Problems

Solution:

1. \(\,\)
   

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2. \(\,\) \begin{align*} && \sin y &= \tfrac12 \sin x \\ \Rightarrow && \frac{\d y}{\d x} \cos y &= \tfrac12 \cos x \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{\cos x}{2 \cos y} \\ &&&= \frac{\cos x}{2 \sqrt{1-\sin^2 y}} \\ &&&= \frac{\cos x}{2 \sqrt{1-\frac14 \sin^2 x}} \\ &&&= \frac{\cos x}{\sqrt{4-\sin^2 x}} \\ \\ && y'' &= \frac{-\sin x \cdot (4-\sin^2 x)^{\frac12} - \cos x \cdot (4-\sin^2 x)^{-\frac12} \cdot 2 \sin x \cos x}{(4-\sin^2 x)} \\ &&&= \frac{-\sin x \cdot (4-\sin^2 x) - \cos x \cdot 2 \sin x \cos x}{(4-\sin^2x)^{\frac32}} \\ &&&= \frac{-\sin x \cdot (4-\sin^2 x) - \sin x (1-\sin^2x)}{(4-\sin^2x)^{\frac32}} \\ &&&= \frac{-3\sin x }{(4-\sin^2x)^{\frac32}} \\ \end{align*}
   

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3. \(\,\)
   

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Solution: The condition is that we match the first and second derivative (as well as passing through the point in question, which is \((\frac{\pi}{4}, 2 - \frac{\pi}{4})\) The gradient is \(y' = -1 + \sec^2 x\), so the value is \(1\). The second derivative is \(y'' = 2 \sec^2 x \tan x\), which is \(4\) If we have a circle, radius \(r\), so \((x-a)^2 + (y-b)^2 = r^2\) then \(2(x-a) + 2(y-b) \frac{\d y}{\d x} = 0\) and \(2 + 2 \left ( \frac{\d y}{\d x} \right)^2 + 2(y-b) \frac{\d^2y}{\d x^2} = 0\). Therefore we must have \(1+1+(2-\frac{\pi}{4}-b)4 = 0 \Rightarrow b =\frac52-\frac{\pi}{4}\) We know that the centre lies on the line \(y = 2-x\), so we must have \(a = \frac{\pi}{4}-\frac12\) and so the centre is \(( \frac{\pi}{4} - \frac12,\frac52 - \frac{\pi}{4})\) and the radius is \(\sqrt{\frac14 + \frac14} = \frac{\sqrt{2}}{2}\)

Solution:

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   \begin{align*} && f(x) &= \exp\left (\tfrac23\sin x \right) \\ && f'(x) &= \exp\left (\tfrac23\sin x \right) \cdot \tfrac23 \cos x \\ && f''(x) &= \left ( \exp\left (\tfrac23\sin x \right) \cdot \tfrac23\right) \left ( \tfrac23 \cos^2 x - \sin x \right) \end{align*} Therefore \(f(x)\) is convex when \(\frac23 \cos^2 x \geq \sin x\). Note that we can find the equality points when \begin{align*} && \sin x &= \frac23 \cos^2 x \\ &&&= \frac23 (1- \sin^2 x) \\ \Rightarrow && 0 &= 2\sin^2 x + 3 \sin x - 2 \\ &&&= (2 \sin x -1) (\sin x+2) \end{align*} ie \(\sin x = \frac12 \Rightarrow x = \frac{\pi}{6}, \frac{5\pi}{6}\). Therefore \(f\) is convex on \([0, \frac{\pi}{6}] \cup [\frac{5\pi}{6}, 2\pi]\)
2. Suppose \(g(x) = \exp \left ( -k \tan x \right)\) then \begin{align*} && g'(x) &= \exp \left ( -k \tan x \right) \cdot (-k \sec^2 x ) \\ && g''(x) &= \left ( -k \exp \left ( -k \tan x \right) \right) \left ( -k\sec^4 x + 2 \sec x \cdot \sec x \tan x\right) \\ &&&= -k \exp \left ( -k \tan x \right) \sec^4 x \left ( -k + 2\sin x \cos x \right) \\ &&&= -k \exp \left ( -k \tan x \right) \sec^4 x \left ( -k + \sin 2x \right) \\ \end{align*} If \(0 < \alpha < \frac{\pi}{4}\) then \(k > 0\) so \(g\) is convex if \(-k + \sin 2x < 0\), ie \(\sin 2x < \sin 2\alpha\), ie on \((0, \alpha)\) and \((\frac{\pi}{2} - \alpha, \frac{\pi}{2})\)

Solution:

1. \begin{align*} && (1+x+x^{2}+\cdots+x^{n})(1-x) &= 1-x+x-x^2+\cdots -x^n+x^n-x^{n+1} \\ &&&= 1-x^{n+1} \\ \Rightarrow && 1+x+x^2+\cdots+x^n &= \frac{1-x^{n+1}}{1-x} \tag{dividing by \(1-x\)} \end{align*}
2. \begin{align*} \frac{\d}{\d x}: && 0+1+2x+\cdots+nx^{n-1} &= \frac{(n+1)(1-x)x^n+(1-x^{n+1})}{(1-x)^2} \\ \Rightarrow && 1-2x+\cdots+(-1)^n n &= \frac{-(n+1)2(-1)^n+(1-(-1)^{n+1})}{4} \\ &&&= \begin{cases} \frac{-(n+1)\cdot2\cdot1+(1-(-1)}{4} & \text{if }n\text{ is even} \\ \frac{-(n+1)\cdot 2 \cdot(-1)+(1-1)}{4} & \text{if }n\text{ is odd}\end{cases} \\ &&&= \begin{cases} \frac{-n}{2} & \text{if }n\text{ is even}\\ \frac{n+1}{2} & \text{if }n\text{ is odd}\end{cases} \\ \end{align*}
3. \begin{align*} x: && x+2x^2+\cdots+nx^{n} &= \frac{(n+1)(1-x)x^{n+1}+x(1-x^{n+1})}{(1-x)^2} \\ &&&= \frac{x+(n+1)x^{n+1}-nx^{n+2}}{(1-x)^2}\\ \frac{\d}{\d x}: && 1^2+2^2x + \cdots + n^2x^{n-1} &= \frac{(1-x)^2(1+(n+1)^2x^{n}-n(n+2)x^{n+1}) +2(1-x)(x+(n+1)x^{n+1}-nx^{n+2})}{(1-x)^4} \\ &&&= \frac{1 + x - (1 + n)^2 x^n + (2 n^2+2n-1) x^{n+1} - n^2 x^{n+2}}{(1-x)^3} \\ \Rightarrow && 1^2-2^2 + \cdots + (-1)^{n-1}n^2 &= \frac{(-1)^n \l - (1 + n)^2- (2 n^2+2n-1) - n^2 \r}{8} \\ &&&= \frac{(-1)^n(-4n^2-4n}{8} \\ &&&= \frac{(-1)^{n-1}(n^2+n)}{2} \end{align*}

Solution:

1. \begin{align*} v &= \mathrm{sech} u \\ &= \frac{1}{\mathrm{cosh } u} \\ &= \frac{1}{\sqrt{1+\mathrm{sinh}^2 u}} \tag{\(u > 0\)} \\ &= \frac{1}{\sqrt{1+\tan^2 \theta}} \\ &= \frac{1}{\sqrt{\mathrm{sec}^2 \theta}} \\ &= \cos \theta \tag{\(0 < \theta < \tfrac{\pi}{2}\)} \end{align*}
2. \begin{align*} && \tan \theta &= \textrm{sinh} u \\ \underbrace{\Rightarrow}_{\frac{\d}{\d u}} && \sec^2 \theta \cdot \frac{\d \theta}{\d u} &= \cosh u \\ \Rightarrow && \frac{\d \theta}{\d u} &=\cosh u \cdot \cos^2 \theta \\ &&&= \frac{1}{v} \cdot v^2 \\ &&&=v \end{align*}
3. \begin{align*} \sin 2 \theta &= 2 \sin \theta \cos \theta \\ &= 2 \sin \theta \cdot \frac{\d \theta}{\d u} \\ &= -2 \frac{\d v}{\d \theta} \cdot \frac{\d \theta}{\d u} \tag{\(\cos \theta = v\)} \\ &= -2 \frac{\d v}{\d u} \end{align*} \begin{align*} && \sin 2 \theta &= -2 \frac{\d v}{\d u} \\ \underbrace{\Rightarrow}_{\frac{\d}{\d u}} && 2 \cos 2 \theta \cdot \frac{\d \theta}{\d u} &= -2 \frac{\d^2 v}{\d u^2} \\ \Rightarrow && \cos 2 \theta &= - \frac{\d^2 v}{\d u^2} \frac{1}{v} \\ &&&= -\frac{\d ^2v}{\d u^2} \cosh u \end{align*}
4. \begin{align*} && \frac{\d u}{\d \theta} &= \frac{1}{v} \\ \Rightarrow && \frac{\d^2 u}{\d \theta^2} &= -\frac{1}{v^2} \frac{\d v}{\d \theta} \\ &&&= \frac{1}{v^2} \sin \theta \\ && \frac{\d v}{\d \theta} &= -\sin \theta \\ \Rightarrow && \frac{\d^2 v}{\d \theta^2} &= -\cos \theta \\ &&&= - v \\ \end{align*} Therefore \begin{align*} \frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}v}{\mathrm{d}\theta^{2}}+\frac{\mathrm{d}v}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta^{2}}+\left(\frac{\mathrm{d}u}{\mathrm{d}\theta}\right)^{2} &= \frac{1}{v} \cdot \left (-v\right) + \left ( - \sin \theta \right ) \cdot \left (\frac{1}{v^2} \sin \theta \right) + \frac{1}{v^2} \\ &= -1 + \frac{1-\sin^2 \theta}{v^2} \\ &= -1 + \frac{\cos^2 \theta}{v^2} \\ &= -1 + 1 \\ &= 0 \end{align*}