If \(ABC\) is a triangle self-polar with respect to a conic \(S\), and if \(\alpha\) is the polar of another point \(A'\) with respect to \(S\), prove that the double points of the involution cut out on \(\alpha\) by conics through \(A, B, C\) and \(A'\) are the intersections of \(\alpha\) with \(S\). \par Hence, or otherwise, prove that, if each of two triangles is self-polar with respect to a conic, their six vertices lie on a conic. \par Deduce that if a triangle is self-polar with respect to a rectangular hyperbola, its circumcircle passes through the centre of the rectangular hyperbola.
The equation of a conic referred to rectangular Cartesian coordinates is \[ S \equiv ax^2+2hxy+by^2+2gx+2fy+c=0; \] if \(u \equiv ax+hy+g\) and \(v \equiv hx+by+f\), prove that
The equation \(x^n+p_1x^{n-1}+p_2x^{n-2}+\dots+p_n=0\) has roots \(\alpha_1, \alpha_2, \dots, \alpha_n\); and \[ S_m = \alpha_1^m + \alpha_2^m + \dots + \alpha_n^m. \] Obtain equations of the type \[ S_m + p_1 S_{m-1} + p_2 S_{m-2} + \dots + p_n S_{m-n} = 0, \text{ etc.,} \] connecting the sums of the powers of the roots with the coefficients \(p_r\). \par If \(\alpha, \beta, \gamma\) are the roots of the cubic equation \(x^3-px-q=0\), shew, by expanding \(\log(1-px^2-qx^3)\) or otherwise, that \[ \alpha^m+\beta^m+\gamma^m = m \sum \frac{(\lambda+\mu-1)!}{\lambda!\mu!}p^\lambda q^\mu, \] where the summation is over all values of \(\lambda\) and \(\mu\) such that \(2\lambda+3\mu=m\).
Two polynomials, \(P\) and \(Q\), have no factor in common. Shew that the maximum and minimum values of \(P/Q\) are the values of \(\lambda\) for which \(P-\lambda Q=0\) has a real root of even multiplicity. \par Shew further that the value \(\lambda\) is a maximum or a minimum according as \[ \{P^{(2v)}(\alpha)Q(\alpha) - P(\alpha)Q^{(2v)}(\alpha)\} \le 0, \] \(\alpha\) being the root of \(P-\lambda Q=0\) of multiplicity \(2v\). \par Find the turning values of \[ \frac{x+2}{x^2+x+2}, \] distinguishing between maxima and minima. \par [The condition that the cubic \(y^3+gy+h=0\) should have a pair of equal roots is that \(4g^3+27h^2=0\).]
``If \(\xi\) is an approximate root of the equation \(f(x)=0\), then in general \(\xi - f(\xi)/f'(\xi)\) is a better approximation.'' \par Discuss this statement graphically, pointing out cases when repeated application of the approximation will give the value of the root to any desired degree of accuracy and cases when it will not. \par If \(a\) is small the equation \(\sin x = ax\) has a root \(\xi\), nearly equal to \(\pi\). Shew that \[ \xi = \pi\left\{1-a+a^2-\left(\frac{\pi^2}{6}+1\right)a^3\right\} \] is a better approximation, if \(a\) is sufficiently small.
Explain the use of the ``angle of friction'' in the determination of the positions of equilibrium of a system containing imperfectly rough rigid bodies in contact. \par A uniform circular cylinder of radius \(a\) rests on a horizontal plane, with its axis parallel to and at a distance \((a+2a\sin 2\theta)\) from a vertical wall, where \(2\theta\) is a given acute angle. A similar cylinder is gently placed in contact with the first cylinder and with the wall, touching each along a generator. The coefficient of friction between the upper cylinder and the wall is \(\mu_1\), between the two cylinders \(\mu_2\), and between the lower cylinder and the plane \(\mu_3\). Shew that equilibrium is not possible unless \(\mu_1 \ge 1\). \par Assuming now \(\mu_1 \ge 1, \mu_2 \ge 1\), shew that equilibrium is possible for all values of \(\theta\) if \(\mu_3 \ge \frac{1}{4}\). If \(\mu_3 < \frac{1}{4}\), determine the greatest value of \(\theta\) for which equilibrium is possible.
A particle of constant mass \(m\) moves on a straight line under a force which is a function of position on the line. If \(v\) denotes the velocity of the particle when its potential energy is \(\Omega\), shew that \[ \frac{1}{2}mv^2 + \Omega \] remains constant throughout the motion. \par If the mass \(m\) of the particle is not constant, but is a function of \(v\), shew that \[ \int mv dv - \int m \dot{v} dv + \Omega \] remains constant. \par An electron, whose mass is \[ \frac{m_0}{\sqrt{(1-v^2/c^2)}}, \] moves from rest under a constant force. Shew that the increase of mass of the electron is proportional to the distance travelled.
Two particles, whose masses are \(m_1\) and \(m_2\), move on a straight line. Prove that the kinetic energy of the system is \[ \frac{1}{2}M\dot{\xi}^2 + \frac{1}{2}\mu\dot{x}^2, \] where \(M=m_1+m_2, \mu=m_1m_2/M, \xi\) is the distance of the centre of gravity from a fixed origin on the line, and \(x\) the distance between the particles. \par If there are no forces other than the mutual attraction between the particles, \(F\), shew that \[ \mu\ddot{x} + F = 0. \] Shew further that, if \(F=\gamma m_1 m_2 / x^2\), and the particles start from rest at distance \(c\) apart, they will collide after a time \(\pi\sqrt{(c^3/8\gamma M)}\).
Shew that the tangential and normal components of acceleration of a point moving on a given curve are \(\ddot{s}\) and \(\dot{s}^2/\rho\). \par A smooth wire is in the form of an arc of the cycloid \(s=4a\sin\psi\) between consecutive cusps, and is fixed with its axis vertical and its ends upwards. A bead slides on the wire. If the bead is let go from rest at a point of the wire, determine its position at any subsequent time. \par If the bead is let go from rest at a cusp, shew that the angular velocity of the direction of motion is constant during each half-period, and that the complete hodograph consists of two circles.
Explain and establish the principle of conservation of linear momentum. \par The base of a solid hemisphere of mass \(M\) and radius \(a\) rests on a horizontal plane. A particle of mass \(m\) is placed on the highest point of the hemisphere and is slightly disturbed. Assuming the surfaces to be smooth, shew that, so long as the particle remains in contact with the hemisphere, the path of the particle in space is an arc of an ellipse. \par Shew also that, if \(\theta\) denotes the angle which the radius through the particle makes with the vertical at time \(t\), then \[ \left(\frac{d\theta}{dt}\right)^2 = \frac{2g(1-\cos\theta)}{a(1-k\cos^2\theta)}, \] where \(k=m/(M+m)\).