Johann Bernoulli posed the following problem in 1696:

If you wanted a ball to roll down a slide

^{1}in theshortestamount of time, what shape should the slide be?

Seeing words like **shortest**, **longest**, **fastest**, etc. in mechanics typically means we need to identify a functional^{2} and **extremize** it. This process is an essential component of the **calculus of variations**, and in the most basic sense, it is similar to finding the minimum/maximum of a function of a single variable in ordinary calculus.

The **brachistochrone** is the problem of identifying the path with the shortest time of descent. The time to travel from point to point is the distance traveled over the velocity. An increment of distance is given by the Pythagorean theorem, such that , while the velocity can be determined by a **conservation of energy**. The sum of kinetic and potential energy for a ball^{3} is where is its mass, and is the acceleration due to gravity. The velocity of the ball is then . Therefore, the functional we wish to extremize is therefore:

To minimize our functional, this term must satisfy the Euler-Lagrange equations

.**Satisfying the Euler-Lagrange equations (Expand)**

Things are a bit easier than they appear at first glance. Since is not a function of , the Euler-Lagrange equations in reduce to the Beltrami identity

.Now, let’s evaluate the partial derivative

Inserting this into the Beltrami identity results in

.This equation can be simplified by multiplying the first term in the parenthesis by one, in the form of , and factoring. Since the result is equal to a constant, we can make the math easier going forward by taking the reciprocal of this result, squaring it, and calling that a new constant:

.By a separation of variables, beginning with , we can now isolate .

Which results in the following ordinary differential equation:

.**Integration via Trigonometric Substitution (Expand)**

Since we have successfully separated the variables, the next step is to integrate both sides. However, integrals of roots are often a headache, and in this particular case, a standard change of variables doesn’t help. What we will use is a trigonometric substitution.

Since the square of the numerator added to the square of the denominator is equal to our unknown constant, we can construct a right triangle with as the hypotenuse (Fig. 1).

Now, we can write

.We can solve for by making use of some trigonometric identities, such that

**Relevant trigonometric identities (Expand)**

.

Since we are looking for the , we first need . Differentiating this with respect to yields

.With this step, we’re almost there, we can now write the differential equation as

.Therefore, the differential becomes

,which can be readily integrated to give

.Since must vanish at , the constant of integration must be . Finally, if we substitute , and let , we arrive the parametric solution.

Integration of both sides does not lead to an equation that can be expressed as a function . However, with a substitution for the angle and the unknown constant, the result can be represented by parametric equations.

The solution to the

brachistochroneis theparametric equations of a cycloid.

The cycloid is the curve traced by a point on a circle rolling in a straight line.

Here is the final result, shown with other curves for comparison: