Brachistochrone problem

The time to travel from point $p_1$ to $p_2$ is given by this integral $$t_{12}=\int_{p_1}^{p_2}\frac{ds}{v}.$$ With $ds=\sqrt{dx^2+dy^2}=\sqrt{1+y'^2}\,dx$ and $v=\sqrt{2g\,y}$, we obtain $$t_{12}=\int_{p_1}^{p_2}\underbrace{\sqrt{\frac{1+y'^2}{2g\,y(x)}}}_{f(y,y')}\,dx,$$ and because the function $f$ is Independent of $x$, we can use the Beltrami identity $$f-y'\frac{\partial f}{\partial y'}=c\tag 1$$ to obtain the solution.

Instead, I want to use a different approach. Instead taking $ds=\sqrt{\dot x^2+\dot y^2}\,dt$ and $v=\sqrt{2g\,y}$ gives $$ t_{12}\mapsto\int_{p_1}^{p_2}\frac{\sqrt{\dot x^2+\dot y^2}\,dt}{\sqrt{2g\,y}},$$ which implies $$f(y,\dot{y},\dot{x})=\sqrt{\frac{\dot{x}^2+\dot{y}^2}{2g\,y}}\equiv\mathcal L(y,\dot{y},\dot{x}).$$ So the first integral is $$\left(\frac{\partial \mathcal L}{\partial{\dot{x}}}\right)^2=c.\tag 2$$ From equation (2), I obtain the equation $$\frac{\dot x^2}{(\dot x^2+\dot y^2)y}=\tilde k^2,$$ implying $$\frac{(\dot x^2+\dot y^2)y}{\dot x^2}=k^2,$$ which reduces to $$(1+y'^2)y(x)=k^{2},$$ to which the solution is $$x(t)=\frac{k^2}{2}[t-\sin(t)],\quad y(t)=\frac{k^2}{2}[1-\cos(t)],$$ which is indeed the solution of brachistochrone problem.

So what is wrong with this approach?

Brachistochrone problem

The time to travel from point $p_1$ to $p_2$ is given by this integral $$t_{12}=\int_{p_1}^{p_2}\frac{ds}{v}.$$ With $ds=\sqrt{dx^2+dy^2}=\sqrt{1+y'^2}\,dx$ and $v=\sqrt{2g\,y}$, we obtain $$t_{12}=\int_{p_1}^{p_2}\underbrace{\sqrt{\frac{1+y'^2}{2g\,y(x)}}}_{f(y,y')}\,dx,$$ and because the function $f$ is Independent of $x$, we can use the Beltrami identity $$f-y'\frac{\partial f}{\partial y'}=c\tag 1$$ to obtain the solution.

Instead, I want to use a different approach. Taking $ds=\sqrt{\dot x^2+\dot y^2}\,dt$ and $v=\sqrt{2g\,y}$ gives $$ t_{12}\mapsto\int_{p_1}^{p_2}\frac{\sqrt{\dot x^2+\dot y^2}\,dt}{\sqrt{2g\,y}},$$ which implies $$f(y,\dot{y},\dot{x})=\sqrt{\frac{\dot{x}^2+\dot{y}^2}{2g\,y}}\equiv\mathcal L(y,\dot{y},\dot{x}).$$ So the first integral is $$\left(\frac{\partial \mathcal L}{\partial{\dot{x}}}\right)^2=c.\tag 2$$ From equation (2), I obtain the equation $$\frac{\dot x^2}{(\dot x^2+\dot y^2)y}=\tilde k^2,$$ implying $$\frac{(\dot x^2+\dot y^2)y}{\dot x^2}=k^2,$$ which reduces to $$(1+y'^2)y(x)=k^{2},$$ to which the solution is $$x(t)=\frac{k^2}{2}[t-\sin(t)],\quad y(t)=\frac{k^2}{2}[1-\cos(t)],$$ which is indeed the solution of brachistochrone problem.

So what is wrong with this approach?

Brachistochrone problem

The time to travel from point $~p_1~$ to $~p_2~$ is give by this Integral

$$t_{12}=\int_{p_1}^{p_2}\,\frac{ds}{v}$$

with $~ds=\sqrt{dx^2+dy^2}=\sqrt{1+y'^2}\,dx\quad ,v=\sqrt{2\,g\,y}~$ we obtain

$$t_{12}=\int_{p_1}^{p_2}\,\underbrace{\sqrt{\frac{1+y'^2}{2\,g\,y(x)}}}_{f(y,y')}\,dx$$

and because the function $~f~$ is Independent of $x$, we can use the Beltrami identity

$$f-y‘\,\frac{\partial f}{\partial y'}=c\tag 1$$

to obtain the solution.

Instead I want to use the following approach

with $~ds=\sqrt{\dot x^2+\dot y^2}\,dt~,v=\sqrt{2\,g\,y}~$ thus

$$ t_{12}\mapsto\int_{p_1}^{p_2}\,\frac{\sqrt{\dot x^2+\dot y^2}\,dt}{\sqrt{2\,g\,y}}\quad\Rightarrow\\ f(y,\dot y~,\dot x)=\sqrt{\frac{\dot x^2+\dot y^2}{2\,g\,y}}\equiv\mathcal L(y,\dot y~,\dot x)$$

so the first integral is:

$$\left(\frac{\partial \mathcal L}{\partial{\dot{x}}}\right)^2=c\tag 2$$

from equation (2) i obtain this equation

$$\frac{\dot x^2}{(\dot x^2+\dot y^2)\,y}=\tilde k^2\quad\Rightarrow \frac{(\dot x^2+\dot y^2)\,y}{\dot x^2}=k^2$$ and the solution is

$$x(t)=\frac{k^2}{2}\,(t-\sin(t))\quad,y(t)=\frac{k^2}{2}\,(1-\cos(t))$$

this is the solution of Brachistochrone problem.

so what is wrong with this approach?

Edit

with

$$\frac{\dot x^2}{(\dot x^2+\dot y^2)\,y}=k\quad\Rightarrow\quad \frac{1}{(1+y'^2)\,y(x)}=k$$

Brachistochrone problem

The time to travel from point $p_1$ to $p_2$ is given by this integral $$t_{12}=\int_{p_1}^{p_2}\frac{ds}{v}.$$ With $ds=\sqrt{dx^2+dy^2}=\sqrt{1+y'^2}\,dx$ and $v=\sqrt{2g\,y}$, we obtain $$t_{12}=\int_{p_1}^{p_2}\underbrace{\sqrt{\frac{1+y'^2}{2g\,y(x)}}}_{f(y,y')}\,dx,$$ and because the function $f$ is Independent of $x$, we can use the Beltrami identity $$f-y'\frac{\partial f}{\partial y'}=c\tag 1$$ to obtain the solution.

Instead, I want to use a different approach. Instead taking $ds=\sqrt{\dot x^2+\dot y^2}\,dt$ and $v=\sqrt{2g\,y}$ gives $$ t_{12}\mapsto\int_{p_1}^{p_2}\frac{\sqrt{\dot x^2+\dot y^2}\,dt}{\sqrt{2g\,y}},$$ which implies $$f(y,\dot{y},\dot{x})=\sqrt{\frac{\dot{x}^2+\dot{y}^2}{2g\,y}}\equiv\mathcal L(y,\dot{y},\dot{x}).$$ So the first integral is $$\left(\frac{\partial \mathcal L}{\partial{\dot{x}}}\right)^2=c.\tag 2$$ From equation (2), I obtain the equation $$\frac{\dot x^2}{(\dot x^2+\dot y^2)y}=\tilde k^2,$$ implying $$\frac{(\dot x^2+\dot y^2)y}{\dot x^2}=k^2,$$ which reduces to $$(1+y'^2)y(x)=k^{2},$$ to which the solution is $$x(t)=\frac{k^2}{2}[t-\sin(t)],\quad y(t)=\frac{k^2}{2}[1-\cos(t)],$$ which is indeed the solution of brachistochrone problem.

So what is wrong with this approach?

Brachistochrone problem

The time to travel from point $~p_1~$ to $~p_2~$ is give by this Integral

$$t_{12}=\int_{p_1}^{p_2}\,\frac{ds}{v}$$

with $~ds=\sqrt{dx^2+dy^2}=\sqrt{1+y'^2}\,dx\quad ,v=\sqrt{2\,g\,y}~$ we obtain

$$t_{12}=\int_{p_1}^{p_2}\,\underbrace{\sqrt{\frac{1+y'^2}{2\,g\,y(x)}}}_{f(y,y')}\,dx$$

and because the function $~f~$ is Independent of $x$, we can use the Beltrami identity

$$f-y\,\frac{\partial f}{\partial y'}=c\tag 1$$

to obtain the solution.

Instead I want to use the following approach

with $~ds=\sqrt{\dot x^2+\dot y^2}\,dt~,v=\sqrt{2\,g\,y}~$ thus

$$ t_{12}\mapsto\int_{p_1}^{p_2}\,\frac{\sqrt{\dot x^2+\dot y^2}\,dt}{\sqrt{2\,g\,y}}\quad\Rightarrow\\ f(y,\dot y~,\dot x)=\sqrt{\frac{\dot x^2+\dot y^2}{2\,g\,y}}\equiv\mathcal L(y,\dot y~,\dot x)$$

so the first integral is:

$$\left(\frac{\partial \mathcal L}{\partial{\dot{x}}}\right)^2=c\tag 2$$

from equation (2) i obtain this equation

$$\frac{\dot x^2}{(\dot x^2+\dot y^2)\,y}=\tilde k^2\quad\Rightarrow \frac{(\dot x^2+\dot y^2)\,y}{\dot x^2}=k^2$$ and the solution is

$$x(t)=\frac{k^2}{2}\,(t-\sin(t))\quad,y(t)=\frac{k^2}{2}\,(1-\cos(t))$$

this is the solution of Brachistochrone problem.

so what is wrong with this approach?

with

$$\frac{\dot x^2}{(\dot x^2+\dot y^2)\,y}=k\quad\Rightarrow\quad \frac{1}{(1+y'^2)\,y(x)}=k$$

Brachistochrone problem

The time to travel from point $~p_1~$ to $~p_2~$ is give by this Integral

$$t_{12}=\int_{p_1}^{p_2}\,\frac{ds}{v}$$

with $~ds=\sqrt{dx^2+dy^2}=\sqrt{1+y'^2}\,dx\quad ,v=\sqrt{2\,g\,y}~$ we obtain

$$t_{12}=\int_{p_1}^{p_2}\,\underbrace{\sqrt{\frac{1+y'^2}{2\,g\,y(x)}}}_{f(y,y')}\,dx$$

and because the function $~f~$ is Independent of $x$, we can use the Beltrami identity

$$f-y‘\,\frac{\partial f}{\partial y'}=c\tag 1$$

to obtain the solution.

Instead I want to use the following approach

with $~ds=\sqrt{\dot x^2+\dot y^2}\,dt~,v=\sqrt{2\,g\,y}~$ thus

$$ t_{12}\mapsto\int_{p_1}^{p_2}\,\frac{\sqrt{\dot x^2+\dot y^2}\,dt}{\sqrt{2\,g\,y}}\quad\Rightarrow\\ f(y,\dot y~,\dot x)=\sqrt{\frac{\dot x^2+\dot y^2}{2\,g\,y}}\equiv\mathcal L(y,\dot y~,\dot x)$$

so the first integral is:

$$\left(\frac{\partial \mathcal L}{\partial{\dot{x}}}\right)^2=c\tag 2$$

from equation (2) i obtain this equation

$$\frac{\dot x^2}{(\dot x^2+\dot y^2)\,y}=\tilde k^2\quad\Rightarrow \frac{(\dot x^2+\dot y^2)\,y}{\dot x^2}=k^2$$ and the solution is

$$x(t)=\frac{k^2}{2}\,(t-\sin(t))\quad,y(t)=\frac{k^2}{2}\,(1-\cos(t))$$

this is the solution of Brachistochrone problem.

so what is wrong with this approach?

Edit

with

$$\frac{\dot x^2}{(\dot x^2+\dot y^2)\,y}=k\quad\Rightarrow\quad \frac{1}{(1+y'^2)\,y(x)}=k$$