Problems

What does it mean to say that a number \(x\) is irrational? Prove by contradiction statements A and B below, where \(p\) and \(q\) are real numbers.

• A: If \(pq\) is irrational, then at least one of \(p\) and \(q\) is irrational.
• B: If \(p+q\) is irrational, then at least one of \(p\) and \(q\) is irrational.

• C: If \(p\) and \(q\) are irrational, then \(p+q\) is irrational.

The variables \(t\) and \(x\) are related by \(t=x+ \sqrt{x^2+2bx+c\;} \,\), where \(b\) and \(c\) are constants and \(b^2 < c\). Show that \[ \frac{\d x}{\d t} = \frac{t-x}{t+b}\;, \] and hence integrate \(\displaystyle \frac1 {\sqrt{x^2+2bx+c}}\,\). Verify by direct integration that your result holds also in the case \(b^2=c\) if \(x+b > 0\) but that your result does not hold in the case \(b^2=c\) if \(x+b < 0\,\).

Prove that, if \(c\ge a\) and \(d\ge b\), then \[ ab+cd\ge bc+ad\,. \tag{\(*\)} \]

1. If \(x\ge y\), use \((*)\) to show that \(x^2+y^2\ge 2xy\,\). If, further, \(x\ge z\) and \(y\ge z\), use \((*)\) to show that \(z^2+xy\ge xz+yz\) and deduce that \(x^2+y^2+z^2\ge xy+yz+zx\,\). Prove that the inequality \(x^2+y^2+z^2\ge xy+yz+zx\,\) holds for all \(x\), \(y\) and \(z\).
2. Show similarly that the inequality \[\frac st +\frac tr +\frac rs +\frac ts +\frac rt +\frac sr \ge 6\] holds for all positive \(r\), \(s\) and \(t\).

A function \(\f(x)\) is said to be convex in the interval \(a < x < b\) if \(\f''(x)\ge0\) for all \(x\) in this interval.

1. Sketch on the same axes the graphs of \(y= \frac23 \cos^2 x\) and \(y=\sin x\) in the interval \(0\le x \le 2\pi\). The function \(\f(x)\) is defined for \(0 < x < 2\pi\) by \[\f(x) = \e^{\frac23 \sin x}. \] Determine the intervals in which \(\f(x)\) is convex.
2. The function \(\g(x)\) is defined for \(0 < x < \frac12\pi\) by \[\g(x) = \e^{-k \tan x}. \] If \(k=\sin 2 \alpha\) and \(0 < \alpha < \frac{1}{4}\pi\), show that \(\g(x)\) is convex in the interval \(0 < x < \alpha\), and give one other interval in which \(\g(x)\) is convex.

The polynomial \(\p(x)\) is given by \[ \ds \p(x)= x^n +\sum\limits_{r=0}^{n-1}a_rx^r\,, \] where \(a_0\), \(a_1\), \(\ldots\) , \(a_{n-1}\) are fixed real numbers and \(n\ge1\). Let \(M\) be the greatest value of \(\big\vert \p(x) \big\vert\) for $\vert x \vert\le 1\(. Then Chebyshev's theorem states that \)M\ge 2^{1-n}$.

1. Prove Chebyshev's theorem in the case \(n=1\) and verify that Chebyshev's theorem holds in the following cases:
   1. \( \p(x) = x^2 - \frac12\,\);
   2. \(\p(x) = x^3 - x \,\).
2. Use Chebyshev's theorem to show that the curve $ \ y= 64x^5+25x^4-66x^3-24x^2+3x+1 \ $ has at least one turning point in the interval \(-1\le x \le 1\).

The function \(\f\) is defined by \[ \f(x) = \frac{\e^x-1}{\e-1}, \ \ \ \ \ x\ge0, \] and the function \(\g\) is the inverse function to \(\f\), so that \(\g(\f(x))=x\). Sketch \(\f(x)\) and \(\g(x)\) on the same axes. Verify, by evaluating each integral, that \[ \int_0^\frac12 \f(x) \,\d x + \int_0^k \g(x) \,\d x = \frac1 {2(\sqrt \e +1)}\,, \] where \(\displaystyle k= \frac 1{\sqrt\e+1}\), and explain this result by means of a diagram.

The point \(P\) has coordinates \((x,y)\) with respect to the origin \(O\). By writing \(x=r\cos\theta\) and \(y=r\sin\theta\), or otherwise, show that, if the line \(OP\) is rotated by \(60^\circ\) clockwise about \(O\), the new \(y\)-coordinate of \(P\) is \(\frac12(y-\sqrt3\,x)\). What is the new \(y\)-coordinate in the case of an anti-clockwise rotation by \(60^\circ\,\)? An equilateral triangle \(OBC\) has vertices at \(O\), \((1,0)\) and \((\frac12,\frac12 \sqrt3)\), respectively. The point \(P\) has coordinates \((x,y)\). The perpendicular distance from \(P\) to the line through \(C\) and \(O\) is \(h_1\); the perpendicular distance from \(P\) to the line through \(O\) and \(B\) is \(h_2\); and the perpendicular distance from \(P\) to the line through \(B\) and \(C\) is \(h_3\). Show that \(h_1=\frac12 \big\vert y-\sqrt3\,x\big\vert\) and find expressions for \(h_2\) and \(h_3\). Show that \(h_1+h_2+h_3=\frac12 \sqrt3\) if and only if \(P\) lies on or in the triangle \(OBC\).

1. The gradient \(y'\) of a curve at a point \((x,y)\) satisfies \[ (y')^2 -xy'+y=0\,. \tag{\(*\)} \] By differentiating \((*)\) with respect to \(x\), show that either \(y''=0\) or \(2y'=x\,\). Hence show that the curve is either a straight line of the form \(y=mx+c\), where \(c=-m^2\), or the parabola \(4y=x^2\).
2. The gradient \(y'\) of a curve at a point \((x,y)\) satisfies \[ (x^2-1)(y')^2 -2xyy'+y^2-1=0\,. \] Show that the curve is either a straight line, the form of which you should specify, or a circle, the equation of which you should determine.

Two identical particles \(P\) and \(Q\), each of mass \(m\), are attached to the ends of a diameter of a light thin circular hoop of radius \(a\). The hoop rolls without slipping along a straight line on a horizontal table with the plane of the hoop vertical. Initially, \(P\) is in contact with the table. At time \(t\), the hoop has rotated through an angle \(\theta\). Write down the position at time \(t\) of \(P\), relative to its starting point, in cartesian coordinates, and determine its speed in terms of \(a\), \(\theta\) and \(\dot\theta\). Show that the total kinetic energy of the two particles is \(2ma^2\dot\theta^2\). Given that the only external forces on the system are gravity and the vertical reaction of the table on the hoop, show that the hoop rolls with constant speed.

Show Solution

---

Solution:

+ - ↺

We can see that the position of \(O\) is \(\begin{pmatrix} a \theta \\ a \end{pmatrix}\) since the hoop is not slipping. \(P\)'s position relative to \(O\) is \(\begin{pmatrix} -a\sin\theta\\a(1-\cos \theta) \end{pmatrix}\), therefore the position of \(P\) is \(\begin{pmatrix} a(\theta-\sin\theta) \\ a(1-\cos \theta) \end{pmatrix}\). We can now calculate \(\mathbf{v}_P = a \begin{pmatrix} (\dot{\theta}-\dot{\theta}\cos\theta) \\ \dot{\theta}\sin \theta \end{pmatrix} = a \dot{\theta} \begin{pmatrix} (1-\cos\theta) \\ \sin \theta \end{pmatrix}\) We can also see that \begin{align*} && |\mathbf{v}_P|^2 &= a^2\dot{\theta}^2 \l \l 1 - \cos \theta \r^2 + \sin^2 \theta \r \\ && &= a^2\dot{\theta}^2 ( 2 - 2\cos \theta) \\ && &= 2a^2\dot{\theta}^2 ( 1 - \cos \theta) \\ && &= a^2\dot{\theta}^2 4 \sin^2 \frac{\theta}{2} \\ \Rightarrow |\mathbf{v}_P| &= 2a \dot{\theta} \left | \sin \frac{\theta}2 \right | \end{align*} Not that the position of \(Q\) is \(\begin{pmatrix} a(\theta+\sin\theta) \\ a(1+\cos \theta) \end{pmatrix}\) Therefore \begin{align*} && |\mathbf{v}_Q|^2 &= a^2\dot{\theta}^2 \l \l 1 + \sin \theta \r^2 + \l 1 + \cos \theta \r^2 \r \\ && &= a^2\dot{\theta}^2 \l \l 1 + \sin \theta \r^2 + \cos^2 \theta \r \\ && &= 2a^2\dot{\theta}^2 \l 1 + \cos \theta \r \\ \end{align*} Therefore \[ \text{K.E.} = \frac12m|\mathbf{v}_P|^2 + |\mathbf{v}_Q|^2 = \frac12m2a^2 \dot{\theta}^2 (1 - \cos \theta + 1-\cos \theta) = 2ma^2 \dot{\theta}^2\] Since there are no external forces acting conservation of energy tells us that kinetic energy is constant, ie \(4ma^2 \dot{\theta}\ddot{\theta} = 0 \Rightarrow \ddot{\theta} = 0\), ie the hoop is rolling with constant speed.

On the (flat) planet Zog, the acceleration due to gravity is \(g\) up to height \(h\) above the surface and \(g'\) at greater heights. A particle is projected from the surface at speed \(V\) and at an angle \(\alpha\) to the surface, where \(V^2 \sin^2\alpha > 2 gh\,\). Sketch, on the same axes, the trajectories in the cases \(g'=g\) and \(g' < g\). Show that the particle lands a distance \(d\) from the point of projection given by \[ d = \left(\frac {V-V'} g + \frac {V'}{ g'} \right) V\sin2\alpha\,, \] where \(V' = \sqrt{V^2-2gh\,\rm{cosec}^2\alpha\,}\,\).