1 problem found
The normal to the curve \(y=\mathrm{f}(x)\) at the point \(P\) with coordinates \((x,\mathrm{f}(x))\) cuts the \(y\)-axis at the point \(Q\). Derive an expression in terms of \(x\), \(\mathrm{f}(x)\) and \(\mathrm{f}'(x)\) for the \(y\)-coordinate of \(Q\). If, for all \(x\), \(PQ=\sqrt{\mathrm{e}^{x^{2}}+x^{2}}\), find a differential equation satisfied by \(\mathrm{f}(x)\). If the curve also has a minimum point \((0,-2)\), find its equation.
Solution: The normal to the curve \(y = f(x)\) has gradient \(-\frac{1}{f'(x)}\) and so has equation: \begin{align*} && \frac{Y - f(x)}{X - x} &= -\frac{1}{f'(x)} \\ \Rightarrow && Y &= -\frac{1}{f'(x)}X + \frac{x}{f'(x)}+f(x) \end{align*} Hence the \(Q\) is \(\displaystyle \left (0, f(x) + \frac{x}{f'(x)} \right)\). \begin{align*} && |PQ|^2 &= x^2 + \frac{x^2}{(f'(x))^2} \\ \Rightarrow && x^2 + e^{x^2} &= x^2 + \frac{x^2}{(f'(x))^2} \\ \Rightarrow && (f'(x))^2 &=x^2 e^{-x^2} \end{align*} Therefore \(f'(x) = \pm x e^{-x^2/2}\). If \(f(x)\) has a minimum at \((0,-2)\) then \(f''(0) > 0\), and \(f''(x) = \pm (e^{-x^2/2} - x^2e^{-x^2/2}) = \pm e^{-x^2/2}(1-x^2)\) so we should take the positive branch of the solution, ie \(f'(x) = xe^{-x^2/2}\). Therefore \(f(x) = - e^{-x^2/2}+C\). Since \(f(0) = -2\) we must have \(-2 = -1 + C\), ie \(C = -1\). Therefore \(f(x) = -1 - e^{-x^2/2}\)