Problems

Filters
Clear Filters

3 problems found

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 \(\sqrt{\frac14 + \frac14} = \frac{\sqrt{2}}{2}\)

View
2003 Paper 3 Q4
D: 1700.0 B: 1516.0
parametric equations tangent recurrence relation trigonometric periodic sequence curve tan identity algebraic manipulation

A curve is defined parametrically by \[ x=t^2 \;, \ \ \ y=t (1 + t^2 ) \;. \] The tangent at the point with parameter \(t\), where \(t\ne0\,\), meets the curve again at the point with parameter \(T\), where \(T\ne t\,\). Show that \[ T = \frac{1 - t^2 }{2t} \mbox { \ \ \ and \ \ \ } 3t^2\ne 1\;. \] Given a point \(P_0\,\) on the curve, with parameter \(t_0\,\), a sequence of points \(P_0 \, , \; P_1 \, , \; P_2 \, , \ldots\) on the curve is constructed such that the tangent at \(P_i\) meets the curve again at \(P_{i+1}\). If \(t_0 = \tan \frac{ 7 } {18}\pi\,\), show that \(P_3 = P_0\) but \(P_1\ne P_0\,\). Find a second value of \(t_0\,\), with \(t_0>0\,\), for which \(P_3 = P_0\) but \(P_1\ne P_0\,\).

View
1990 Paper 1 Q3
D: 1516.0 B: 1484.0
differentiation from first principles curve unit normal parametric gradient polynomial vectors geometric proof

Given a curve described by \(y=\mathrm{f}(x)\), and such that \(y\geqslant0\), a push-off of the curve is a new curve obtained as follows: for each point \((x,\mathrm{f}(x))\) with position vector \(\mathbf{r}\) on the original curve, there is a point with position vector \(\mathbf{s}\) on the new curve such that \(\mathbf{s-r}=\mathrm{p}(x)\mathbf{n},\) where \(\mathrm{p}\) is a given function and \(\mathbf{n}\) is the downward-pointing unit normal to the original curve at \(\mathbf{r}\).

  1. For the curve \(y=x^{k},\) where \(x>0\) and \(k\) is a positive integer, obtain the function \(\mathrm{p}\) for which the push-off is the positive \(x\)-axis, and find the value of \(k\) such that, for all points on the original curve, \(\left|\mathbf{r}\right|=\left|\mathbf{r-s}\right|\).
  2. Suppose that the original curve is \(y=x^{2}\) and \(\mathrm{p}\) is such that the gradient of the curves at the points with position vectors \(\mathbf{r}\) and \(\mathbf{s}\) are equal (for every point on the original curve). By writing \(\mathrm{p}(x)=\mathrm{q}(x)\sqrt{1+4x^{2}},\) where \(\mathrm{q}\) is to be determined, or otherwise, find the form of \(\mathrm{p}\).

Solution:

  1. Suppose we have \(y = x^k\), then the tangent at \((t,t^k)\) has gradient \(\frac{\d y}{\d x} = kx^{k-1} = kt^{k-1}\) and the normal has gradient \(-\frac1k x^{1-k}\). For the push-off to be the positive \(x\)-axis, we need \(p(x)\) to be the length of the line. The line will have the equation: \begin{align*} && \frac{y - t^k}{x - t} &= -\frac1k t^{1-k} \\ y = 0: && x-t &= \frac{kt^k}{t^{1-k}} \\ && x& =t + kt^{2k-1} \end{align*} The distance from \((t,t^k)\) to \((t+kt^{2k-1},0)\) is \(\sqrt{t^{2k} + k^2t^{4k-2}} = t^k \sqrt{1+k^2t^{2k-2}} = y \sqrt{1+k^2 \frac{y^2}{x^2}} = \frac{y}{x} \sqrt{x^2 + k^2y^2}\), ie \(p(x) = \frac{y}{x}\sqrt{x^2+k^2y^2}\) If \(\left|\mathbf{r}\right|=\left|\mathbf{r-s}\right|\), then we need \(\sqrt{x^2+y^2} = \frac{y}{x}\sqrt{x^2+ky^2}\), but clearly this is satisfied when \(k = 1\).
  2. The points are \((t, t^2)\) and the normal has graident \(-\frac{1}{2t}\), the normal vector is \(\frac{1}{\sqrt{1+\frac{1}{4t^2}}}\binom{1}{-\frac1{2t}} = \frac{1}{\sqrt{4t^2+1}} \binom{2t}{-1}\). If we write \(p(x) = q(x)\sqrt{4x^2+1}\) then the the new points are at \(\binom{t+q(t)2t}{t^2-\q(t)}\) an the gradient will be \(\frac{2t-q'(t)}{1+q'(t)2t+2q(t)}\). We need it to be the case that \begin{align*} && 2t &= \frac{2t-q'(t)}{1+2q(t)+2tq'(t)} \\ \Rightarrow && 4tq(t) &= -q'(t)(1+4t^2) \\ \Rightarrow && \frac{q'(t)}{q(t)} &= -\frac{4t}{1+4t^2} \\ \Rightarrow && \ln q(t) &= -\frac12 \ln(1+4t^2)+C \\ \Rightarrow && q(t) &= A(1+4t^2)^{-1/2} \\ \Rightarrow && p(x) &= A \end{align*} So the push-off's are constants.

View