Question on analytic mechanics

[hagopbul]

TL;DR

There this example in analytical mechanics which I am not able to understand

Hello :

There is this example ,on analytical mechanics , which I am not able to understand why he solve it in that way

We have a bead moving on a parabola that it's equation is y = b*x^2

Find the equation of motion
Constrain equations :
y=b*x^2
z=0
The bead have only one coordinate call it x ,x(dot)=d(x)/dt, y(dot)=d(y)/dt , z(dot) = d(z)/dt
x(dot)^2 = x(dot)*x(dot)
Kinetic energy:
T = 1/2*m*(x(dot)^2+y(dot)^2+z(dot)^2)
=1/2*mx(dot)^2*(1+4*b^2*x^2)
Here is my question how he reached this above line
Why y(dot)^2 = 4*b^2*x^2*x(dot)^2
Shouldn't y(dot)^2 = 4*b^2*x^2

I am reviewing this class and just not able to find why it is like that

Regards
H

 

[Orodruin]

Because the dot denotes derivative wrt time and not wrt x. It is just the chain rule of derivatives: $dy/dt = (dy/dx)(dx/dt)$

 

[Gavran]

The kinetic energy is $$ \frac12m|\vec v|^2 $$ where $\vec v$ is a vector quantity which can be expressed in Cartesian coordinate system as $v_x\hat i+v_y\hat j+v_z\hat k$.
$|\vec v|^2$ is $v_x^2+v_y^2+v_z^2$ and the kinetic energy can be expressed as $$ \frac12m(v_x^2+v_y^2+v_z^2) $$ which is the three dimensional generalization of the kinetic energy. $v_x$, $v_y$ and $v_z$ are one-dimensional velocities along the $x$, $y$ and $z$ directions and they can be expressed as $dx/dt=\dot x$, $dy/dt=\dot y$ and $dz/dt=\dot z$, respectively.
$dy/dx=2bx$ is not a velocity and $$ \frac12m(\frac{dy}{dx})^2=\frac12m4b^2x^2 $$ can not be kinetic energy.

 

[Albertus Magnus]

In the case of the bead constrained on the parabola, the only forces present is the force of constraint. The motion is two dimensional, however, the constraint $y-bx^2=0$ reduces the configuration space to one dimension, thus we should expect one equation of motion. This problem can be handled with the technique of Lagrange multipliers or one can simply eliminate one degree of freedom by incorporating the constraint explicitly.
The kinetic energy is given by:$$T={m\over 2}({\dot x}^2+{\dot y}^2)={m\over 2}{\dot x}^2(1+4b^2x^2),$$ where the constraint has been used to eliminate $y$. Since there is no potential or other external forces present the Lagrangian is such that $L=T$. From the Euler-Lagrange equations we have that:
$$m\ddot x=\dot p={\partial L\over\partial x}=4mb^2x{\dot x}^2.$$ Finding the equation of motion is an easy step form this point.