Sliding On and Off of a Logarithmic Surface

There is no motion in the $z$ -direction. the position vector of the particle can be given by $\mathbf{r} \; = \; \binom{x}{\ln x}$ and hence $\dot{\mathbf{r}} \; = \; \binom{1}{x^{-1}}\dot{x} \hspace{2cm} \ddot{\mathrm{r}} \; = \; \binom{1}{x^{-1}}\ddot{x} + \binom{0}{-x^{-2}}\dot{x}^2$ The particle is subject to gravity and (while in contact with the surface) a normal reaction $\mathbf{N}$ perpendicular to $\dot{\mathbf{r}}$ , and hence parallel to $\binom{x^{-1}}{-1}$ . If we write $\mathbf{N} = mK\binom{x^{-1}}{-1}$ , the equation of motion of the particle is $m\ddot{\mathbf{r}} \; = \; mg\binom{0}{-1} + mK\binom{x^{-1}}{-1}$ Thus $\begin{aligned} \ddot{\mathbf{r}} \cdot \binom{1}{x^{-1}} & = \; -gx^{-1} \\ (1 + x^{-2})\ddot{x} - x^{-3}\dot{x}^2 + gx^{-1} & = \; 0 \\ \frac{d}{dx}\left[\tfrac12(1 + x^{-2})\dot{x}^2 + g\ln x\right] & = \; 0 \end{aligned}$ Given the initial conditions of this problem, we deduce that $\begin{aligned} \tfrac12(1 + x^{-2})\dot{x}^2 + g\ln x & = \; 50 \\ \dot{x}^2 & = \; \frac{20(5 - \ln x)}{1 + x^{-2}} \end{aligned}$ Moreover, $\begin{aligned} \ddot{r} \cdot \binom{x^{-1}}{-1} & = \; g + (1 + x^{-2})X \\ (1 + x^{-2})X & = \; x^{-2}\dot{x}^2 - g \; = \; \frac{10}{x^2 + 1}\big(9 - 2\ln x - x^2\big) \end{aligned}$ and so the particle first leaves the surface when $9 - 2\ln x - x^2 = 0$ , which happens when $x = \boxed{2.65468}$ .