\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)

2019.3.7 Question 7

1. When \(a = b\), \begin {align*} y^2 (y^2 - a^2) & = x^2 (x^2 - a^2) \\ x^4 - y^4 - a^2 x^2 + a^2 y^2 & = 0 \\ (x^2 + y^2 - a^2) (x^2 - y^2) & = 0 \\ (x^2 + y^2 - a^2) (x + y) (x - y) & = 0, \end {align*}

   so the Devil’s Curve in this case consists of the line \(x + y = 0\), the line \(x - y = 0\), and the circle \(x^2 + y^2 = a^2\).

   The curve is shown as follows.

   

2. When \(a = 2\) and \(b = \sqrt {5}\), \[ y^2 (y^2 - 5) = x^2 (x^2 - 4). \]

   1. Rearrangement gives us \[ (x^2)^2 - 4x^2 - y^2 (y^2 - 5) = 0, \] and considering the discriminant, we have \[ (-4)^2 + 4 y^2 (y^2 - 5) \geq 0, \] i.e. \[ \left (y^2 - 1\right ) \left (y^2 - 4\right ) \geq 0. \]

      This gives \(y^2 \leq 1\) or \(y^2 \geq 4\), and in the case where \(y \geq 0\), this must give \(0 \leq y \leq 1\) or \(y \geq 2\), as desired.

   2. When the curve is very close to the origin, we must have \(x^4, y^4 \ll x^2, y^2\), and hence \(4x^2 \approx 5y^2\), which means \(y \approx \frac {2}{\sqrt {5}}x\).

      When the curve is very far from the origin, we must have \(x^4, y^4 \gg x^2, y^2\), and hence \(x^4 \approx y^4\), which means \(y \approx x\).

   3. Using implicit differentiation, we have \begin {align*} y^2 (y^2 - 5) & = x^2 (x^2 - 4) \\ (4y^3 - 10y) \DiffFrac {y}{x} & = 4x^3 - 8x \\ (2y^2 - 5)y \DiffFrac {y}{x} & = 2x(x^2 - 2). \end {align*}

      When \(\DiffFrac {y}{x} = 0\), the tangent to the curve is parallel to the \(x\)-axis, and hence \[ 2x(x^2 - 2) = 0, \] giving \(x = 0\) or \(x = \sqrt {2}\).

      For \(x = 0\), \(y^2 (y^2 - 5) = 0\), and therefore \(y = 0\) or \(y = \sqrt {5}\). The case where \(y = 0\) does not necessarily give that \(\DiffFrac {y}{x} = 0\), but the case where \(y = \sqrt {5}\) does.

      For \(x = \sqrt {2}\), \(y^2 (y^2 - 5) = -4\), \(y = 2\) or \(y = 1\). Both cases give \(\DiffFrac {y}{x} = 0\).

      So the tangent to the curve is parallel to the \(x\)-axis at points \[ \left (0, \sqrt {5}\right ), \left (\sqrt {2}, 1\right ), \left (\sqrt {2}, 2\right ). \]

      We must have \[ (2y^2 - 5)y = 2x(x^2 - 2) \DiffFrac {x}{y}, \] and when \(\DiffFrac {x}{y} = 0\), the tangent to the curve is parallel to the \(y\)-axis.

      This gives \((2y^2 - 5)y = 0\), and hence \(y = 0\) or \(y = \sqrt {\frac {5}{2}}\).

      For \(y = 0\), \(x = 0\) or \(x = 2\). The case \(x = 0\) does not necessarily give \(\DiffFrac {x}{y} = 0\), but the case where \(x = 2\) does.

      For \(y = \sqrt {\frac {5}{2}}\), \(x^2 (x^2 - 4) = - \frac {25}{4}\), and hence \[ 4x^4 - 16x^2 + 25 = 4 (x^2 - 2)^2 + 9 = 0, \] which is not possible.

      Hence, the tangent to the curve is parallel to the \(y\)-axis only at \((2, 0)\).

   Therefore, from the analysis in the previous parts, the curve looks as follows for \(x \geq 0\) and \(y \geq 0\):

   

3. All \(x\) terms in the curve is in \(x^2\), so the graph is symmetric in the \(y\)-axis since \(x^2 = (-x)^2\). Similarly, the graph is symmetric in the \(x\)-axis as well. Hence, the complete graph looks as follows.