1 problem found
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 \(\sqrt{\frac14 + \frac14} = \frac{\sqrt{2}}{2}\)