Bead on Exponential Wire

The coordinates of the particle at any general instant are $(x_1,y_1)=(x, \mathrm{e}^x)$ . The x and y velocity components respectively at that instant are $(\dot{x}_1,\dot{y}_1)=(\dot{x},\mathrm{e}^x \dot{x})$ . Therefore, the kinetic energy of the bead is:

$\mathcal{T} = \frac{m}{2} \left(\dot{x}_1^2 + \dot{y}_1^2\right)=\frac{(1+\mathrm{e}^{2x})\dot{x}^2}{2}$

The potential energy at the same instant is:

$\mathcal{V} = mgy_1 = 10\mathrm{e}^x$

Applying conservation of energy:

$\mathcal{T} +\mathcal{V} =\mathcal{T}_{initial} +\mathcal{V}_{initial} \implies \frac{(1+\mathrm{e}^{2x})\dot{x}^2}{2} + 10\mathrm{e}^x = 10\mathrm{e}^2$

Rearranging:

$\dot{x}^2 = \frac{20\left(\mathrm{e}^2 - \mathrm{e}^x\right)}{1 + \mathrm{e}^{2x}} \implies \dot{x} = -\sqrt{\frac{20\left(\mathrm{e}^2 - \mathrm{e}^x\right)}{1 + \mathrm{e}^{2x}}}$

The negative sign is introduced because as time increases $x$ decreases. Finally, separating the variables and integrating:

$T = \int_{0}^{2} \sqrt{\frac{1 + \mathrm{e}^{2x}}{20\left(\mathrm{e}^2 - \mathrm{e}^x\right)}} \ dx \approx 1.1748 \ \mathrm{seconds}$

I used Lagrangian mechanics.

Kinetic energy:

$T = \frac{1}{2} m ( \dot{x}^2 + \dot{y}^2) = \frac{1}{2} m \dot{x}^2 ( 1 + e^{2x} )$

Potential energy:

$V = m g y = m g e^x$

Lagrangian:

$L = T - V = \frac{1}{2} m \dot{x}^2 ( 1 + e^{2x} ) - m g e^x$

Equation of motion:

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x}}} = \frac{\partial{L}}{\partial{x}}$

Evaluating results in the following expression for the second time derivative:

$\ddot{x} = \frac{- \dot{x}^2 e^{2x} - g e^x}{1 + e^{2x}}$

And then numerical integration takes care of the rest. I have shown the explicit Euler technique below. The "k" subscript denotes the present processing interval, and the "k-1" subscript denotes the previous processing interval.

$x_k = x_{k-1} + \dot{x}_{k-1} \Delta t \\ \dot{x}_k = \dot{x}_{k-1} + \ddot{x}_{k-1} \Delta t \\ \ddot{x}_k = \frac{- \dot{x}_k^2 e^{2x_k} - g e^{x_k}}{1 + e^{2x_k}}$