Going down the parabolic slide

The Lagrangian of the system is the kinetic energy minus the potential:

$\displaystyle L = \frac{1}{2} \dot{x}^2 +\frac{1}{2} (2x \dot x)^2 - 10x^2$

Solving the Euler-Lagrange equation for $\ddot{x}$ ;

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

it yields:

$\displaystyle \ddot{x} = \frac{-(20x + 4x \dot{x}^2)}{1 + 4x^2}$

I felt the best way to find the time taken for the object to reach point $(0,0)$ from point ( $1, 1)$ when released from $(2, \sqrt{2})$ is to find the velocity at point ( $1, 1)$ by simulating from $(2, \sqrt{2})$ to ( $1, 1)$ , and then simulating (with time = 0 again) with the new initial velocity from ( $1, 1)$ to $(0,0)$ . I used the Explicit Euler numerical method to do this; by numerically integrating the acceleration to find the position and ultimately the time it takes to reach the bottom. Here's the code:

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import math

x = math.sqrt(2) #initialisation
xDot =0


deltaT = 10**-6 #timestep

while x >= 1:
  xDotDot = -1*(20*x + 4*x*xDot**2)/(1+4*x**2) #x acceleration
  xDot += xDotDot*deltaT #xDot numerical extrapolation
  x += xDot*deltaT  #x numerical extrapolation

#with the new initial velocity, we can numerically simulate from (1,1) to (0,0)

time = 0 #initialisation of time

while x >= 0:
  xDotDot = -1*(20*x + 4*x*xDot**2)/(1+4*x**2) #x acceleration
  xDot += xDotDot*deltaT #xDot numerical extrapolation
  x += xDot*deltaT #x numerical extrapolation

  time += deltaT #updating time value

print("time: " +str(time))

#results: time = 0.26637699999979114

my name is still not updated in the solver's list , still , just to show my approach here is my solution