Problems

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Solution: \begin{align*} && y' &= 2ax+b \\ \Rightarrow && m &= 2ax_t+b \\ \Rightarrow && x_t &= \frac{m-b}{2a} \end{align*} Therefore we must have \begin{align*} mx_t &= 2ax_t^2+bx_t \\ y - mx &= ax_t^2+bx_t+c - mx_t \\ &= ax_t^2+bx_t+c - (2ax_t^2+bx_t) \\ &= c - ax_t^2 \\ &= c-a\left (\frac{m-b}{2a} \right)^2 \\ &= c - \frac{(m-b)^2}{4a} \end{align*} They will have a common tangent if and only if the constant terms are equal, ie \begin{align*} && c_1 - \frac{(m-b_1)^2}{4a_1} &= c_2 - \frac{(m-b_2)^2}{4a_2} \\ \Leftrightarrow && (c_1-c_2) &= \frac{(m-b_1)^2}{4a_1} -\frac{(m-b_2)^2}{4a_2} \\ \Leftrightarrow && 4a_1a_2(c_1-c_2) &= a_2(m-b_1)^2-a_1(m-b_2)^2 \\ &&&= (a_2-a_1)m^2+2(a_1b_2-a_2b_1)m+a_2b_1^2-a_1b_2^2 \end{align*} as required. Treating this as a polynomial in \(m\), we can see that the two curves will have exactly one common tangent iff \(\Delta = 0\), ie: \begin{align*} && 0 &= \Delta \\ &&&= (2(a_1b_2-a_2b_1))^2 - 4 (a_2-a_1)(4a_1 a_2(c_2-c_1)+ a_2 b_1^2-a_1 b_2 ^2) \\ &&&= 4a_1^2b_2^2-8a_1a_2b_1b_2+4a_2b_1^2 - 4a_2^2b_1^2-4a_1^2b_2^2 + 4a_1a_2(b_1^2+b_2^2)-16(a_2-a_1)a_1a_2(c_2-c_1) \\ &&&=-8a_1a_2b_1b_2+4a_1a_2(b_1^2+b_2^2)-16(a_2-a_1)a_1a_2(c_2-c_1) \\ &&&=a_1a_2(4(b_1-b_2)^2-16(a_2-a_1)(c_2-c_1)) \\ &&&= 4a_1a_2((b_2-b_1)^2 - 4(a_2-a_1)(c_2-c_1) \end{align*} But this is just the discriminant of the difference, ie equivalent to the two parabolas just touching. (Assuming \(a_1-a_2 \neq 0\) and we do end up with a quadratic). If \(a_1 = a_2 = a\) then we need exactly one solution to \(2a(b_1-b_2)m +4a^2(c_2-c_1)+a(b_1^2-b_2^2) = 0\), ie \(b_1 \neq b_2\).

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Solution: If \(p \geq 0\) then the derivative is \(x^2+p \geq 0\) and in particular the function is increasing. Therefore it will have exactly \(1\) real root (as for very large negative \(x\) it is negative, and vice-versa fo positive \(x\)). \begin{align*} && \frac{\d y}{\d x} &= \frac{\dot{y}}{\dot{x}} \\ &&&= \frac{2a}{2at} \\ &&&= \frac{1}{t} \\ \text{eq of normal} && \frac{k-2at}{h-at^2} &= -t \\ \Rightarrow && k-2at &= at^3-th \\ && 0 &= at^3+(2a-h)t-k \end{align*} Since \(a > 0\) this is the same constraint as the first part, in particular \(2a-h \geq 0 \Leftrightarrow 2a \geq h\). If exactly two normals can be drawn to \(C\) we must have that our equation has a repeated root, ie \begin{align*} && 0 &= at^3+(2a-h)t-k\\ && 0 &= 3at^2+2a-h\\ \Rightarrow && 0 &= 3at^3+ 3(2a-h)t-3k \\ && 0 &= 3at^3+(2a-h)t \\ \Rightarrow && 0 &= 2(2a-h)t-3k \\ \Rightarrow && t &= \frac{3k}{2(2a-h)} \\ \Rightarrow && 0 &= 3a \left (\frac{3k}{2(2a-h)} \right)^2+2a-h \\ && 0 &= 27ak^2+4(2a-h)^3 \end{align*}

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Solution:

\begin{align*} && y' &= \frac{2xy}{x^2-1} \\ \Rightarrow && \int \frac{1}{y} \d y &= \int \frac{2x}{x^2-1} \d x \\ \Rightarrow && \ln |y| &= \ln |x^2-1| + C \\ \Rightarrow && y &= A(x^2-1) \end{align*} which is a family of parabolas each passing through \((\pm1, 0)\) and with a vertex on the \(y\)-axis. The curve we seek must satisfy \begin{align*} && y' &= \frac{1-x^2}{2xy} \\ \Rightarrow && \int2 y \d y &= \int \left ( \frac{1}{x} - x \right) \d x \\ \Rightarrow && y^2 &= \ln x - \tfrac12 x^2 + C \\ (1,1): && 1 &= -\tfrac12+C \\ \Rightarrow && C &= \frac32 \\ \Rightarrow && y^2 &= \tfrac32 + \ln x - \tfrac12 x^2 \end{align*}

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Solution: The position of the particle is projected at angle \(\theta\) is \((x,y) = (v \cos \theta t, v \sin \theta t - \frac12 g t^2)\), ie \(t = \frac{x}{v \cos \theta}\), \begin{align*} && y &= x\tan \theta -\frac12 g \frac{x^2}{v^2} \sec^2 \theta \\ && y &= x \tan \theta -\frac{1}{4h} (1+\tan^2 \theta) x^2 \\ && 0 &= \frac{1}{4h} x^2\tan^2 \theta - x \tan \theta + \frac{x^2}{4h} +y \\ \Delta \geq 0: && 0 &\leq \Delta = x^2-4\frac{x^2}{4h}\left (\frac{x^2}{4h}+y \right) \\ &&&=1-\frac{1}{4h^2}(x^2+4hy) \\ \Rightarrow && x^2+4hy &\leq 4h^2 \\ \Rightarrow && x^2 &\leq 4h(h-y) \end{align*} We are assuming that there are no forces acting other than gravity (eg air resistance)

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Solution:

1. The line \(PO\) has gradient \(\frac{p^2}{p} = p\) and teh line \(QO\) has gradient \(q\), therefore we must have that \(pq = -1\). Therefore, \(R\) is the point \begin{align*} && R &= \left ( \frac{p-\frac{1}{p}}{2}, \frac{p^2+\frac{1}{p^2}}{2} \right) \\ &&&= \left ( \frac12\left ( p - \frac{1}{p} \right),2\left (\frac12 \left(p-\frac{1}{p}\right) \right)^2+1 \right) \\ &&&= \left ( t, 2t^2+1\right) \end{align*} So we are looking at another parabola.
   

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2. The tangents are \(y = 2px+c\), ie \(p^2 = 2p^2+c\), ie \(y = 2px -p^2\) so we have \begin{align*} && y - 2px &= -p^2 \\ && y - 2qx &= -q^2 \\ \Rightarrow && (2p-2q)x &= p^2-q^2 \\ \Rightarrow && x &= \frac12 (p+q)\\ && y &= p(p+q)-p^2 \\ && y &= pq = -1 \end{align*} Therefore \(x = \frac12(p - \frac1p), y= -1\), so we have the line \(y = -1\) (the directrix)
   

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Solution: First notice that \(y = \cos 2 \theta = 1 - 2\sin^2 \theta = 1- 2x^2\), therefore \(P\) is lies on that parabola.

The arc length is \begin{align*} && s &= \int_{-\pi/2}^{\pi/2} \sqrt{\left ( \frac{\d x}{\d \theta} \right)^2+\left ( \frac{\d y}{\d \theta} \right)^2} \d \theta\\ && &= \int_{-\pi/2}^{\pi/2} \sqrt{\cos^2 \theta+16 \sin^2 \theta \cos^2 \theta } \d \theta\\ && &= \int_{-\pi/2}^{\pi/2} \cos \theta\sqrt{1+16 \sin^2 \theta} \d \theta\\ u = \sin \theta, \d u = \cos \theta \d \theta && &= \int_{u=-1}^{u=1} \sqrt{1+16 u^2} \d u\\ 4u = \sinh v, 4\d u = \cosh v: && &= \int_{v=-\sinh^{-1} 4}^{v=\sinh^{-1} 4} \sqrt{1+\sinh^2 v} \tfrac14\cosh v \d v\\ && &= \frac14 \int_{-\sinh^{-1} 4}^{\sinh^{-1} 4} \cosh^2 v \d v\\ && &= \frac18 \int_{-\sinh^{-1} 4}^{\sinh^{-1} 4} (1 + \cosh 2v) \d v\\ && &= \frac14 \sinh^{-1} 4 + \frac18\left [ \frac12\sinh 2v \right]_{-\sinh^{-1} 4}^{\sinh^{-1} 4}\\ && &= \frac14 \sinh^{-1} 4 + \frac18\left [ \sinh v \sqrt{1 + \sinh^2 v} \right]_{-\sinh^{-1} 4}^{\sinh^{-1} 4}\\ && &= \frac14 \sinh^{-1} 4 + \left (\frac18 \cdot 4 \sqrt{17} \right) - \left (\frac18 \cdot (-4) \sqrt{17} \right)\\ && &= \frac14 \sinh^{-1} 4 + \sqrt{17}\\ \end{align*} The volume of revolution is \begin{align*} && V &=\pi \int_{-1}^1 (2-2x^2)^2 \d x \\ &&&= \pi \left [4x-\frac83x^3+\frac45x^5 \right]_{-1}^1 \\ &&&= \pi \left ( 8-\frac{16}3+\frac85 \right) \\ &&&= \frac{64}{15}\pi \end{align*}

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Solution:

The curves will intersect when: \begin{align*} && \frac{1}{1+\cos \theta} &= 4 (1 + \cos \theta) \\ \Rightarrow && 1 + \cos \theta &= \pm \frac{1}{2} \\ \Rightarrow && \cos \theta &= -\frac12 \\ \Rightarrow && \theta &= \pm \frac{2\pi}{3}, \end{align*} Therefore we can measure the two sides of the boundaries. For the cardioid it will be: \begin{align*} s &= \int_{-2\pi/3}^{2 \pi /3} \sqrt{r^2 + \left ( \frac{\d r}{\d \theta} \right)^2} \d \theta \\ &= \int_{-2\pi/3}^{2 \pi /3}\sqrt{r^2 + \left ( \frac{\d r}{\d \theta} \right)^2} \d \theta \\ &= \int_{-2\pi/3}^{2 \pi /3}\sqrt{16(1 + \cos \theta)^2 + 16 \sin^2 \theta} \d \theta \\ &= 4\int_{-2\pi/3}^{2 \pi /3}\sqrt{2 + 2 \cos \theta} \d \theta \\ &= 8\int_{-2\pi/3}^{2 \pi /3}\sqrt{\cos^2 \frac{\theta}{2}} \d \theta \\ &= 8\int_{-2\pi/3}^{2 \pi /3}|\cos \frac{\theta}{2}| \d \theta \\ &= 16\int_{\pi}^{2\pi/3}(-\cos \frac{\theta}{2}) \d \theta + 8\int_{-\pi}^{\pi}\cos \frac{\theta}{2} \d \theta \\ &= 16 \cdot \left [ 2\sin \frac{\theta}{2}\right]_{\pi}^{2\pi/3}+ 8 \cdot 4 \\ &= 16 \cdot (\sqrt{3}-2) + 8 \cdot 4 \\ &= 16\sqrt{3} \end{align*} For the parabola we have that \(\sqrt{x^2+y^2} + x = 1 \Rightarrow x^2 + y^2 = 1 - 2x + x^2 \Rightarrow y^2 = 1-2x\). So we can parameterise our parabola as \(y = t, x = \frac{1-t^2}2\). And we are interested in the points \(t = -\sqrt{3}\) and \(t =\sqrt3\) \begin{align*} &&s &= \int_{-\sqrt3}^\sqrt3 \sqrt{\left ( \frac{\d x}{\d t} \right)^2 + \left ( \frac{\d y}{\d t} \right)^2 } \d t \\ &&&= \int_{-\sqrt3}^\sqrt3 \sqrt{t^2+1^2} \d t \\ \sinh u = t, \frac{\d t}{\d u} = \cosh u&&&= \int_{-\sinh^{-1} \sqrt3}^{\sinh^{-1}\sqrt3} \cosh^2 u \d u \\ &&&= \left [\frac12 u + \frac14 \sinh(2u) \right ]_{-\sinh^{-1} \sqrt3}^{\sinh^{-1}\sqrt3} \\ &&&= \sinh^{-1} \sqrt{3} + 2\sqrt{3} \\ &&&= \ln(2 + \sqrt{3}) + 2\sqrt{3} \end{align*} Therefore the total distance is as required.

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Solution: \(AG = r\), therefore the area is: \begin{align*} A &= [AHG] - 2*[ABC] + [CDE] \\ &= \frac12 \pi r^2 - \pi h^2 + \frac12 \pi (r-2h)^2 \\ &= \frac12 \pi \l r^2 - 2h^2 + r^2 -4rh+4h^2 \r \\ &= \frac12 \pi \l 2r^2 -4rh + 2h^2\r \\ &= \pi (r-h)^2 \end{align*} This is the same area as a circle radius \(r-h\) But \(HD = r + (r-2d) = 2(r-d)\), ie the circle with diameter \(HD\) has radius \(r-h\) as required. Suppose \(A = (-h, 0), C = (h, 0), B = (0, h)\) then our parabola is \(y = \frac1{h}(h^2-x^2)\)

The area of \(ABC\) would then be \(\int_{-h}^h \frac{1}{h}(h^2-x^2) \d x = \frac1{h} \left [ h^2x - \frac{x^3}{3} \right] = \frac1{h} \l 2h^3-2\frac{h^3}{3} \r = \frac{4}{3}h^2\) so we have: \begin{align*} A &= [AHG] - 2*[ABC] +[CDE] \\ &= \frac43 r^2-\frac83h^2+\frac43(r-2h)^2 \\ &= \frac43 \l r^2 -2h^2+r^2-4rh+4h^2) \\ &= \frac43 (r-h)^2 \end{align*}

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Solution:

1. Suppose our curve is \(r(\theta)\), then \(y = r \sin \theta, x = r \cos \theta\) and \begin{align*} && \frac{\d y}{\d \theta} &= \frac{\d r}{\d \theta} \sin \theta + r \cos \theta \\ && \frac{\d x}{\d \theta} &= \frac{\d r}{\d \theta} \cos \theta - r \sin \theta \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{\d y}{\d \theta} \Bigg / \frac{\d x}{\d \theta} \\ &&&= \frac{\frac{\d r}{\d \theta} \sin \theta + r \cos \theta}{\frac{\d r}{\d \theta} \cos \theta - r \sin \theta} \\ &&&= \frac{\frac{\d r}{\d \theta} \tan\theta + r }{\frac{\d r}{\d \theta} - r \tan\theta} \end{align*} as required.
2. Set up a system of polar coordinates such that the origin is at \(O\) and all points in the plane containing \(L\) are represented by \((r, \theta)\). The constraint we have is that the angle of the normal, is \(\frac12 \theta\). Let \(\tan \tfrac12 \theta = t\), then \(\tan \theta = \frac{2t}{1-t^2}\) \begin{align*} && \tan \frac12 \theta &= -\frac{\frac{\d r}{\d \theta} - r \tan\theta}{\frac{\d r}{\d \theta} \tan\theta + r } \\ \Rightarrow && t &= -\frac{r'-r\frac{2t}{1-t^2}}{r' \frac{2t}{1-t^2}+r} \\ &&&= \frac{2tr-(1-t^2)r'}{2tr'+(1-t^2)r} \\ \Rightarrow && (2t^2+1-t^2)r' &= (2t-t+t^3)r \\ && (1+t^2)r' &= t(t^2+1) r \\ \Rightarrow && r' &= t r \\ \Rightarrow && \frac{\d r}{\d \theta} &= \tan \tfrac12 \theta r \\ \Rightarrow && \int \frac1r \d r &= \int \tan \frac12 \theta \d \theta \\ && \ln r &= -2\ln \cos \tfrac12 \theta+C \\ \Rightarrow && r\cos^2 \frac12 \theta &= C \\ \Rightarrow && r + r\cos \theta &= D \\ \Rightarrow && r &= D-x \\ \Rightarrow && x^2 + y^2 &= D^2 - 2Dx + x^2 \\ \Rightarrow && y^2 &= D^2-2Dx \end{align*} Therefore it is a parabola

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Solution:

1. Suppose \(w = u+ vi\) then \(w^2 - 2w = u^2-v^2-2u+(2uv-2v)i\) so to be purely real we must have \(2uv-2v = 2v(u-1) = 0\) ie either \(v = 0\) or \(u = 1\). Therefore the locus is the real axis and the line \(1 + ti\):
   

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2. If \(w^2 = x+yi\) then we must have \(x = u^2-v^2\) and \(y = 2uv\), so either \(v = 0, y = 0, x = u^2-2u \geq -1\) or \(u = 1, x = 1-v^2, y = 2v\) which is a parabola:
   

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If \(t = u+iv\) then \(t^2-2t = u^2-v^2-2u + (2uv-2v)i\). For this to be purely imaginary, we need \(u^2-v^2 - 2u = 0 \Rightarrow (u-1)^2-v^2 = 1\), ie points on a hyperbola. Then \(p = u^2-v^2\) and \(q = 2uv\). We can parameterise our hyperbola as \(u = 1 \pm \cosh s, v = \sinh s\) and so \(p = 1 + 2 \cosh s\) and \(q = \sinh 2s\) or \(q = \pm (p-1) \sqrt{(\frac{p-1}{2})^2-1}\) where \(p \geq 3\)

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Solution: The path can be broken down into \(5\) sections. 1. The section from \((-R,0)\) to \((-a,0)\) which will have distance \(R-a\) and is unchangeable. 2. The distance from \((-a,0)\) to \((-b, \frac{a^2-b^2}{2a})\) whose distance we will calculate shortly. 3. The distance from \((-b, \frac{a^2-b^2}{2a})\) to \((b, \frac{a^2-b^2}{2a})\) which will have distance \(\pi b\). 4. The distance from \((b, \frac{a^2-b^2}{2a})\) to \((a,0)\) which will have the same distance as 2. 5. The distance from \((a,0)\) to \((R,0)\) which will have distance \(R-a\) and we have no control over. \begin{align*} \text{distance 2.} &= \int_b^a \sqrt{1 + \left ( \frac{x}{a}\right)^2 } \d x \end{align*} We want to minimize the total, by varying \(b\), so it makes sense to differentiate and set to zero. \begin{align*} &&0&= -2\sqrt{1+\frac{b^2}{a^2}} + \pi \\ \Rightarrow && \frac{\pi^2}{4} &= 1 + \frac{b^2}{a^2} \\ \Rightarrow && b &= a \sqrt{\frac{\pi^2}{4}-1} \end{align*} Since \(\pi \approx 3\) this point is outside our range \(0 \leq b \leq a\), and our derivative is always positive. Therefore the distance is always increasing and the king would be better off going around the hill as soon as he arrives at it.