Problems

Solution:

1. 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 \(a\).
   

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2. 1. 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 \(y\)).
   2. 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 \(y \approx x\)
   3. \(\,\) \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 \((2, 0)\)
   

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3. \(\,\)
   

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Solution:

1. \begin{align*} (3+2\sqrt{5})^3 &= 3^3 + 3 \cdot 3^2 \cdot 2\sqrt{5} + 3 \cdot 3 \cdot (2 \sqrt{5})^2 + (2\sqrt{5})^3 \\ &= 27 + 180 + (54+40)\sqrt{5} \\ &= 207 + 94\sqrt{5} \end{align*}
2. \begin{align*} && (c-d\sqrt{2})^3 &= c^3+6cd -(3c^2d+2d^3)\sqrt{2} \\ \Rightarrow && 99 &= c(c^2+6d^2) \\ && 70 &= d(3c^2+2d^2) \\ \Rightarrow && c & \mid 99, d \mid 70 \\ && c &= 3, d = 2 \end{align*}
3. \begin{align*} && 0 &= x^6 - 198x^3 + 1 \\ \Rightarrow && 0 &= (x^3-99)^2+1-99^2 \\ \Rightarrow && x^3 &= 99 \pm \sqrt{99^2-1} \\ &&&= 99 \pm 10 \sqrt{98} \\ &&&= 99 \pm 70 \sqrt{2} \\ \Rightarrow && x &= 3 \pm 2 \sqrt{2} \end{align*}

Solution: \begin{align*} && 0 &= V\sin \alpha t-\frac12 gt^2 \\ \Rightarrow && t &= \frac{2V \sin \alpha}{g} \\ && R &= V \cos \alpha \, t \\ &&&= \frac{2V^2 \sin \alpha \cos \alpha}{g} \\ &&&= \frac{V^2 \sin 2 \alpha}{g} \end{align*} Therefore the max distance is \(\frac{V^2}{g}\), therefore we cannot hit the target if \(R > \frac{V^2}{g} \Rightarrow gR > V^2\). We have \(\beta = \omega t \Rightarrow t = \frac{\beta}{\omega}\) \begin{align*} && \sin \alpha &= \frac{gt}{2V} \\ && \cos \alpha &= \frac{R}{Vt} \\ \Rightarrow && 1 &= \left (\frac{gt}{2V} \right)^2 + \left ( \frac{R}{Vt} \right)^2 \\ &&&= \left (\frac{g\beta}{2V \omega} \right)^2 + \left ( \frac{R\omega}{V\beta} \right)^2 \\ &&&= \frac{g^2 \beta^2}{4 V^2 \omega^2} + \frac{R^2 \omega^2}{V^2 \beta ^2} \\ \Rightarrow && 4V^2 \omega^2 \beta^2 &= g^2 \beta^4 + 4R^2 \omega^4 \\ \Rightarrow && 0 &= g^2 \beta^4 - 4\omega^2 V^2 \beta^2+4R^2\omega^4 \end{align*} This (quadratic) equation in terms of \(\beta^2\) has two solution if \(\Delta = 16\omega^4V^4-16g^2R^2\omega^4 =16\omega^4(V^4-g^2R^2) > 0\) since \(V^2 > gR\). Since \(\beta > 0\) there are exactly two solutions, once we have values for \(\beta\). First notice, \begin{align*} && \beta_1^2 + \beta_2^2 &= \frac{4\omega^2V^2}{g^2} \\ && \beta_1^2\beta_2^2 &= \frac{4R^2\omega^4}{g^2} \end{align*} Then notice the positions of \(P_1\) and \(P_2\) are \((R\cos \beta_1 , R\sin \beta_1)\) and \((R\cos \beta_2, R\sin \beta_2)\). \begin{align*} && d^2 &= R^2\left ( \cos \beta_1 - \cos \beta_2 \right)^2 + R^2 \left ( \sin \beta_1 - \sin \beta_2 \right)^2 \\ &&&= 2R^2 - 2R^2(\cos \beta_1 \cos \beta_2 + \sin \beta_1 \sin \beta_2) \\ &&&= 2R^2-2R^2\cos(\beta_1 - \beta_2) \\ &&&= 2R^2 \left (1-\cos(\sqrt{(\beta_1-\beta_2)^2} \right ) \\ &&&= 2R^2 \left (1 - \cos\left ( \sqrt{\frac{4\omega^2 V^2}{g^2} - \frac{4R\omega^2}{g}} \right) \right) \\ &&&= 2R^2 \left (1 - \cos\left (\frac{2\omega}{g} \sqrt{V^2 - Rg} \right) \right) \\ &&&= 4 R^2 \sin^2 \left (\frac{\omega}{g} \sqrt{V^2 - Rg} \right) \end{align*} which gives us the required result.