] {\tiny $x^2 + y^2 = a^2$};
% \draw[thick, red, dashed, domain=\xl:\xu, samples =100]
% plot (\x, {\functionf(\x)});
\filldraw ({\a}, 0) circle (1.5pt) node[below] {$a$};
\filldraw ({-\a}, 0) circle (1.5pt) node[below] {$-a$};
\filldraw (0,{\a}) circle (1.5pt) node[below] {$a$};
\filldraw (0,{-\a}) circle (1.5pt) node[below] {$-a$};
\end{scope}
% 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}
\item \begin{enumerate}
\item Since $x^4 - 4x^2 - y^2(y^2-5)= 0$, we must have $0 \leq \Delta = 16 + 4y^2(y^2-5) \Rightarrow y^4-5y^2+4 = (y^2-4)(y^2-1) \geq 0$, therefore $0 \leq y \leq 1$ or $y \geq 2$ (since we are only considering positive values of $y$).
\item When $(x, y) \approx 0$ the equation is more like $4x^2 \approx 5y^2$ or $y \approx \frac{2}{\sqrt{5}}x$
If $|x|, |y|$ are very large, it is more like $x^4 \approx y^4$, ie $y \approx x$
\item $\,$ \begin{align*}
&& (2y(y^2-5)+y^2(2y))y' &= 2x(x^2-4)+2x^3 \\
\Rightarrow && (4y^3-10y)y' &= 4x^3-8x
\end{align*}
Therefore the gradient is parallel to the $x$-axis when $x = 0, x = \sqrt{2}$. We need $x = 0, y \neq 0$, ie $y = \sqrt{5}$, so $(0, \sqrt{5})$ and $(\sqrt{2}, 0)$
It is parallel to the $y$-axis when $y = 0$ or $y = \sqrt{\frac52}$, ie $(2, 0)$
\end{enumerate}
\begin{center}
\begin{tikzpicture}
\def\a{2};
\def\functionf(#1){abs((#1)-\c};
\def\disc(#1){4*(#1)^4 - 20*(#1)^2 + 15}
\def\xl{-1};
\def\xu{5};
\def\yl{-1};
\def\yu{5};
% 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);
% --- Outer Branches (Exist for all x) ---
% \addplot[thick, blue, domain=0:3.5] {sqrt((5 + sqrt(\disc(\x)/2)))};
% \addplot[thick, blue, domain=-3.5:3.5] {-sqrt((5 + sqrt(\disc))/2)};
% --- Inner Loop (Exists only for -2 <= x <= 2) ---
% \addplot[thick, red, domain=-2:2] {sqrt((5 - sqrt(\disc))/2)};
% \addplot[thick, red, domain=-2:2] {-sqrt((5 - sqrt(\disc))/2)};
\draw[thick, blue, smooth, domain=0:2, samples =100]
plot (\x, {sqrt(5 - sqrt(25 - 16 *\x^2 + 4 *\x^4))/sqrt(2)});
\draw[thick, blue, smooth, domain=0:\xu, samples =100]
plot (\x, {sqrt(5 + sqrt(25 - 16 *\x^2 + 4 *\x^4))/sqrt(2)});
\draw[dashed, red] (\xl, 1) -- (\xu, 1);
\draw[dashed, red] (\xl, 2) -- (\xu, 2);
\filldraw (2, 0) circle (1.5pt) node[below] {$2$};
% \filldraw ({-\a}, 0) circle (1.5pt) node[below] {$-a$};
\filldraw (0,1) circle (1.5pt) node[right] {$1$};
\filldraw (0,2) circle (1.5pt) node[right] {$2$};
\end{scope}
% 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}
\item $\,$
\begin{center}
\begin{tikzpicture}
\def\a{2};
\def\functionf(#1){abs((#1)-\c};
\def\disc(#1){4*(#1)^4 - 20*(#1)^2 + 15}
\def\xl{-5};
\def\xu{5};
\def\yl{-5};
\def\yu{5};
% 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);
% --- Outer Branches (Exist for all x) ---
% \addplot[thick, blue, domain=0:3.5] {sqrt((5 + sqrt(\disc(\x)/2)))};
% \addplot[thick, blue, domain=-3.5:3.5] {-sqrt((5 + sqrt(\disc))/2)};
% --- Inner Loop (Exists only for -2 <= x <= 2) ---
% \addplot[thick, red, domain=-2:2] {sqrt((5 - sqrt(\disc))/2)};
% \addplot[thick, red, domain=-2:2] {-sqrt((5 - sqrt(\disc))/2)};
\draw[thick, blue, smooth, domain=-2:2, samples =100]
plot (\x, {sqrt(5 - sqrt(25 - 16 *(\x)^2 + 4 *(\x)^4))/sqrt(2)});
\draw[thick, blue, smooth, domain=\xl:\xu, samples =100]
plot (\x, {sqrt(5 + sqrt(25 - 16 *(\x)^2 + 4 *(\x)^4))/sqrt(2)});
\draw[thick, blue, smooth, domain=-2:2, samples =100]
plot (\x, {-sqrt(5 - sqrt(25 - 16 *(\x)^2 + 4 *(\x)^4))/sqrt(2)});
\draw[thick, blue, smooth, domain=\xl:\xu, samples =100]
plot (\x, {-sqrt(5 + sqrt(25 - 16 *(\x)^2 + 4 *(\x)^4))/sqrt(2)});
\draw[dashed, red] (\xl, 1) -- (\xu, 1);
\draw[dashed, red] (\xl, 2) -- (\xu, 2);
\draw[dashed, red] (\xl, -1) -- (\xu, -1);
\draw[dashed, red] (\xl, -2) -- (\xu, -2);
\filldraw (2, 0) circle (1.5pt) node[below] {$2$};
% \filldraw ({-\a}, 0) circle (1.5pt) node[below] {$-a$};
\filldraw (0,1) circle (1.5pt) node[right] {$1$};
\filldraw (0,2) circle (1.5pt) node[right] {$2$};
\filldraw (0,-1) circle (1.5pt) node[right] {$-1$};
\filldraw (0,-2) circle (1.5pt) node[right] {$-2$};
\end{scope}
% 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}
\end{questionparts}
Attempted by two thirds, the mean score was only about one third marks. Part (i) was not very well answered with many appearing to guess one or both of the solutions without managing to factorise, or equivalent. By contrast, part (ii)(a) was generally well‐answered. Most candidates saw how to do this correctly, though a few tried to treat it as a polynomial in y and take that discriminant which got them nowhere. In part (b), most candidates did not realise what was expected of them so, for example, many wrote x → ∞, y → ∞ or found the x and y intercepts. Almost all candidates saw what was needed for part (c), however, many made small algebraic errors or didn't check all the cases; the most common error was failure to eliminate the origin. The sketch was generally badly answered, often owing to errors in previous parts or failure to use correctly the information already obtained. Candidates did appreciate what was needed for part (iii) and did this well.