Bead on Quartic Wire

Coordinates of the particle at a general time:

$(x_1,y_1) = (x,x^4) \implies (\dot{x}_1,\dot{y}_1) = (\dot{x},4x^3\dot{x})$

The kinetic and gravitational potential energies are:

$\mathcal{T} =\frac{m}{2}\left(\dot{x}_1^2 + \dot{y}_1^2\right)= \frac{\dot{x}^2(1+16x^6)}{2} \ ; \ \mathcal{V} = 10x^4$

Applying conservation of energy:

$\mathcal{T}+\mathcal{V}= \mathcal{T}_{initial}+\mathcal{V}_{initial}$ $\frac{\dot{x}^2(1+16x^6)}{2} + 10x^4 = \frac{1}{1000}$

Rearranging gives:

$\dot{x} = \sqrt{\frac{1-10000x^4}{500(1+16x^6)}}$

Separating variables and integrating from $x=-0.1$ to $x=0$ gives a quarter of the time period. Therefore:

$T = 4 \int_{-0.1}^{0} \sqrt{\frac{{500(1+16x^6)}}{1-10000x^4}} \ dx \approx 11.726$