\(% 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}}\)

2023.3.2 Question 2

1. If the two curves meet at \(\theta = \alpha \), then \(\alpha \) must satisfy that \[ k (1 + \sin \alpha ) = k + \cos \alpha . \]

   Subtracting \(k\) on both sides, we have \[ k \sin \alpha = \cos \alpha , \] and since \(k > 1 > 0\), \(\sin \alpha \) and \(\cos \alpha \) cannot be simultaneously zero, they must both be non-zero. Dividing through both sides by \(\cos \alpha \) gives \[ k \tan \alpha = 1 \] and hence \[ \tan \alpha = \frac {1}{k} \] as desired.

   The curves are as follows.

   

2. The area of \(A\) is given by \begin {align*} [A] & = \frac {1}{2} \cdot \int _{0}^{\alpha } \left (k (1 + \sin \theta )\right )^2 \Diff \theta \\ & = \frac {k^2}{2} \cdot \int _{0}^{\alpha } \left (1 + 2 \sin \theta + \sin ^2 \theta \right ) \Diff \theta \\ & = \frac {k^2}{2} \cdot \int _{0}^{\alpha } \left (1 + 2 \sin \theta + \frac {1 - \cos 2\theta }{2}\right ) \Diff \theta \\ & = \frac {k^2}{2} \cdot \left [\frac {3}{2} \cdot \theta - 2 \cos \theta - \frac {1}{4} \sin 2\theta \right ]_{0}^{\alpha } \\ & = \frac {k^2}{2} \cdot \left [\left (\frac {3}{2}\alpha - 2 \cos \alpha - \frac {1}{4} \sin 2\alpha \right ) - \left (0 - 2 - \frac {1}{4} \cdot 0\right )\right ] \\ & = \frac {k^2}{2} \left [\frac {3}{2}\alpha - 2 \cos \alpha - \frac {1}{2} \sin \alpha \cos \alpha + 2\right ] \\ & = \frac {k^2}{4} \left (3 \alpha - \sin \alpha \cos \alpha \right ) + k^2 \left (1 - \cos \alpha \right ). \end {align*}
3. The area of \(B\) is given by \begin {align*} [B] & = \frac {1}{2} \cdot \int _{\alpha }^{\pi } \left (k + \cos \theta \right )^2 \Diff \theta \\ & = \frac {1}{2} \cdot \int _{\alpha }^{\pi } \left (k^2 + 2k \cos \theta + \cos ^2 \theta \right ) \Diff \theta \\ & = \frac {1}{2} \cdot \int _{\alpha }^{\pi } \left (k^2 + 2k \cos \theta + \frac {1 + \cos 2\theta }{2}\right ) \Diff \theta \\ & = \frac {1}{2} \cdot \left [\left (k^2 + \frac {1}{2}\right )\theta + 2k \sin \theta + \frac {\sin 2\theta }{4}\right ]_{\alpha }^{\pi } \\ & = \frac {1}{2} \cdot \left [\left (\left (k^2 + \frac {1}{2}\right )\pi + 2k \cdot 0 + \frac {0}{4}\right ) - \left (\left (k^2 + \frac {1}{2}\right )\alpha + 2k \cdot \sin \alpha + \frac {\sin 2\alpha }{4}\right )\right ] \\ & = \frac {1}{2} \cdot \left [\left (k^2 + \frac {1}{2}\right ) \left (\pi - \alpha \right ) - 2k \sin \alpha - \frac {\sin \alpha \cos \alpha }{2}\right ] \\ & = \frac {1}{4} \cdot \left (2k^2 \pi + \pi - 2k^2 \alpha - \alpha - 4k \sin \alpha - \sin \alpha \cos \alpha \right ). \end {align*}

4. \(T\) is given by \begin {align*} T & = \frac {1}{2} \cdot \int _{0}^{\pi } \left (k + \cos \theta \right )^2 \Diff \theta \\ & = \frac {1}{2} \cdot \left [\left (k^2 + \frac {1}{2}\right )\theta + 2k \sin \theta + \frac {\sin 2\theta }{4}\right ]_{0}^{\pi } \\ & = \frac {1}{4} \cdot \left (2k^2 \pi + \pi - 2k^2 \cdot 0 - 0 - 4k \cdot 0 - 0 \cdot 1\right ) \\ & = \frac {\pi \left (2k^2 + 1\right )}{4}. \end {align*}

   As \(k \to \infty \), \(\frac {1}{k} = \tan \alpha \to 0^{+}\), and therefore, \[ \alpha , \sin \alpha , \tan \alpha \approx \frac {1}{k} \] and \[ \cos \alpha \approx 1 - \frac {1}{2k^2}. \]

   Therefore, considering only terms with the highest power of \(k\) \begin {align*} [A] & = \frac {k^2}{4} \left (3 \alpha - \sin \alpha \cos \alpha \right ) + k^2 \left (1 - \cos \alpha \right ) \\ & \approx \frac {k^2}{4} \left (3 \left (\frac {1}{k}\right ) - \left (\frac {1}{k}\right ) \left (1 - \frac {1}{2k^2}\right )\right ) + k^2 \left (1 - \left (1 - \frac {1}{2k^2}\right )\right ) \\ & = \frac {k^2}{4} \left (\frac {2}{k} + \frac {1}{2k^3} + 1\right ) \\ & \approx \frac {k}{2}, \end {align*}

   and \begin {align*} [B] & = \frac {1}{4} \cdot \left (2k^2 \pi + \pi - 2k^2 \alpha - \alpha - 4k \sin \alpha - \sin \alpha \cos \alpha \right ) \\ & = \frac {1}{4} \cdot \left (2k^2 \pi + \pi - 2k^2 \cdot \frac {1}{k} - \frac {1}{k} - 4k \cdot \frac {1}{k} - \frac {1}{k} \cdot \left (1 - \frac {1}{2k^2}\right )\right ) \\ & = \frac {1}{4} \cdot \left (2k^2 \pi + \pi - 2k - \frac {1}{k} - 4 - \frac {1}{k} + \frac {1}{2k^3}\right ) \\ & \approx \frac {k^2 \pi }{2}. \end {align*}

   Therefore, \[ R = [A] + [B] \approx \frac {k^2 \pi }{2} \]

   Hence, \begin {align*} \lim _{k \to \infty } \frac {R}{T} & = \lim _{k \to \infty } \frac {\frac {k^2 \pi }{2}}{\frac {\pi \left (2k^2 + 1\right )}{4}} \\ & = \lim _{k \to \infty } \frac {2k^2}{2k^2 + 1} \\ & = 1 \end {align*}

   as desired.

   Similarly, \(S\) is given by \begin {align*} S & = \frac {1}{2} \cdot \int _{0}^{\pi } \left (k (1 + \sin \theta )\right )^2 \Diff \theta \\ & = \frac {k^2}{2} \cdot \left [\frac {3}{2} \cdot \theta - 2 \cos \theta - \frac {1}{4} \sin 2\theta \right ]_{0}^{\pi } \\ & = \frac {k^2}{4} \left (3 \pi - \sin \pi \cos \pi \right ) + k^2 \left (1 - \cos \pi \right ) \\ & = \frac {k^2}{4} \cdot 3\pi + 2k^2 \\ & = \left (2 + \frac {3\pi }{4}\right ) k^2. \end {align*}

   Hence, \begin {align*} \lim _{k \to \infty } \frac {R}{S} & = \lim _{k \to \infty } \frac {\frac {k^2 \pi }{2}}{\left (2 + \frac {3\pi }{4}\right ) k^2} \\ & = \lim _{k \to \infty } \frac {4 \pi }{8 + 3 \pi }. \end {align*}