Problems

1. Sketch the curve \(C_1\) with equation \[ \left(y^2 + (x-1)^2 - 1\right)\left(y^2 + (x+1)^2 - 1\right) = 0. \]
2. Consider the curve \(C_2\) with equation \[ \left(y^2 + (x-1)^2 - 1\right)\left(y^2 + (x+1)^2 - 1\right) = \tfrac{1}{16}. \]
   1. Show that the line \(y = k\) meets the curve \(C_2\) at points for which \[ x^4 + 2(k^2 - 2)x^2 + \left(k^4 - \tfrac{1}{16}\right) = 0. \] Hence determine the number of intersections between curve \(C_2\) and the line \(y = k\) for each positive value of \(k\).
   2. Determine whether the points on curve \(C_2\) with the greatest possible \(y\)-coordinate are further from, or closer to, the \(y\)-axis than those on curve \(C_1\).
   3. Show that it is not possible for both \(y^2 + (x-1)^2 - 1\) and \(y^2 + (x+1)^2 - 1\) to be negative, and deduce that curve \(C_2\) lies entirely outside curve \(C_1\).
   4. Sketch the curves \(C_1\) and \(C_2\) on the same axes.

The curves \(C_1\) and \(C_2\) both satisfy the differential equation \[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{kxy - y}{x - kxy},\] where \(k = \ln 2\). All points on \(C_1\) have positive \(x\) and \(y\) co-ordinates and \(C_1\) passes through \((1,\,1)\). All points on \(C_2\) have negative \(x\) and \(y\) co-ordinates and \(C_2\) passes through \((-1,\,-1)\).

1. Show that the equation of \(C_1\) can be written as \((x-y)^2 = (x+y)^2 - 2^{x+y}\). Determine a similar result for curve \(C_2\). Hence show that \(y = x\) is a line of symmetry of each curve.
2. Sketch on the same axes the curves \(y = x^2\) and \(y = 2^x\), for \(x \geqslant 0\). Hence show that \(C_1\) lies between the lines \(x + y = 2\) and \(x + y = 4\). Sketch curve \(C_1\).
3. Sketch curve \(C_2\).

1. Sketch, on \(x\)-\(y\) axes, the set of all points satisfying \(\sin y = \sin x\), for \(-\pi \le x \le \pi\) and \(-\pi \le y \le \pi\). You should give the equations of all the lines on your sketch.
2. Given that \[ \sin y = \tfrac12 \sin x \] obtain an expression, in terms of \(x\), for \(y'\) when \(0\le x \le \frac12 \pi\) and \(0\le y \le \frac12 \pi\), and show that \[ y'' = - \frac {3\sin x}{(4-\sin^2 x)^{\frac32}} \;. \] Use these results to sketch the set of all points satisfying \(\sin y = \tfrac12 \sin x\) for \(0 \le x \le \frac12 \pi\) and \(0 \le y \le \frac12 \pi\). Hence sketch the set of all points satisfying \(\sin y = \tfrac12 \sin x\) for \(-\pi\! \le \! x \! \le \! \pi\) and \mbox{\( -\pi \, \le\, y\, \le\, \pi\,\)}.
3. Without further calculation, sketch the set of all points satisfying \(\cos y = \tfrac12 \sin x\) for \(- \pi \le x \le \pi\) and \( -\pi \le y \le \pi\).

1. If \[{\mathrm f}(x)=\tan^{-1}x+\tan^{-1}\left(\frac{1-x}{1+x}\right),\] find \({\mathrm f}'(x)\). Hence, or otherwise, find a simple expression for \({\mathrm f}(x)\).
2. Suppose that \(y\) is a function of \(x\) with \(0 < y < (\pi/2)^{1/2}\) and \[x=y\sin y^{2}\] for \(0 < x < (\pi/2)^{1/2}\). Show that (for this range of \(x\)) \[\frac{{\mathrm d}y}{{\mathrm d}x}= \frac{y}{x+2y^2\sqrt{y^{2}-x^{2}}}.\]

Let \[\mathrm{f}(t)=\frac{\ln t}t\quad\text{ for }t>0.\] Sketch the graph of \(\mathrm{f}(t)\) and find its maximum value. How many positive values of \(t\) correspond to a given value of \(\mathrm f(t)\)? Find how many positive values of \(y\) satisfy \(x^y=y^x\) for a given positive value of \(x\). Sketch the set of points \((x,y)\) which satisfy \(x^y=y^x\) with \(x,y>0\).

Sketch the curve \(y^{2}=1-\left|x\right|\). A rectangle, with sides parallel to the axes, is inscribed within this curve. Show that the largest possible area of the rectangle is \(8/\sqrt{27}\). Find the maximum area of a rectangle similarly inscribed within the curve given by \(y^{2m}=\left(1-\left|x\right|\right)^{n}\), where \(m\) and \(n\) are positive integers, with \(n\) odd.

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Solution:

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Suppose one of the \(x\) coordinates is \(t > 0\), then the coordinates are \(y = \pm \sqrt{1-t}, x = \pm t\). The area will be \(A = 2t \cdot 2 \sqrt{1-t}\). To maximise this, \begin{align*} && \frac{\d A}{\d t} &= 4 \sqrt{1-t} - 2t(1-t)^{-\frac12} \\ &&&= \frac{4(1-t) - 2t}{\sqrt{1-t}} \\ &&&= \frac{4-6t}{\sqrt{1-t}} \end{align*} Therefore there is a stationary point at \(t = \frac23\). Since we know the area is \(0\) when \(t = 0, 1\) we can see this must be a maximum for the area. Therefore the area is \(\displaystyle 4 \frac23 \sqrt{1-\frac23} = \frac{8}{3\sqrt{3}} = \frac{8}{\sqrt{27}}\). For this similar problem, using a similar approach we find \(y = \pm (1- t)^{n/2m}, x = \pm t\) and so the area is \(A = 4 t \cdot (1-t)^{n/2m}\). \begin{align*} && \frac{\d A}{\d t} &= 4(1-t)^{n/2m} - 4t \frac{n}{2m} (1-t)^{\frac{n}{2m} - 1} \\ &&&= (1-t)^{\frac{n}{2m}-1} \left ( 4(1-t) - \frac{2n}{m} t\right) \\ &&&= (1-t)^{\frac{n}{2m}-1} \left ( 4 - (4 + \frac{2n}{m})t \right) \\ \end{align*} Therefore \(\displaystyle t = \frac{2m}{2m+n}\) and \(\displaystyle A = 4\cdot \frac{2m}{2m+n} \cdot (1 - \frac{2m}{2m+n})^{n/2m} = \frac{8m}{2m+n} \cdot \left ( \frac{n}{2m+n}\right)^{n/2m}\)