Climbing Gauss Hill

We start with the fact that:

$E_{tot} \ = \ E_0 \ + \ mg \ y(x_0) \ = \ E_0 \ + \ mge^{-x_0^2}$ $\Rightarrow E_{tot} \ = \ \frac{1}{2}mv^2 \ + \ mg \ y(x) \ = \ \frac{1}{2}mv^2 \ + \ mg \ e^{-x^2} \ = \ \ E_0 \ + \ mge^{-x_0^2}$

Now, let us rearrange the equation to solve for $v$ :

$\frac{1}{2}mv^2 \ + \ mg \ e^{-x^2} \ = \ \ E_0 \ + \ mge^{-x_0^2} \ \Rightarrow \ v \ = \ \sqrt{2g(\frac{E_0}{mg}\ + \ e^{-x_0^2} - \ e^{-x^2})}$

Now, we re-define $v$ by expanding out the derivative of the curve-length along which our mass is travelling:

$\Big( \frac{ds}{dx} \Big)^2 \ = \ \Big( \frac{dy}{dx} \Big)^2 \ + \ \Big( \frac{dx}{dx} \Big)^2 \ \Rightarrow \ \frac{ds}{dx} \ = \ \sqrt{\Big( \frac{dy}{dx} \Big)^2 \ + \ 1}$ $v \ = \ \frac{ds}{dt} \ = \ \frac{ds}{dx} \ \frac{dx}{dt} \ = \ \sqrt{\Big( \frac{dy}{dx} \Big)^2 \ + \ 1} \ \frac{dx}{dt} \ = \ \sqrt{2g(\frac{E_0}{mg} \ + \ e^{-x_0^2} - \ e^{-x^2})}$

And then we simply do a bit more re-arranging and substitution:

$\frac{dy}{dx} \ = \ \frac{d}{dx} \ e^{-x^2} \ = \ -2xe^{-x^2}$ $\Rightarrow \ \sqrt{\Big( \frac{dy}{dx} \Big)^2 \ + \ 1} \ \frac{dx}{dt} \ = \ \sqrt{4x^2e^{-2x^2} \ + \ 1} \ \frac{dx}{dt} \ = \ \sqrt{2g(\frac{E_0}{mg} \ + \ e^{-x_0^2} - \ e^{-x^2})}$ $\Rightarrow \ \frac{dx}{dt} \ = \ \sqrt{\frac{2g(\frac{E_0}{mg} \ + \ e^{-x_0^2} - \ e^{-x^2})}{4x^2e^{-2x^2} \ + \ 1}}$

Now, we know that if we evaluate the integral of $\frac{1}{\frac{dx}{dt}} \ = \ \frac{dt}{dx}$ from $x_0$ to $0$ , we will get the total time it takes for our mass to climb to the top of the hill!

$t \ = \ \displaystyle\int_{x_0}^{0} \ \frac{dt}{dx} \ dx \ = \ \displaystyle\int_{x_0}^{0} \ \sqrt{\frac{4x^2e^{-2x^2} \ + \ 1}{2g(\frac{E_0}{mg} \ + \ e^{-x_0^2} - \ e^{-x^2})}} \ dx \ = \ \displaystyle\int_{-2}^{0} \ \sqrt{\frac{4x^2e^{-2x^2} \ + \ 1}{20(1.01 \ + \ e^{-(-2)^2} - \ e^{-x^2})}} \ dx$

This is a fairly nasty integral, but we can evaluate it numerically to get an answer of $t \ \approx \ 0.9213374 \ \text{s}$ !

The initial total energy is:

$E_o + mgy_{i} = E_{i}$

Where $y_{i} = e^{-4}$ . The subscript $i$ denotes 'initial'. This gives:

$E_{i} = mg\left(1.01 + e^{-4}\right)$

Now, the coordinates of the bead at any general position are:

$r_p = \left(x,y\right) = \left(x,e^{-x^2}\right)$

The kinetic energy of the particle at this instant is:

$T = \frac{1}{2}m\left(\dot{x}^2 + \dot{y}^2\right) = \frac{\dot{x}^2}{2}\left(1+4x^2e^{-2x^2}\right)$

The potential energy is:

$V = mge^{-x^2} = 10e^{-x^2}$

Applying conservation of energy:

$E_{i} = T + V = \frac{\dot{x}^2}{2}\left(1+4x^2e^{-2x^2}\right)+10e^{-x^2}$

Solving for $\dot{x}$ gives an expression of the form:

$\dot{x} = \frac{dx}{dt} = \sqrt{\frac{2(E_i - 10e^{-x^2})}{(1+4x^2e^{-2x^2})}}$

Therefore y separating the variables and integrating:

$t_o = \int_{-2}^{0} \left(\dfrac{\sqrt{4x^2\mathrm{e}^{-2x^2}+1}}{\sqrt{20\left(\mathrm{e}^{-4}+\frac{101}{100}\right)-20\mathrm{e}^{-x^2}}}\right)dx$

Numerically integrating gives:

$\boxed{t_o = 0.9213 s}$