Question

Please make sure to explain, make sure answer is correct and neat. Thank you!

3. Show that the time taken by a particle moving along a curve y = y(x) with velocity, ds dt center on the x-axis. Hint: t = Jo x x from the point (0,0) to the point (1,1) is minimum if the curve is a circle having its 1 ds

Answer 1

Let the particle moves from 'O' to 'P' along the path y=y(x)

We consider a infinitesimal segment ds on the path y(x).

Futher , we assume the particle's velocity is constant while travelling the path ds.

From geometry we find $ds= \sqrt{dx^2+dy^2}= [ \sqrt{1+(\frac{dy}{dx})^2}]dx=dx \sqrt{1+y'^2}$

Where $y'= \frac{dy}{dx}$

Given the particle's velocity $v=\frac{ds}{dt}=x$

So the time taken by the particle to travel the distance ds is given by $dt=\frac{ds}{v}= \frac{\sqrt{[1+y'^2}]dx}{x}$

So the total time taken by the particle to travle from 'O' to 'P' along the path y=y(x) is given by

$t= \int_{O}^{P}dt=\int_{O}^{P} \frac{ \sqrt{[1+y'^2}]dx}{x}$

Or

$t= \int_{O}^{P}fdx$ [ where $f=\frac{ \sqrt{1+y'^2}}{x}$ ]

Now from Least Action principle, we know that 't' will be minimum if it satisfies  Euler-Lagrange differential equation i.e

$\frac{\partial f}{\partial y}-\frac{d}{dx}(\frac{\partial f}{\partial y'})=0$...........................(i0

Now we find

$\frac{\partial f}{\partial y}=0$

So equation (i) reduces to

$\frac{d}{dx}(\frac{\partial f}{\partial y'})=0$

Therefore we can write

$\frac{\partial f}{\partial y'}= K$, [ Where K is a constant]

Differentiating f with respect to y' , we get

$\frac{\partial f}{\partial y'}=\frac{y'}{x \sqrt{1+y'^2}}$

Hence we can write

$\frac{y'}{x \sqrt{1+y'^2}}=K$

Or

$\frac{y'^2}{x^2 (1+y'^2)}=K^2$

Or

Or

Or

$y'^2 =\frac{K^2x^2}{(1-K^2x^2)}$

Or

$y' =\frac{Kx}{\sqrt{(1-K^2x^2)}}$

Or

$\frac{dy}{dx} =\frac{Kx}{\sqrt{(1-K^2x^2)}}$

or

$dy =\frac{Kxdx}{\sqrt{(1-K^2x^2)}}$

Integrating both sides we get

$\int dy =\int \frac{Kxdx}{\sqrt{(1-K^2x^2)}}$

Let $1-K^2x^2=v^2\Rightarrow -K^2xdx=vdv$

So we get

$y=-\frac{1}{K}\int \frac{vdv}{v}= -\frac{v}{K}+L$ [ 'L' is a constant ]

Or

$y=-\frac{\sqrt{1-K^2x^2}}{K}+L$

Squaring both sides we get

$K^2(y-L)^2=1-K^2x^2$

Or

$K^2[(y-L)^2+x^2]=1$.................................

Or

$[(y-L)^2+x^2]=P$.................[ where $P=1/K^2$

Now From the given points we find when x=0 , y=0

$L=P$

And when x=1, y=1

$L=2$, So $K=2$

So the equation of the trajectory is

$x^2+(y-2)^2=2$

So the time taken by the particle is minimum if the curve is a circle and the center lies on y axis