Question

In the brachistochrone problem, show that if the particle is given an initial velocity${\mathbf{v}}_{\mathbf{0}}\mathbf{\ne }\mathbf{0}$, the path of minimum time is still a cycloid.

Short answer

It is proved that if the particle is given an initial velocity ${\mathrm{v}}_0\ne 0$, the path of minimum time is still a cycloid.

Step-by-step solution

Given Information

It is given that initial velocity of a particle is $v_0\ne 0$.

Definition of Calculus

In the same way that geometry is the study of shape and algebra is the study of generalisations of arithmetic operations, calculus, sometimes known as infinitesimal calculus or "the calculus of infinitesimals," is the mathematical study of continuous change.. Differentiation and integration are the two main branches.

Find the velocity

Formula of energy conservation law states,

$\frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}^2-mgy$

Velocity at any $\mathrm{y}$coordinate is given by, $v=\sqrt{2gy+v_0^2}$.

Minimize the time integral

Minimize the integral of time to find the path of minimum time.

$$\begin{gathered} \int dt=\int \frac{ds}{v} \\ =\int \frac{ds}{\sqrt{2gy+v_0^2}} \\ =\int \frac{\sqrt{1+y{\text{'}}^2}}{\sqrt{2gy+v_0^2}}dx \end{gathered}$$

Substitute,

$$\begin{gathered} dx=x\text{'}dy \\ y\text{'}=\frac{1}{x\text{'}} \end{gathered}$$

Solve further,

$$\begin{gathered} \int \frac{\sqrt{1+y{\text{'}}^2}}{\sqrt{2gy+v_0^2}}dx=\int \frac{\sqrt{1+y{\text{'}}^2}}{\sqrt{2gy+v_0^2}}x\text{'}dy \\ =\int \frac{\sqrt{1+x{\text{'}}^{-2}}}{\sqrt{2gy+v_0^2}}x\text{'}dy \\ =\int \frac{\sqrt{x{\text{'}}^2+1}}{\sqrt{2gy+v_0^2}}dy \end{gathered}$$

Euler’s Equation

Let, $F=\int \frac{\sqrt{x{\text{'}}^2+1}}{\sqrt{2gy+v_0^2}}dy$

Write the Euler’s equation and find it’s derivatives.

$$\begin{gathered} \frac{d}{dy}\frac{\partial F}{\partial x\text{'}}-\frac{\partial F}{\partial x}=0 \\ \frac{\partial F}{\partial x\text{'}}=\frac{x\text{'}}{\sqrt{2gy+v_0^2}\sqrt{x{\text{'}}^2+1}} \\ \frac{\partial F}{\partial x}=0 \end{gathered}$$

Obtain the first Euler integral.

$$\begin{gathered} \frac{d}{dy}\left[\frac{x\text{'}}{\sqrt{2gy+v_0^2}\sqrt{x{\text{'}}^2+1}}\right]=0 \\ \frac{x\text{'}}{\sqrt{2gy+v_0^2}\sqrt{x{\text{'}}^2+1}}=C \\ x{\text{'}}^2=\frac{C^2\left(2gy-v_0^2\right)}{1-C^2\left(2gy-v_0^2\right)} \\ x\text{'}=\sqrt{\frac{C^2\left(2gy-v_0^2\right)}{1-C^2\left(2gy-v_0^2\right)}} \end{gathered}$$

Check conditions of initial velocity

Substitute $u=2gy-v_0^2$ in the above differential equation.

$dx=\sqrt{\frac{C^{2}u}{1-C^{2}u}}\frac{1}{2g}du$

As the above differential equation has no initial velocity so it won’t affect the result, so it is a cycloid.

Therefore, it is proved that if the particle is given an initial velocity $v_0\ne 0$, the path of minimum time is still a cycloid.