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2010 Paper 2 Q1
D: 1600.0 B: 1516.0
differentiation osculating circle radius of curvature tangent second derivative trigonometric curve

Let \(P\) be a given point on a given curve \(C\). The \textit{osculating circle} to \(C\) at \(P\) is defined to be the circle that satisfies the following two conditions at \(P\): it touches \(C\); and the rate of change of its gradient is equal to the rate of change of the gradient of \(C\). Find the centre and radius of the osculating circle to the curve \(y=1-x+\tan x\) at the point on the curve with \(x\)-coordinate \(\frac14 \pi\).

Solution: The condition is that we match the first and second derivative (as well as passing through the point in question, which is \((\frac{\pi}{4}, 2 - \frac{\pi}{4})\) The gradient is \(y' = -1 + \sec^2 x\), so the value is \(1\). The second derivative is \(y'' = 2 \sec^2 x \tan x\), which is \(4\) If we have a circle, radius \(r\), so \((x-a)^2 + (y-b)^2 = r^2\) then \(2(x-a) + 2(y-b) \frac{\d y}{\d x} = 0\) and \(2 + 2 \left ( \frac{\d y}{\d x} \right)^2 + 2(y-b) \frac{\d^2y}{\d x^2} = 0\). Therefore we must have \(1+1+(2-\frac{\pi}{4}-b)4 = 0 \Rightarrow b =\frac52-\frac{\pi}{4}\) We know that the centre lies on the line \(y = 2-x\), so we must have \(a = \frac{\pi}{4}-\frac12\) and so the centre is \(( \frac{\pi}{4} - \frac12,\frac52 - \frac{\pi}{4})\) and the radius is \(\sqrt{\frac14 + \frac14} = \frac{\sqrt{2}}{2}\)

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2007 Paper 3 Q4
D: 1700.0 B: 1484.0
parametric equations curve sketching tangent-normal radius of curvature tractrix differentiation trigonometric geometric proof

A curve is given parametrically by \begin{align*} x&= a\big( \cos t +\ln \tan \tfrac12 t\big)\,,\\ y&= a\sin t\,, \end{align*} where \(0 < t < \frac12 \pi\) and \(a\) is a positive constant. Show that \(\ds \frac{\d y}{\d x} = \tan t\) and sketch the curve. Let \(P\) be the point with parameter \(t\) and let \(Q\) be the point where the tangent to the curve at \(P\) meets the \(x\)-axis. Show that \(PQ=a\). The {\sl radius of curvature}, \(\rho\), at \(P\) is defined by \[ \rho= \frac {\big(\dot x ^2+\dot y^2\big)^{\frac32}} {\vert \dot x \ddot y - \dot y \ddot x\vert \ \ } \,, \] where the dots denote differentiation with respect to \(t\). Show that \(\rho =a\cot t\). The point \(C\) lies on the normal to the curve at \(P\), a distance \(\rho\) from \(P\) and above the curve. Show that \(CQ\) is parallel to the \(y\)-axis.

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1988 Paper 3 Q14
D: 1700.0 B: 1484.0
circular motion bead on wire normal reaction radius of curvature parabola energy conservation projectile motion constraint motion

A small heavy bead can slide smoothly in a vertical plane on a fixed wire with equation \[ y=x-\frac{x^{2}}{4a}, \] where the \(y\)-axis points vertically upwards and \(a\) is a positive constant. The bead is projected from the origin with initial speed \(V\) along the wire.

  1. Show that for a suitable value of \(V\), to be determined, a motion is possible throughout which the bead exerts no pressure on the wire.
  2. Show that \(\theta,\) the angle between the particle's velocity at time \(t\) and the \(x\)-axis, satisfies \[ \frac{4a^{2}\dot{\theta}^{2}}{\cos^{6}\theta}+2ga(1-\tan^{2}\theta)=V^{2}. \]

Solution:

  1. The condition that the bead exerts no pressure on the wire is equivalent to the condition that the wire exerts no force on the bead. (Newton's Third Law). This is equivalent to the bead being projected under gravity. Notice that the initial projection is at \(45^{\circ}\) since \(\frac{dy}{dx}|_{x=0} = 1\). The position of the particle (under gravity) at time \(t\) is \(x = \frac{1}{\sqrt{2}}Vt\) and \(y = \frac{1}{\sqrt{2}}Vt - \frac12 gt^2 = x - \frac{1}{2}g \frac{2x^2}{V^2} = x - \frac{g}{V^2}x^2\). Therefore they follow the same trajectory if \(\frac{g}{V^2} = \frac{1}{4a} \Leftrightarrow V = 2\sqrt{ag}\)
  2. First note that the wire does no work as it is perpendicular to the velocity, so it is fine to use conservation of momentum. If we take our \(0\) GPE level to be be \(x = 0\), then we notice the initial energy is \(\frac12mV^2\). Secondly, notice that \(\tan \theta = \frac{\d y}{\d x} = 1- \frac{x}{2a} \Rightarrow x = 2a - 2a \tan \theta\) \begin{align*} y &= 2a(1-\tan \theta) - \frac{4a^2(1-\tan \theta)^2}{4a}\\ &= (1-\tan \theta)(2a-a(1-\tan \theta)) \\ &= a(1-\tan \theta)(1+\tan \theta) \\ &= a(1-\tan^2 \theta) \end{align*} GPE \(mga(1-\tan^2 \theta)\). To calculate the kinetic energy, notice that \(\dot{x} = v \cos \theta \dot{\theta}\) and \(\dot{x} = -2a\sec^2 \theta\dot{\theta} \Rightarrow v = -\frac{2a\dot{\theta} }{\cos^{3} \theta}\). Therefore, energy at time \(t\) is: \begin{align*} && \frac12 m V^2 &= \frac12 m \l - \frac{2a\dot{\theta}}{\cos^3 \theta} \r^2 + mga(1-\tan^2 \theta) \\ \Rightarrow && V^2 &= \frac{4a^2\dot{\theta}^2}{\cos^6 \theta} + 2ag(1-\tan^2 \theta) \end{align*}

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