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1913 Paper 1 Q104
D: 1500.0 B: 1500.0

Solve the equations: \begin{align*} (y-c)(z-b) &= a^2, \\ (z-a)(x-c) &= b^2, \\ (x-b)(y-a) &= c^2. \end{align*}

1913 Paper 1 Q104
D: 1500.0 B: 1500.0

Shew how to reduce the general equation of a conic, referred to rectangular axes, \[ S = ax^2+2hxy+by^2+2gx+2fy+c=0, \] to simpler forms. Consider the various types of simplification that are possible, and the discrimination of cases which arise for central, non-central, and degenerate conics. Shew, by considering the tangential equation \[ Al^2+Bm^2+Cn^2+2Fmn+2Gnl+2Hlm = 0 \] of the conic, or otherwise, that the coordinates of the foci are given by the equations \[ Cx^2-2Gx+A=\kappa, \] \[ Cy^2-2Fy+B=\kappa, \] where \(\kappa\) is either root of the quadratic \[ C\kappa^2-(a+b)\Delta\kappa+\Delta^2=0. \]

1913 Paper 1 Q105
D: 1500.0 B: 1500.0

Two equal heavy cylinders of radius \(a\) are placed in contact in a smooth fixed cylinder of radius \(b\) (\(>2a\)) as shewn in the figure; a third equal cylinder is placed gently on top of them, the axes of all the cylinders being horizontal; shew that the two lower cylinders will not separate if \(b[Diagram showing two cylinders at the bottom of a larger cylinder, with a third on top.]

1913 Paper 1 Q105
D: 1500.0 B: 1500.0

Shew that \(3x^3+25x=70\) has a single real root; find its value correct to two places of decimals; and verify your result by using logarithmic tables.

1913 Paper 1 Q105
D: 1500.0 B: 1500.0

Discuss the nature of the contact of two given curves at a common point. Apply your results to shew that if the coordinates of a point on a curve are given functions of a parameter \(t\), (i) the coordinates \((\xi, \eta)\) of the centre of the osculating circle at the point \((x,y)\) are given by the equations \[ \frac{\xi-x}{-\dot{y}} = \frac{\eta-y}{\dot{x}} = \frac{\dot{x}^2+\dot{y}^2}{\dot{x}\ddot{y}-\ddot{x}\dot{y}}, \] (ii) the equation of the diameter of the parabola of closest contact which passes through the point is % The notation here is very condensed. Representing it with determinants. \[ \left| \begin{matrix} \ddot{x} & \dddot{y} \\ \dot{x} & \dot{y} \end{matrix} \right| - \left| \begin{matrix} \dot{x} & \dddot{y} \\ \dot{x} & \dot{y} \end{matrix} \right| (\xi-x, \eta-y) = 3 \left| \begin{matrix} \dot{x} & \ddot{y} \\ \dot{x} & \dot{y} \end{matrix} \right| \left| \begin{matrix} \dot{x} & \xi-x \\ \dot{y} & \eta-y \end{matrix} \right|, \] where dots indicate differentiation with regard to the parameter.

1913 Paper 1 Q106
D: 1500.0 B: 1500.0

A bullet is fired through three screens placed at equal intervals of \(a\) feet, and the times of passing the screens are recorded by chronograph to be \(t_1, t_2\) and \(t_3\) respectively after the bullet left the gun. Assuming the retardation to be uniform, shew that its value is \[ \frac{(t_3-2t_2+t_1)2a}{(t_3-t_2)(t_2-t_1)(t_3-t_1)}. \]

1913 Paper 1 Q106
D: 1500.0 B: 1500.0

Sketch the curve \(y = \dfrac{x}{(x+1)(x+2)}\) and determine the maximum and minimum values of its ordinate.

1913 Paper 1 Q106
D: 1500.0 B: 1500.0

State the principle of Virtual Work. Prove it (1) for forces acting at a point; (2) for forces acting on a system of connected particles; and extend it to the case of a rigid body under the action of given forces, pointing out the assumptions and limitations of the principle. Two corners of a regular pentagon of light freely jointed rods are connected by a string in tension, and equilibrium is maintained by another string also in tension, connecting one corner to the middle point of the opposite side, the strings being perpendicular to each other. Shew that their tensions are in the ratio \(2\sin\frac{1}{5}\pi : 1\).

1913 Paper 1 Q107
D: 1500.0 B: 1500.0

An electric train starts with an acceleration of 3 ft. per sec. per sec., but the acceleration diminishes uniformly with the time until it becomes zero 30 seconds from the start. Find the velocity attained and the distance travelled in the first 20 seconds of the run.

1913 Paper 1 Q107
D: 1500.0 B: 1500.0

Shew that in any triangle \[ 4Rr(a\cos B + b\cos C + c\cos A) = abc - (a-b)(b-c)(c-a). \]