Problems

A particle is projected vertically upwards with speed \(u\). The motion is subject to gravity and to a resistance per unit mass of \(Kg\) times the speed, \(K\) being a constant. Find the greatest height above the point of projection to which the particle will rise, and shew that the total distance travelled by the particle from projection until the speed is again \(u\) is given by \(\frac{1}{K^2g} \log\frac{1}{1-K^2u^2}\), where \(Ku<1\). What happens if \(Ku \ge 1\)?

A rough rigid wire rotates in a horizontal plane with constant angular velocity \(\omega\) about a vertical axis through a point \(O\) of itself. A bead, which can slide on the wire, is released from relative rest at a distance \(a\) from \(O\). Shew that at any time subsequently, the distance \(r\) of the bead from \(O\) satisfies the equation \[ \frac{d^2r}{dt^2} + 2\mu\omega \frac{dr}{dt} = \omega^2 r, \] \(\mu\) being the coefficient of friction. Prove that after a time \(t\), the velocity of the bead is \[ a\omega e^{-\mu\omega t} \left\{\cosh(n\omega t) + \frac{\mu}{n}\sinh(n\omega t)\right\}^{\frac{1}{2}}, \] (Note: The expression from the scan seems different from the OCR. Let's re-read the scan) It appears to be \(a\omega e^{-\mu\omega t} \{\cosh 2n\omega t + \frac{\mu}{n} \sinh 2n\omega t \}^{\frac{1}{2}}\). I will use this.

Two masses \(m_1\) and \(m_2\) are supported by a light inextensible string slung over a rough pulley of radius \(a\) which can turn freely about a fixed horizontal axis, its moment of inertia about the axis being \(I\). A mass \(m\) is rigidly attached to the pulley at a point distant \(h(

Solve the simultaneous equations: \begin{align*} 4x + 2y - z &= 0, \\ 5x + y - 2z &= 0, \\ 4x^2 - 4(1-a)y^2+a^2z^2+3x+3y-4az+4a(1-a) &= 0. \end{align*} Determine whether there are values of \(a\) for which the equations have (i) only one set of (finite) solutions; (ii) an infinite number of sets of solutions.

The quartic equation \[ 4x^4 + \lambda x^3 + 35x^2 + \mu x + 4 = 0 \] has its roots in geometric progression. What real values might be taken for the ratio of the progression?

Two determinants \(|a_{rs}|, |b_{rs}|\), each of the fourth order, are given by the relations \begin{alignat*}{2} a_{1s} &= x_s^2+y_s^2; \quad & a_{2s} = -2x_s; \quad a_{3s} = -2y_s; \quad a_{4s} = 1; \\ b_{1s} &= 1; \quad & b_{2s} = x_s; \quad b_{3s} = y_s; \quad b_{4s} = x_s^2+y_s^2, \end{alignat*} for \(s=1,2,3,4\). By evaluating the product \(|a_{rs}||b_{rs}|\), prove that the distances \(l_{pq}\) (\(p,q=1,2,3,4\)) connecting four concyclic points satisfy the relation \(|l_{pq}^2|=0\).

Express \(\sin 7\theta\) in terms of \(\sin\theta\), and determine the values of \(\theta\) for which \(7\sin\theta > \sin 7\theta\). Illustrate your answer graphically for values of \(\theta\) between \(0\) and \(2\pi\).

Express \(\tan 5\theta\) in terms of \(\tan\theta\). By considering the values of \(\theta\) for which \(\tan 5\theta = \sqrt{3}\), solve the equation \[ x^4 - 6\sqrt{3}x^3 + 8x^2 + 2\sqrt{3}x - 1 = 0, \] giving your answers in trigonometrical form. Prove that \[ \tan\frac{\pi}{15} \tan\frac{2\pi}{15} \tan\frac{4\pi}{15} \tan\frac{8\pi}{15} + 1 = 0. \]

1. If \(f(u)\) is a function of \(u=ax^2+2hxy+by^2\), and \(f(u)\), when expressed in terms of \(x\) and \(y\), takes the form \(g(x,y)\), prove that \[ x\frac{\partial g}{\partial x} + y\frac{\partial g}{\partial y} = 2u \frac{df}{du}. \]
2. If \(f(u,v)\) is a function of \(u=(x\sin\alpha - y\cos\alpha)^2\) and \(v=(x\cos\alpha+y\sin\alpha)^3\), and \(f(u,v)\), when expressed in terms of \(x\) and \(y\), takes the form \(g(x,y)\), prove that \[ x\frac{\partial g}{\partial y} - y\frac{\partial g}{\partial x} = 2u\frac{\partial f}{\partial u} + 2v\frac{\partial f}{\partial v}, \] and that \[ \frac{\partial^2 g}{\partial x^2} + \frac{\partial^2 g}{\partial y^2} = 2\left(\frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}\right) + 4\left(u \frac{\partial^2 f}{\partial u^2} + v^2 \frac{\partial^2 f}{\partial v^2}\right). \]

Find the third differential coefficient of \(\sin x/x\), and deduce, or find otherwise, the limit as \(x \to 0\) of \[ \frac{(6-3x^2)\sin x - x(6-x^2)\cos x}{x^4}. \] Prove also that the function \(\sin x/x\) has a minimum between \(x=\pi\) and \(x=\frac{3}{2}\pi\), and that a closer approximation to the position of the minimum is \(x = \frac{3\pi}{2} - \frac{2}{3\pi}\).