Question

A classical problem in the calculus of variations is to find the shape of a curve such that a bead, under the influence of gravity, will slide from point $\mathit{A}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$to point $\mathit{B}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{y}}_{\mathbf{1}}\mathbf{)}$in the least time. See Figure 3.R.3. It can be shown that a nonlinear differential for the shape $\mathit{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$of the path is $\mathit{y}\mathbf{[}\mathbf{1}\mathbf{+}{\left(y\text{'}\right)}^{\mathbf{2}}\mathbf{]}\mathbf{=}\mathit{k}$, where $\mathit{k}$is a constant. First solve for k $\mathit{d}\mathit{x}$in terms of $\mathit{y}$and $\mathit{d}\mathit{y}$, and then use the substitution $\mathit{y}\mathbf{=}\mathit{k}\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }$ to obtain a parametric form of the solution. The curve turns out to be a cycloid.

FIGURE3.R.3 Sliding bead in Problem 12

Short answer

The solution is$y=k{\sin }^2\theta ,x=k\left(\theta -\frac{\sin 2\theta }{2}\right)$

Step-by-step solution

 Definition of curve

A curve is an abstract term used to describe the path of a continuously moving point. Such a path is usually generated by an equation.

 Evaluate the given equation

Rearrange the given equation $y\left[1+{\left(y\text{'}\right)}^2\right]=k$as follows:

$$\begin{gathered} y\left[1+{\left(y\text{'}\right)}^2\right]=k \\ y\left[1+{\left(\frac{dy}{dx}\right)}^2\right]=k \\ y+y{\left(\frac{dy}{dx}\right)}^2=k \\ y{\left(\frac{dy}{dx}\right)}^2=k-y \end{gathered}$$

Simplify further as follows:

$$\begin{gathered} {\left(\frac{dy}{dx}\right)}^2=\frac{k-y}{y} \\ \frac{dy}{dx}=\sqrt{\frac{k-y}{y}} \\ \sqrt{\frac{y}{k-y}}dy=dx \end{gathered}$$

Substitute dy and y into the equation

Since $y=k{\sin }^2\theta$, then it follows $dy=2k\sin \theta \cos \theta d\theta$.Substitute $k{\sin }^2\theta$for $y$and $2k\sin \theta \cos \theta d\theta$for $dy$into $\sqrt{\frac{y}{k-y}}dy=dx$and simplify as follows:

$$\begin{gathered} dx=\sqrt{\frac{k{\sin }^2\theta }{k-k{\sin }^2\theta }}2k\sin \theta \cos \theta d\theta \\ dx=\sqrt{\frac{k{\sin }^2\theta }{k{\cos }^2\theta }}2k\sin \theta \cos \theta d\theta \\ dx=\frac{\sin \theta }{\cos \theta }2k\sin \theta \cos \theta d\theta \\ dx=\sin \theta 2k\cos \theta d\theta \end{gathered}$$

Simplify further as follows:

$$\begin{gathered} dx=2k{\sin }^2\theta d\theta \\ dx=2k\left(\frac{1}{2}-\frac{1}{2}\cos 2\theta \right)d\theta \\ dx=k\left(1-\cos 2\theta \right)d\theta \end{gathered}$$

Integrate to obtain value of x

Now integrate the resulting equation as:

$$\begin{gathered} \int dx=\int k\left(1-\cos 2\theta \right)d\theta \\ x=\left(\theta -\frac{\sin 2\theta }{2}\right)+C \end{gathered}$$

Since $x=0$for $\theta =0$. Substitute 0 for $\theta$and 0 for $x$it to obtain the value of $c$

$$\begin{gathered} 0=k\left(0-\frac{\sin 2.0}{2}\right)+C \\ C=0 \end{gathered}$$

Therefore the parametric form of the solution is, $y=k{\sin }^2\theta ,x=k\left(\theta -\frac{\sin 2\theta }{2}\right)$.