Through the story in the previous section, we have gained a basic understanding of the Brachistochrone problem. It requires us to find an unknown function that satisfies a certain extremum condition. Since the function itself is unknown, this type of problem is referred to as the “Calculus of Variations” (or “Pan-analysis”). It was also mentioned that Bernoulli ingeniously derived the answer using Fermat’s Principle. Now, let us revisit history, trace Bernoulli’s solution, and explore further applications.
For computational convenience, we invert the Brachistochrone and set the starting point at the origin. During the descent, gravitational potential energy is converted into kinetic energy; therefore, at point (x, y), we have \frac{1}{2} mv^2 = mgy \Rightarrow v = \sqrt{2gy}. Since we are purely investigating the shape of the curve, we set g = 0.5, such that v = \sqrt{y}. The distance traveled at point (x, y) is ds = \sqrt{dy^2 + dx^2} = \sqrt{\dot{y}^2 + 1} dx, so the time is dt = \frac{ds}{v} = \frac{\sqrt{\dot{y}^2 + 1} dx}{\sqrt{y}}. Thus, the Brachistochrone problem is to find the function that minimizes: t = \int_0^{x_2} \frac{\sqrt{\dot{y}^2 + 1} dx}{\sqrt{y}}
Following Bernoulli’s reasoning, we treat the Brachistochrone as a light path and divide it horizontally into infinitely small segments of equal width. Each segment must satisfy the law of refraction (Snell’s Law). As shown in the figure, we have: \frac{v}{\cos\theta} = \frac{v'}{\cos\theta'} = \dots = \text{Const}
Where Const represents a constant. Since \cos\theta = \frac{1}{\sqrt{1 + \tan^2 \theta}} = \frac{1}{\sqrt{1 + \dot{y}^2}}
It follows that: v^2(1 + \dot{y}^2) = \text{Const} \tag{1}
Please note that we have not yet used the condition v = \sqrt{y}, which means Equation (1) is not limited to this specific problem. So, what types of problems can Equation (1) be applied to? The answer is: problems where the expression for velocity v explicitly contains only y. In other words, when the expression for v explicitly contains only y, the function that minimizes t = \int_{x_1}^{x_2} \frac{\sqrt{\dot{y}^2 + 1} dx}{v} is calculated by Equation (1).
Why is this the case? This requires recalling our knowledge of optics. To apply the law of refraction, one must first construct a “normal line”. How is a “normal line” constructed? It is perpendicular to the tangent plane of the interface between two different media. The different behavior of light in two media is manifested as a difference in propagation speed. Therefore, in mathematical terms, the “interface between two different media” is an equi-velocity surface.
If v = v(x, y), then clearly, the equation for each equi-velocity surface (which degenerates into an equi-velocity line in 2D) can be expressed as v_c = v(x, y), where v_c is a constant representing speed. Differentiating both sides yields (this involves partial derivatives, which readers unfamiliar with the concept may skip): 0 = \frac{\partial v}{\partial x} + \frac{\partial v}{\partial y} \dot{y} = 0 \tag{2} This allows us to find the slope of the velocity tangent plane (line) at point (x, y). For cases where v explicitly contains only y, \dot{y} is identically zero, meaning every normal line is perpendicular to the x-axis. This is why we can follow Bernoulli’s idea of “dividing it horizontally into infinitely small segments”; otherwise, this step would not hold. For cases where v does not purely depend on y, one can still use Equation (2) to find the slope and then perform the rotation; this is cumbersome but effective. However, if v also contains derivative terms such as \dot{y}, this approach fails.
Returning to the Brachistochrone problem, substituting v = \sqrt{y} into (1), we get: y(1 + \dot{y}^2) = C_1 \dot{y} = \sqrt{\frac{C_1}{y} - 1}, which implies x = \int \sqrt{\frac{y}{C_1 - y}} dy.
Let y = C_1(\frac{1}{2} - \frac{1}{2} \cos 2t) = C_1 \sin^2 t. Substituting this in, we get x = \int 2C_1 \sin^2 t dt = C_1(t - \frac{1}{2} \sin 2t + C_2). According to the initial conditions, the curve passes through (0, 0), yielding C_2 = 0. The other constant can be determined from the other endpoint. Thus, the Brachistochrone is: \begin{aligned} x &= C_1(t - \frac{1}{2} \sin 2t) \\ y &= C_1(\frac{1}{2} - \frac{1}{2} \cos 2t) \end{aligned}
Evidently, this is a cycloid, also known as a trochoid or an isochrone curve.
Definition of a Cycloid
The cycloid is one of the many fascinating curves in mathematics. It is defined as follows: the trajectory traced by a fixed point on the circumference of a circle as it rolls slowly along a straight line.
The parametric equations are: x = r(t - \sin t), \quad y = r(1 - \cos t)
By the 17th century, people discovered that the cycloid possesses the following properties:
Its length is equal to 4 times the diameter of the rolling circle. Particularly interesting is that its length is a rational number independent of \pi.
The area under the arch is three times the area of the rolling circle.
The point on the circle tracing the cycloid has varying speeds—in fact, at specific points, it is even stationary.
When marbles are released from different points in a cycloid-shaped container, they reach the bottom at the same time (isochronism).
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